## Wednesday, 3 April 2013

### WEDNESDAY, 3 APRIL 2013

Today is the $93^{rd}$ day of the year.

$93 = 3 \times 31$

$93$ is a member of the following, two primitive Pythagorean triples:
$(93, 4324, 4325)$
$(93, 476, 485)$

$93_{10} = 1011101_{2}$ which means that $93$ is palindromic in binary. Not surprisingly there is a list of such numbers in OEIS and it is A006995.

In contrast to yesterday's Lazy Caterer number, $93$ is a Cake Number, see A000125 and Wikipedia.

The definition of Euler's Idoneal Numbers (also known as suitable, or convenient numbers) is:

The collection of the positive integers $D$ such that any integer expressible in only one way as $x^2 ± Dy^2$ (where $x^2$ is relatively prime to $Dy^2$) is a prime, prime power, or twice one of these.
Fortunately, there is an equivalent statement which says:
A positive integer $n$ is idoneal if and only if it cannot be written as $ab + bc + ac$ for distinct positive integer $a, b$, and $c$
The smallest possible values for $a, b$, and $c$ that meet the requirements of being distinct and positive are $1, 2$, and $3$.
So, $ab + bc + ac = (1 \times 2) + (2 \times 3) + (3 \times 1) = 2 + 6 + 3 = 11$.
Thus all numbers up to $11$ are Ideonal numbers.
The next smallest is the set $1, 2$, and $4$ which gives $ab + bc + ac = (1 \times 2) + (2 \times 4) + (4 \times 1) = 2 + 8 + 4 = 14$.
The next calculation is for the set $1, 2$, and $5$ which gives $ab + bc + ac = (1 \times 2) + (2 \times 5) + (5 \times 1) = 2 + 10 + 5 = 17$.
This means that we now know that the set of Ideonal numbers starts $1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 15, 26$.
The Ideonal numbers become more sparse. In fact there is only one Ideonal number in the nineties and it is $93$, see A000926.

## Tuesday, 2 April 2013

### TUESDAY, 2 APRIL 2013

Today is the $92^{nd}$ day of the year.

$92 = 2^2 \times 23$

$92$ is a Lazy Caterer number, see A000124. It is the maximum number of pieces of a pizza (not all the same size) that can be created with $13$ cuts.

$92$ is a member of the following, two primitive Pythagorean triples:
$(525, 92, 533)$
$(2115, 92, 2117)$

There $92$ ways of placing $8$ queens on a standard chess board so that none of the queens are under attack from another queen, see A000170.

$92$ is a Pentagonal number, see A049452.

## Monday, 1 April 2013

### MONDAY, 1 APRIL 2013

Today is the $91^{st}$ day of the year and since $91 \approx \frac {365} {4}$ then we are one quarter of the way through the year.

$91 = 7 \times 13$

$91 = 1 + 2 + 3 + 4 + 5 + 6 + 7= 8 + 9 + 10 + 11 + 12 + 13$ which means that $91$ is a triangular number, see A000217.

$91 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 = 1 + 4 + 9 + 16 + 25 + 36$ which means that $91$ is a square pyramidal number, see A000330.

$91$ is a member of the following two primitive, Pythagorean triples:
$(602, 912, 1092)$
$(912, 41402, 41412)$

Repeatedly, square the digits of a number and sum them.
$9^2 + 1^2 = 81 + 1 = 82$
$8^2 + 2^2 = 64 + 4 = 68$
$6^2 + 8^2 = 36 + 64 = 100$
$1^2 + 0^2 + 0^2 = 1 + 0 + 0 = 1$
Since the result is the number $1$ then $91$ is, by definition, a Happy Number, see A007770.

$\frac {91! + 2} {2} = \frac {135,200,152,767,840,296,255,166,568,759,495,142,147,586,866,476,906,677,791,741,734, 597,153,670,771,559,994,765,685,283,954,750,449,427,751,168,336,768,008,192,000,000, 000,000,000,000,000 + 2}{2} = \frac {135,200,152,767,840,296,255,166,568,759,495,142,147,586,866,476,906,677,791,741,734, 597,153,670,771,559,994,765,685,283,954,750,449,427,751,168,336,768,008,192,000,000, 000,000,000,000,002}{2} = 67,600,076,383,920,148,127,583,284,379,747,571,073,793,433,238,453,338,895,870,867, 298,576,835,385,779,997,382,842,641,977,375,224,713,875,584,168,384,004,096,000,000, 000,000,000,000,001 \approx. 6.76 \times 10^{139}$ and that is prime, see A082672.
In much the same way as one might be amazed that the invention of bread making ever came about then I cannot fail to be amazed that
1) We know that this 140 digit number is prime and
2) We know that this 140 digit number is $\frac {91! + 2}{2}$ and
3) The two facts have been associated with each other.

## Saturday, 30 March 2013

### SUNDAY, 31 MARCH 2013

Today is the $90^{th}$ day of the year.

$90 = 2 \times 3^2 \times 5$

$90 = 3^2 + 9^2$
$90 = 1^2 + 5^2 + 8^2$
$90 = 4^2 + 5^2 + 7^2$

All primitive Pythagorean triples can be derived from the following:
$$a = k^2 - l^2, b = 2kl, c = k^2 + l^2$$
$for k, l \in \mathbb{N} with k > l > 0, (k,l) = 1 and k \not\equiv l mod 2$
What this is saying is that choose any old integers $k$ and $l$ such that
• $k$ is larger than $l$;
• $k$ and $l$ are co-prime (have no common factors other than one);
• If $k$ is odd then $l$ is even or vice-versa.
then $a, b and c$  as derived above will form a primitive Pythagorean triple and all primitive Pythagorean triples are of this form.
For proof of this see www.math.uconn.edu/~kconrad/blurbs/ugradnumthy/pythagtriple.pdf.
Recalling that an odd number squared is odd and an even number squared is even and that one of $k$ and $l$ is odd and the other is even then we know that $k^2 - l^2$ is either an odd minus an even or en even minus an odd. Which ever is the case the result is odd. Similarly, $k^2 + l^2$ is odd. This means that all primitive Pythagorean triples have one odd-length leg and one odd-length hypotenuse. The remaining leg, the one that is $2kl$, is, of course, even. However, we know that one of $k$ and $l$ is even also so that means that leg $b$ is a multiple of $4$.
The corollary of all this is that no number, $n$, that is even but not a multiple of $4$ i.e. where $n \equiv 2 (mod 4)$, can be a member of a primitive Pythagorean triple. which in turn means that $90 \equiv 2 (mod 4)$ is not a member of any primitive Pythagorean triple.

### SATURDAY, 30 MARCH 2013

Today is the $89^{th}$ day of the year.

$89$ is prime and the eleventh member of the Fibonacci sequence.

$89$ is a member of two primitive Pythagorean triples.
$(39, 80, 89)$
$(89, 3960, 3961)$

$5^2 + 8^2 = 25 + 64 = 89$
$2^2 + 2^2 + 9^2 = 4 + 4 + 81 = 89$
$2^2 + 6^2 + 7^2 = 4 + 36 + 49 = 89$
$3^2 + 4^2 + 8^2 = 9 + 16 + 64 = 89$

$(2 \times 89) + 1 = 179$ which is prime and, therefore, makes $89$ a Sophie Germain prime, see A005384.
$(2 \times 179) + 1 = 359$ which is prime
$(2 \times 359) + 1 = 719$ which is prime
$(2 \times 719) + 1 = 1,439$ which is prime
$(2 \times 1,439) + 1 = 2,879$ which is prime
$(2 \times 2,879) + 1 = 5,759 = 13 * 443$ which is not prime.
This gives us a Cunningham chain of $(89, 179, 359, 719, 1439, 2879)$ of length $6$. In fact $89$ is the smallest prime that starts a chain of this length, see A005602.

$2^{89} -1 = 618,970,019,642,690,137,449,562,111$ which is prime and, therefore, means that $89$ is a Mersenne Prime, see A000043.

T is the $89^{th}$ letter of the following, never-ending sentence:
"T is the first, fourth, eleventh, sixteenth, twenty-fourth, twenty-ninth, thirty-third, thirty-fifth, thirty-ninth,  ... letter in this sentence, not counting spaces or commas"
As beautifully illustrated in A005224, the sentence begins like this:

1234567890 1234567890 1234567890 1234567890 1234567890 Tisthefirs tfourthele venthsixte enthtwenty fourthtwen
tyninththi rtythirdth irtyfiftht hirtyninth fortyfifth
fortyseven thfiftyfir stfiftysix thfiftyeig hthsixtyse
condsixtyf ourthsixty ninthseven tythirdsev entyeighth
eightiethe ightyfourt heightynin thninetyfo urthninety
ninthonehu ndredfourt honehundre deleventho nehundreds
ixteenthon ehundredtw entysecond onehundred twentysixt
honehundre dthirtyfir stonehundr edthirtysi xthonehund

## Friday, 29 March 2013

### FRIDAY, 29 MARCH 2013

Today is the $88^{th}$ day of the year.

$88 = 2^3 \times 11$

$88$ is a member of two primitive Pythagorean triples
$(88, 105, 137)$
$(88, 1935, 1937)$

$88$ is an Erdos–Woods number which is defined as: there exists a positive integer $a$ such that in the sequence $(a, a + 1, …, a + k)$ of consecutive integers, each of the elements has a common factor, other than one, with one of the endpoints.
For example, consider $a = 2,184$ and $k = 16$ which gives us the set of $17$ numbers from $2,184$ to $2,200$ inclusive:
$2,184$ has a common factor of $2,184$ with $2,184$
$2,185$ has a common factor of $5$ with $2,200$
$2,186$ has a common factor of $2$ with $2,184$
$2,187$ has a common factor of $3$ with $2,184$
$2,188$ has a common factor of $4$ with $2,184$
$2,189$ has a common factor of $11$ with $2,200$
$2,190$ has a common factor of $6$ with $2,184$
$2,191$ has a common factor of $7$ with $2,184$
$2,192$ has a common factor of $8$ with $2,184$
$2,193$ has a common factor of $3$ with $2,184$
$2,194$ has a common factor of $2$ with $2,184$
$2,195$ has a common factor of $5$ with $2,200$
$2,196$ has a common factor of $4$ with $2,200$
$2,197$ has a common factor of $13$ with $2,184$

$2,198$ has a common factor of $2$ with $2,200$
$2,199$ has a common factor of $3$ with $2,184$

$2,200$ has a common factor of $2,200$ with $2,200$

So, there is a number $a$ such that every single one of the numbers in the set $(a, a + 1, …, a + 88)$ has a common factor greater than $1$ with one of $a$ or $a + 88$, see A059756.
I can neither find what this $a$ is nor how this sequence has been calculated (Note that the first $1,052$ members of the sequence can be found here). However, sequence A05975 does gives the first few values for $a$. Thus we learn that the ninth member has $k = 70$ and $a = 13,151,117,479,433,859,435,440$. So the thirteenth member with $k = 88$ must be an even larger number.

## Thursday, 28 March 2013

### Update

On Tuesday, 26 March I mentioned that I was surprised that $\frac {4^n - 1} {3}$ was a natural number or to put it slightly more mathematically that $\forall n \in \mathbb{N}, \frac {4^n - 1} {3} = k$ for some $k \in \mathbb{N}$. It isn't quite as surprising as I first thought.

Assume that $\frac {4^n - 1}{3} = k$ then $4^n - 1 = 3k$
Consider $4^{n + 1} - 1 = m$
then $m - 3k = (4^{n + 1} - 1) - (4^n - 1)$
$m - 3k = 4^{n + 1} - 1 - 4^n + 1$
$m - 3k = 4^{n + 1} - 4^n$
$m - 3k = 4^n(4 - 1)$
$m - 3k = 4^n \times 3$
Therefore, $m - 3k$ is a multiple of $3$ and thus $m$ is a multiple of $3$, which means that $4^{n + 1} - 1$ is a multiple of $3$.
So, we have shown that if $4^n - 1$ is a multiple of three then $4^{n + 1} - 1$ is a multiple of $3$. Since we know that when $n = 1$, $4^n - 1 = 3$ which is a multiple of $3$ then all calculations of the form $4^n - 1$ are a multiple of $3$.
Thus, $\frac {4^n - 1}{3} \in \mathbb{N} \forall n \in \mathbb{N}$

There is, of course, nothing remarkable about the choice of $4$ and $3$ in the above proof. It could equally well have been $p \in \mathbb{N}$ giving the following theorem
$\frac {p^n - 1}{p - 1} \in \mathbb{N} \forall n, p \in \mathbb{N}$

### THURSDAY, 28 MARCH 2013

Today is the $87^{th}$ day of the year.

$87 = 3 \times 29$

$2^2 + 3^2 + 5^2 + 7^2 = 4 + 9 + 25 + 49 = 87$ which means that $87$ is the sum of the first four prime numbers squared.

$87$ is a member of the following two primitive Pythagorean triples:
$(87, 416, 425)$
$(87, 3785, 3785)$

According to the Prime Curios website $87^2 + 3^2 + 29^2 = 8,719$ and $87^2 - 3^2 - 29^2 = 6,719$  are both prime.
According to OEIS, $6 \times 87 +1 = 523$ and $6 \times 87 - 1 = 521$ are both primes, see A002822. In fact $521$ and$523$ are twin primes because they differ by two, see A001359 and A006512.

## Tuesday, 26 March 2013

### Tuesday, 26 MARCH 2013

Today is the $85^{th}$ day of the year.

$85 = 5 \times 17$

$85 = 2^2 + 9^2 = 6^2 + 7^2$, so it is the sum of two squares in two different ways.

$85$ is the member of the following four primitive Pythagorean triples:

$(13, 84, 85)$
$(77, 36, 85)$
$(85, 132, 157)$
$(85, 3612, 3613)$

$85$ is a Joke Number or a Smith Number. A Joke or Smith Number is one where the sum of the digits of the number, $8 + 5 = 13$, is equal to the sum of the digits of the factors of that number $5 + 1 + 7 = 13$, see A006753.

$\frac {4^4 - 1} {3} = \frac {256 - 1} {3} = \frac {255} {3} = 85$ so $85$ is a member of the sequence $\frac {4^n -1} {3}$, see A002450. What I find surprising is that this sequence implies that all powers of 4 less one are a multiple of three.

It transpires that there are $85$ different ways partitions of $29$ into at most three parts, see A001399.
The $85$ partitions are:

$1: [0, 0, 29]$
$2: [0, 1, 28]$
$3: [0, 2, 27]$
$4: [0, 3, 26]$
$5: [0, 4, 25]$
$6: [0, 5, 24]$
$7: [0, 6, 23]$
$8: [0, 7, 22]$
$9: [0, 8, 21]$
$10: [0, 9, 20]$
$11: [0, 10, 19]$
$12: [0, 11, 18]$
$13: [0, 12, 17]$
$14: [0, 13, 16]$
$15: [0, 14, 15]$
$16: [1, 1, 27]$
$17: [1, 2, 26]$
$18: [1, 3, 25]$
$19: [1, 4, 24]$
$20: [1, 5, 23]$
$21: [1, 6, 22]$
$22: [1, 7, 21]$
$23: [1, 8, 20]$
$24: [1, 9, 19]$
$25: [1, 10, 18]$
$26: [1, 11, 17]$
$27: [1, 12, 16]$
$28: [1, 13, 15]$
$29: [1, 14, 14]$
$30: [2, 2, 25]$
$31: [2, 3, 24]$
$32: [2, 4, 23]$
$33: [2, 5, 22]$
$34: [2, 6, 21]$
$35: [2, 7, 20]$
$36: [2, 8, 19]$
$37: [2, 9, 18]$
$38: [2, 10, 17]$
$39: [2, 11, 16]$
$40: [2, 12, 15]$
$41: [2, 13, 14]$
$42: [3, 3, 23]$
$43: [3, 4, 22]$
$44: [3, 5, 21]$
$45: [3, 6, 20]$
$46: [3, 7, 19]$
$47: [3, 8, 18]$
$48: [3, 9, 17]$
$49: [3, 10, 16]$
$50: [3, 11, 15]$
$51: [3, 12, 14]$
$52: [3, 13, 13]$
$53: [4, 4, 21]$
$54: [4, 5, 20]$
$55: [4, 6, 19]$
$56: [4, 7, 18]$
$57: [4, 8, 17]$
$58: [4, 9, 16]$
$59: [4, 10, 15]$
$60: [4, 11, 14]$
$61: [4, 12, 13]$
$62: [5, 5, 19]$
$63: [5, 6, 18]$
$64: [5, 7, 17]$
$65: [5, 8, 16]$
$66: [5, 9, 15]$
$67: [5, 10, 14]$
$68: [5, 11, 13]$
$69: [5, 12, 12]$
$70: [6, 6, 17]$
$71: [6, 7, 16]$
$72: [6, 8, 15]$
$73: [6, 9, 14]$
$74: [6, 10, 13]$
$75: [6, 11, 12]$
$76: [7, 7, 15]$
$77: [7, 8, 14]$
$78: [7, 9, 13]$
$79: [7, 10, 12]$
$80: [7, 11, 11]$
$81: [8, 8, 13]$
$82: [8, 9, 12]$
$83: [8, 10, 11]$
$84: [9, 9, 11]$
$85: [9, 10, 10]$

## Monday, 25 March 2013

### MONDAY, 25 MARCH 2013

Today is the $84^{th}$ day of the year.

$84 = 2^2 \times 3 \times 7$

$84$ is a member of the following primitive Pythagorean triples:
$(13, 84, 85)$
$(187, 84, 205)$
$(437, 84, 445)$
$(1763, 84, 1765)$

$1^2 + 3^2 + 5^2 + 7^2 = 1 + 9 + 25 + 49 = 84$ making $84$ a member of the sequence $\sum_{k=1}^n (2n-1)^2 = \frac {n(2n - 1)(2n + 1)}{3}$ which is the sum of the odd numbers squared, see A000447.

## Sunday, 24 March 2013

### SUNDAY, 24 MARCH 2013

Today is the $83^{rd}$ day of the year.

$83$ is a prime number. In fact $(2 \times 83) + 1 = 167$ is prime which means that $83$ is a Sophie Germain prime, see A005384.
Given that
$20$ is not prime and $(2 \times 20) + 1 = 41$
$41$ is prime and $(2 \times 41) + 1 = 83$
$83$ is prime and $(2 \times 83) + 1 = 167$
$167$ is prime and $(2 \times 167) + 1 = 335$
$335$ is not prime
Then we have a Cunningham Chain of length $3$ i.e. $(41, 83, 167)$, see A059762.

Take every number less than or equal to $83$, write it down in base $25$, something like this:

$1_{10} = 1_{25}$
$2_{10} = 2_{25}$
$3_{10} = 3_{25}$
$4_{10} = 4_{25}$
$5_{10} = 5_{25}$
$6_{10} = 6_{25}$
$7_{10} = 7_{25}$
$8_{10} = 8_{25}$
$9_{10} = 9_{25}$
$10_{10} = a_{25}$
$11_{10} = b_{25}$
$12_{10} = c_{25}$
$13_{10} = d_{25}$
$14_{10} = e_{25}$
$15_{10} = f_{25}$
$16_{10} = g_{25}$
$17_{10} = h_{25}$
$18_{10} = i_{25}$
$19_{10} = j_{25}$
$20_{10} = k_{25}$
$21_{10} = l_{25}$
$22_{10} = m_{25}$
$23_{10} = n_{25}$
$24_{10} = o_{25}$
$25_{10} = 10_{25}$
$26_{10} = 11_{25}$
$27_{10} = 12_{25}$
$28_{10} = 13_{25}$
...
$59_{10} = 29_{25}$
$60_{10} = 2a_{25}$
$61_{10} = 2b_{25}$
$62_{10} = 2c_{25}$
$63_{10} = 2d_{25}$
$64_{10} = 2e_{25}$
$65_{10} = 2f_{25}$
$66_{10} = 2g_{25}$
$67_{10} = 2h_{25}$
$68_{10} = 2i_{25}$
$69_{10} = 2j_{25}$
$70_{10} = 2k_{25}$
$71_{10} = 2l_{25}$
$72_{10} = 2m_{25}$
$73_{10} = 2n_{25}$
$74_{10} = 2o_{25}$
$75_{10} = 30_{25}$
$76_{10} = 31_{25}$
$77_{10} = 32_{25}$
$78_{10} = 33_{25}$
$79_{10} = 34_{25}$
$80_{10} = 35_{25}$
$81_{10} = 36_{25}$
$82_{10} = 37_{25}$
$83_{10} = 38_{25}$

Now, concatenate each of the base $25$ values starting with $1$ and reversing each value before it is concatenated giving:
$1,234,567,89a,bcd,efg,hij,klm,no0,111,213,141,516,171,819,1a1,b1c,1d1,e1f,1g1,h1i,1j1,k1l,1m1,n1o,102,122,232,425,262,728,292,a2b,2c2,d2e,2f2,g2h,2i2,j2k,2l2,m2n,2o2,031,323,334,353,637,383$
Convert this $142$ digit number in base $25$ to decimal and the result is divisible by $8$.
Not in itself remarkable but given that the number $83$ is the lowest number that can generate a multiple of $8$ in this fashion is remarkable, see A029518.

Consider, the sum of the squares of all the numbers up to and including $83$. This is $194,054$ which is divisible by $83$, see A007310.

 Square Total $1^2 = 1$ $1$ $2^2 = 4$ $5$ $3^2 = 9$ $14$ $4^2 = 16$ $30$ $5^2 = 25$ $55$ $6^2 = 36$ $91$ $7^2 = 49$ $140$ $8^2 = 64$ $204$ $9^2 = 81$ $285$ $10^2 = 100$ $385$ $11^2 = 121$ $506$ $12^2 = 144$ $650$ $13^2 = 169$ $819$ $14^2 = 196$ $1015$ $15^2 = 225$ $1240$ $16^2 = 256$ $1496$ $17^2 = 289$ $1785$ $18^2 = 324$ $2109$ $19^2 = 361$ $2470$ $20^2 = 400$ $2870$ $21^2 = 441$ $3311$ $22^2 = 484$ $3795$ $23^2 = 529$ $4324$ $24^2 = 576$ $4900$ $25^2 = 625$ $5525$ $26^2 = 676$ $6201$ $27^2 = 729$ $6930$ $28^2 = 784$ $7714$ $29^2 = 841$ $8555$ $30^2 = 900$ $9455$ $31^2 = 961$ $10416$ $32^2 = 1024$ $11440$ $33^2 = 1089$ $12529$ $34^2 = 1156$ $13685$ $35^2 = 1225$ $14910$ $36^2 = 1296$ $16206$ $37^2 = 1369$ $17575$ $38^2 = 1444$ $19019$ $39^2 = 1521$ $20540$ $40^2 = 1600$ $22140$ $41^2 = 1681$ $23821$ $42^2 = 1764$ $25585$ $43^2 = 1849$ $27434$ $44^2 = 1936$ $29370$ $45^2 = 2025$ $31395$ $46^2 = 2116$ $33511$ $47^2 = 2209$ $35720$ $48^2 = 2304$ $38024$ $49^2 = 2401$ $40425$ $50^2 = 2500$ $42925$ $51^2 = 2601$ $45526$ $52^2 = 2704$ $48230$ $53^2 = 2809$ $51039$ $54^2 = 2916$ $53955$ $55^2 = 3025$ $56980$ $56^2 = 3136$ $60116$ $57^2 = 3249$ $63365$ $58^2 = 3364$ $66729$ $59^2 = 3481$ $70210$ $60^2 = 3600$ $73810$ $61^2 = 3721$ $77531$ $62^2 = 3844$ $81375$ $63^2 = 3969$ $85344$ $64^2 = 4096$ $89440$ $65^2 = 4225$ $93665$ $66^2 = 4356$ $98021$ $67^2 = 4489$ $102510$ $68^2 = 4624$ $107134$ $69^2 = 4761$ $111895$ $70^2 = 4900$ $116795$ $71^2 = 5041$ $121836$ $72^2 = 5184$ $127020$ $73^2 = 5329$ $132349$ $74^2 = 5476$ $137825$ $75^2 = 5625$ $143450$ $76^2 = 5776$ $149226$ $77^2 = 5929$ $155155$ $78^2 = 6084$ $161239$ $79^2 = 6241$ $167480$ $80^2 = 6400$ $173880$ $81^2 = 6561$ $180441$ $82^2 = 6724$ $187165$ $83^2 = 6889$ $194054$

## Saturday, 23 March 2013

### SATURDAY, 23 MARCH 2013

Today is the $82^{nd}$ day of the year.

$82 = 2 \times 41$. Since $2$ and $41$ are both prime then $82$ is a semi-prime, see A001358.

$82 = 3^2 + 3^2 + 8^2$

The following calculations all result in prime numbers and all, amazingly, represent sequences that are in OEIS:
A097480: $(2 \times 82) - 15 = 149$
A097363: $(2 \times 82) - 13 = 151$
A089192: $(2 \times 82) - 7 = 157$
A006254: $(2 \times 82) - 1 = 163$
A067076: $(2 \times 82) + 3 = 167$
A155722: $(2 \times 82) + 9 = 173$
A089559: $(2 \times 82) + 15 = 179$
A173059: $(2 \times 82) + 17 = 181$
A095278: $(4 \times 82) + 3 = 331$
A033868: $(7 \times 82) - 11 = 563$
A105133: $(8 \times 82) + 5 = 661$
A007811: $(10 \times 82) + 1 = 821$
A007811: $(10 \times 82) + 3 = 823$
A007811: $(10 \times 82) + 7 = 827$
A007811: $(10 \times 82) + 9 = 829$
A127575: $(16 \times 82) + 15 = 1,327$
A201816: $(90 \times 82) + 13 = 7,393$
A198382: $(90 \times 82) + 37 = 7,417$
A027861: $82^2 + (82 + 1)^2 = 13,613$
A090563: $(5 \times 82^2) + (5 \times 82) + 1 = 34,031$
A125881: $82^3 + 82^2 - 1 = 558,091$
A000068: $82^4 + 1 = 45,212,177$
A139065: $\frac {7 + 82!}{7}$
A139063: $\frac {6 + 82!}{6}$
A007749: $82!! - 1$

I cannot calculate the last three. In fact I doubt anyone can calculate the last one because it would just take too long to calculate the factorial of
$475,364,333,701,284,174,842,138,206,989,404,946,643,813,294,067,993,328,617,160,934,076,743,994,734,899,148,613,007,131,808,479,167,119,360,000,000,000,000,000,000$

## Monday, 18 March 2013

### MONDAY, 18 MARCH 2013

Today is the $77^{th}$ day of the year.

$77 = 7 \times 11$ which, since $7$ and $11$ are both prime makes $77$ a semi-prime.

$77$ occurs in two primitive Pythagorean triples.
$(77, 36, 85)$
$(77, 2964, 2965)$

$77 = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19$ which means that $77$ is the sum of the first $8$ prime numbers, see A007504.

$77! + 1 = 145,183,092,028,285,869,634,070,784,086,308,284,983,740,379,224,208,358,846,781,574,688,061,991,349,156,420,080,065,207,861,248,000,000,000,000,000,001$, which, apparently, is prime. This number consists of $114$ digits. I can understand someone discovering primes. I can understand somebody calculating $77!$. I find it amazing that somebody having calculated of those facts then checked to see if the other one was true. Amazed or not, somebody has checked more than just $77$, see A002981.

Binary Partitions were mentioned on Friday, 15 March 2013 following a short discourse on the more general topic of partitions. To repeat what is stated there:
A partition of a number is a way of writing a number as the sum of positive integers. If two sums contain the same digits and differ only in their order then they are considered the same partition. The number of partitions for a given number $n$ is what we will consider.
As an example consider the number $4$, this can be partitioned in $5$ different ways:
$1) 4$
$2) 3+1$
$3) 2+2$
$4) 2+1+1$
$5) 1+1+1+1$

A little inspection shows that, for the first thirteen numbers, the number of partitions are (see A000041)
$0: 1$
$1: 1$
$2: 2$
$3: 3$
$4: 5$
$5: 7$
$6: 11$
$7: 15$
$8: 22$
$9: 30$
$10: 42$
$11: 56$
$12: 77$

So there are $77$ ways of paritioning the number 12, here they are:
$[1] [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]$
$[2] [2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]$
$[3] [2, 2, 1, 1, 1, 1, 1, 1, 1, 1]$
$[4] [2, 2, 2, 1, 1, 1, 1, 1, 1]$
$[5] [2, 2, 2, 2, 1, 1, 1, 1]$
$[6] [2, 2, 2, 2, 2, 1, 1]$
$[7] [2, 2, 2, 2, 2, 2]$
$[8] [3, 1, 1, 1, 1, 1, 1, 1, 1, 1]$
$[9] [3, 2, 1, 1, 1, 1, 1, 1, 1]$
$[10] [3, 2, 2, 1, 1, 1, 1, 1]$
$[11] [3, 2, 2, 2, 1, 1, 1]$
$[12] [3, 2, 2, 2, 2, 1]$
$[13] [3, 3, 1, 1, 1, 1, 1, 1]$
$[14] [3, 3, 2, 1, 1, 1, 1]$
$[15] [3, 3, 2, 2, 1, 1]$
$[16] [3, 3, 2, 2, 2]$
$[17] [3, 3, 3, 1, 1, 1]$
$[18] [3, 3, 3, 2, 1]$
$[19] [3, 3, 3, 3]$
$[20] [4, 1, 1, 1, 1, 1, 1, 1, 1]$
$[21] [4, 2, 1, 1, 1, 1, 1, 1]$
$[22] [4, 2, 2, 1, 1, 1, 1]$
$[23] [4, 2, 2, 2, 1, 1]$
$[24] [4, 2, 2, 2, 2]$
$[25] [4, 3, 1, 1, 1, 1, 1]$
$[26] [4, 3, 2, 1, 1, 1]$
$[27] [4, 3, 2, 2, 1]$
$[28] [4, 3, 3, 1, 1]$
$[29] [4, 3, 3, 2]$
$[30] [4, 4, 1, 1, 1, 1]$
$[31] [4, 4, 2, 1, 1]$
$[32] [4, 4, 2, 2]$
$[33] [4, 4, 3, 1]$
$[34] [4, 4, 4]$
$[35] [5, 1, 1, 1, 1, 1, 1, 1]$
$[36] [5, 2, 1, 1, 1, 1, 1]$
$[37] [5, 2, 2, 1, 1, 1]$
$[38] [5, 2, 2, 2, 1]$
$[39] [5, 3, 1, 1, 1, 1]$
$[40] [5, 3, 2, 1, 1]$
$[41] [5, 3, 2, 2]$
$[42] [5, 3, 3, 1]$
$[43] [5, 4, 1, 1, 1]$
$[44] [5, 4, 2, 1]$
$[45] [5, 4, 3]$
$[46] [5, 5, 1, 1]$
$[47] [5, 5, 2]$
$[48] [6, 1, 1, 1, 1, 1, 1]$
$[49] [6, 2, 1, 1, 1, 1]$
$[50] [6, 2, 2, 1, 1]$
$[51] [6, 2, 2, 2]$
$[52] [6, 3, 1, 1, 1]$
$[53] [6, 3, 2, 1]$
$[54] [6, 3, 3]$
$[55] [6, 4, 1, 1]$
$[56] [6, 4, 2]$
$[57] [6, 5, 1]$
$[58] [6, 6]$
$[59] [7, 1, 1, 1, 1, 1]$
$[60] [7, 2, 1, 1, 1]$
$[61] [7, 2, 2, 1]$
$[62] [7, 3, 1, 1]$
$[63] [7, 3, 2]$
$[64] [7, 4, 1]$
$[65] [7, 5]$
$[66] [8, 1, 1, 1, 1]$
$[67] [8, 2, 1, 1]$
$[68] [8, 2, 2]$
$[69] [8, 3, 1]$
$[70] [8, 4]$
$[71] [9, 1, 1, 1]$
$[72] [9, 2, 1]$
$[73] [9, 3]$
$[74] [10, 1, 1]$
$[75] [10, 2]$
$[76] [11, 1]$
$[77] [12]$

## Sunday, 17 March 2013

### SUNDAY, 17 MARCH 2013

Today is the $76^{th}$ day of the year.

$76 = 2^2 \times 19$

$76_{10} = 44_{18}$ which means it is a Brazilian Number, see A125134.

$76$ occurs in two primitive Pythagorean triples.
$(357, 76, 365)$
$(1443, 76, 1445)$

$76 * 76 = 5,776$, since $5,776$ end with the digits $76$ this makes it automorphic, see A003226.
Consider, $76^3 = 76 \times 76^2 = 76 \times 5,776 = 76 \times (5,700 + 76) = (76 \times 57 \times 100) + (76 \times 76)$
The first term is a multiple of $100$ which means that it ends in $00$ whilst the second term is $5,776$ which ends in $76$
Thus we can infer that the sum of these terms and, therefore $76^3$, ends in the digits $76$.
A similar argument applies to all other powers of $76$.

Follow up to THURSDAY, 14 MARCH 2013
In the blog for last Thursday I posed the question "Can anyone find a proof for the hypothesis that $8^n - 1$ is always divisible by $7$?".
I offer the following as an answer.

Assume that $\frac {8^n - 1} {7}$ is an integer for some value of $n$ then
$8^n - 1 = 7k$ for some integer value $k$.
$\therefore 8^n = 7k + 1$
Multiplying both sides by 8 gives
$(8 \times 8^n) = 8 \times (7k + 1)$
Subtract one from both sides gives
$8^{n+1} - 1 = (8 \times 7k) + 8 -1 = (8 \times 7k) + 7$
$\therefore 8^{n+1} - 1 = 7 \times (8k + 1)$
Since $k$ is an integer then $8k + 1$ is an integer which means that $8^{n+1} - 1$ is a multiple of $7$.
However, we can see that for $n = 1$, $8^n - 1$ becomes $8 -1 = 7$ which is clearly divisible by $7$.
Thus, by induction, we have the desired result that $\frac {8^n - 1} {7}$ is always a whole number.

## Saturday, 16 March 2013

### SATURDAY, 16 MARCH 2013

Today is the $75^{th}$ day of the year.

$75 = 3 \times 5^2$

According to A036378 there are $75$ primes between $2^9 = 512$ and $2^{10} = 1024$
The $75$ primes are:
________ ________ ________
$1) 521$$2) 523$$3) 541$
$4) 547$$5) 557$$6) 563$
$7) 569$$8) 571$$9) 577$
$10) 587$$11) 593$$12) 599$
$13) 601$$14) 607$$15) 613$
$16) 617$$17) 619$$18) 631$
$19) 641$$20) 643$$21) 647$
$22) 653$$23) 659$$24) 661$
$25) 673$$26) 677$$27) 683$
$28) 691$$29) 701$$30) 709$
$31) 719$$32) 727$$33) 733$
$34) 739$$35) 743$$36) 751$
$37) 757$$38) 761$$39) 769$
$40) 773$$41) 787$$42) 797$
$43) 809$$44) 811$$45) 821$
$46) 823$$47) 827$$48) 829$
$49) 839$$50) 853$$51) 857$
$52) 859$$53) 863$$54) 877$
$55) 881$$56) 883$$57) 887$
$58) 907$$59) 911$$60) 919$
$61) 929$$62) 937$$63) 941$
$64) 947$$65) 953$$66) 967$
$67) 971$$68) 977$$69) 983$
$70) 991$$71) 997$$72) 1009$
$73) 1013$$74) 1019$$75) 1021$

Consider writing down a sequence using the following rules
1) Write down the first odd non-negative integer, $1$
2) Write down the next two even numbers, $2, 4$
3) Write down the next three odd numbers, $5, 7, 9$
4) Write down the next four even numbers $10, 12, 14, 16$
5) Write down the next five odd numbers $17, 19, 21, 23, 25$
6) Well, you get the idea.

The first $47$ members of this sequence looks like this:
$1, 2, 4, 5, 7, 9, 10, 12, 14, 16, 17, 19, 21, 23, 25, 26, 28, 30, 32, 34, 36, 37, 39, 41, 43, 45, 47, 49, 50, 52, 54, 56, 58, 60, 62, 64, 65, 67, 69, 71, 73, 75, 77, 79, 81$

As you can see $75$ is the $42^{nd}$ member, of what is known as the Connell Sequence, see A001614.

The formula for this sequence is
$a(n) = 2n - \lfloor \frac {1 + \sqrt {8n - 7} } {2} \rfloor$ where $\lfloor \rfloor$ indicates the floor function.

Thus we can calculate the $75^{th}$ member of this function:
$a(75) = (2 \times 75) - \lfloor \frac {1 + \sqrt {(8 \times 75) - 7} } {2} \rfloor$

$a(75) = 150 - \lfloor \frac {1 + \sqrt {600 - 7} } {2} \rfloor$

$a(75) = 150 - \lfloor \frac {1 + \sqrt {593} } {2} \rfloor$

$a(75) = 150 - \lfloor \frac {1 + 24.35159 } {2} \rfloor$

$a(75) = 150 - \lfloor \frac {25.35159 } {2} \rfloor$

$a(75) = 150 - \lfloor 12.67579 \rfloor$

$a(75) = 150 - 12$

$\underline {\underline {a(75) = 138}}$

As an aside note that the number at the end of each sub-sequence above is the square of the index of the sub-sequence, i.e the $5^{th}$ sub-sequence ends in $25$.

## Friday, 15 March 2013

### FRIDAY, 15 MARCH 2013

Today is the $74^{th}$ day of the year.

$74 = 2 \times 37$ which makes $74$ a semi-prime, see A001358.

$74 = 5^2 + 7^2$ which means that $74$ is the sum of two squares, see A000404.

$74^2 + 1 = 5476 +1 = 5477$ which is prime, see A005574.

The partition of a number is a way of writing a number as the sum of positive integers. If two sums contain the same digits and differ only in their order then they are considered the same partition. The number of partitions for a given number $n$ is what we will consider.
As an example consider the number $4$, this can be partitioned in $5$ different ways:
$1) 4$
$2) 3+1$
$3) 2+2$
$4) 2+1+1$
$5) 1+1+1+1$

A little inspection shows that, for the first ten numbers, the number of partitions are
$0: 1$
$1: 1$
$2: 2$
$3: 3$
$4: 5$
$5: 7$
$6: 11$
$7: 15$
$8: 22$
$9: 30$
The sequence $1, 1, 2, 3, 5, 7, 11, 15, 22, 30$ is sequence A000041.

Partitions can be further restricted, for example, by only allowing powers of two to be used in the sum. Thus for $5$ we are reduced to the following four binary partitions:
$1) 4$
$2) 2+2$
$3) 2+1+1$
$4) 1+1+1+1$

Not surprisingly the sequence of the number of binary partitions is also in OEIS and is sequence A018819.
This sequence tells us that $74$ is the number ofbinary partitions for $22$ and $23$.
The $74$ binary partitions for $22$ are:
$[1] [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1]$
$[2] [2,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1]$
$[3] [2,2,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1]$
$[4] [2,2,2,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1]$
$[5] [2,2,2,2,1,1,1,1,1,1,1,1,1,1,1,1,1,1]$
$[6] [2,2,2,2,2,1,1,1,1,1,1,1,1,1,1,1,1]$
$[7] [2,2,2,2,2,2,1,1,1,1,1,1,1,1,1,1]$
$[8] [2,2,2,2,2,2,2,1,1,1,1,1,1,1,1]$
$[9] [2,2,2,2,2,2,2,2,1,1,1,1,1,1]$
$[10] [2,2,2,2,2,2,2,2,2,1,1,1,1]$
$[11] [2,2,2,2,2,2,2,2,2,2,1,1]$
$[12] [2,2,2,2,2,2,2,2,2,2,2]$
$[13] [4,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1]$
$[14] [4,2,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1]$
$[15] [4,2,2,1,1,1,1,1,1,1,1,1,1,1,1,1,1]$
$[16] [4,2,2,2,1,1,1,1,1,1,1,1,1,1,1,1]$
$[17] [4,2,2,2,2,1,1,1,1,1,1,1,1,1,1]$
$[18] [4,2,2,2,2,2,1,1,1,1,1,1,1,1]$
$[19] [4,2,2,2,2,2,2,1,1,1,1,1,1]$
$[20] [4,2,2,2,2,2,2,2,1,1,1,1]$
$[21] [4,2,2,2,2,2,2,2,2,1,1]$
$[22] [4,2,2,2,2,2,2,2,2,2]$
$[23] [4,4,1,1,1,1,1,1,1,1,1,1,1,1,1,1]$
$[24] [4,4,2,1,1,1,1,1,1,1,1,1,1,1,1]$
$[25] [4,4,2,2,1,1,1,1,1,1,1,1,1,1]$
$[26] [4,4,2,2,2,1,1,1,1,1,1,1,1]$
$[27] [4,4,2,2,2,2,1,1,1,1,1,1]$
$[28] [4,4,2,2,2,2,2,1,1,1,1]$
$[29] [4,4,2,2,2,2,2,2,1,1]$
$[30] [4,4,2,2,2,2,2,2,2]$
$[31] [4,4,4,1,1,1,1,1,1,1,1,1,1]$
$[32] [4,4,4,2,1,1,1,1,1,1,1,1]$
$[33] [4,4,4,2,2,1,1,1,1,1,1]$
$[34] [4,4,4,2,2,2,1,1,1,1]$
$[35] [4,4,4,2,2,2,2,1,1]$
$[36] [4,4,4,2,2,2,2,2]$
$[37] [4,4,4,4,1,1,1,1,1,1]$
$[38] [4,4,4,4,2,1,1,1,1]$
$[39] [4,4,4,4,2,2,1,1]$
$[40] [4,4,4,4,2,2,2]$
$[41] [4,4,4,4,4,1,1]$
$[42] [4,4,4,4,4,2]$
$[43] [8,1,1,1,1,1,1,1,1,1,1,1,1,1,1]$
$[44] [8,2,1,1,1,1,1,1,1,1,1,1,1,1]$
$[45] [8,2,2,1,1,1,1,1,1,1,1,1,1]$
$[46] [8,2,2,2,1,1,1,1,1,1,1,1]$
$[47] [8,2,2,2,2,1,1,1,1,1,1]$
$[48] [8,2,2,2,2,2,1,1,1,1]$
$[49] [8,2,2,2,2,2,2,1,1]$
$[50] [8,2,2,2,2,2,2,2]$
$[51] [8,4,1,1,1,1,1,1,1,1,1,1]$
$[52] [8,4,2,1,1,1,1,1,1,1,1]$
$[53] [8,4,2,2,1,1,1,1,1,1]$
$[54] [8,4,2,2,2,1,1,1,1]$
$[55] [8,4,2,2,2,2,1,1]$
$[56] [8,4,2,2,2,2,2]$
$[57] [8,4,4,1,1,1,1,1,1]$
$[58] [8,4,4,2,1,1,1,1]$
$[59] [8,4,4,2,2,1,1]$
$[60] [8,4,4,2,2,2]$
$[61] [8,4,4,4,1,1]$
$[62] [8,4,4,4,2]$
$[63] [8,8,1,1,1,1,1,1]$
$[64] [8,8,2,1,1,1,1]$
$[65] [8,8,2,2,1,1]$
$[66] [8,8,2,2,2]$
$[67] [8,8,4,1,1]$
$[68] [8,8,4,2]$
$[69] [16,1,1,1,1,1,1]$
$[70] [16,2,1,1,1,1]$
$[71] [16,2,2,1,1]$
$[72] [16,2,2,2]$
$[73] [16,4,1,1]$
$[74] [16,4,2]$

The $74$ binary partitions for $23$ are:
$[1] [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1]$
$[2] [2,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1]$
$[3] [2,2,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1]$
$[4] [2,2,2,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1]$
$[5] [2,2,2,2,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1]$
$[6] [2,2,2,2,2,1,1,1,1,1,1,1,1,1,1,1,1,1]$
$[7] [2,2,2,2,2,2,1,1,1,1,1,1,1,1,1,1,1]$
$[8] [2,2,2,2,2,2,2,1,1,1,1,1,1,1,1,1]$
$[9] [2,2,2,2,2,2,2,2,1,1,1,1,1,1,1]$
$[10] [2,2,2,2,2,2,2,2,2,1,1,1,1,1]$
$[11] [2,2,2,2,2,2,2,2,2,2,1,1,1]$
$[12] [2,2,2,2,2,2,2,2,2,2,2,1]$
$[13] [4,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1]$
$[14] [4,2,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1]$
$[15] [4,2,2,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1]$
$[16] [4,2,2,2,1,1,1,1,1,1,1,1,1,1,1,1,1]$
$[17] [4,2,2,2,2,1,1,1,1,1,1,1,1,1,1,1]$
$[18] [4,2,2,2,2,2,1,1,1,1,1,1,1,1,1]$
$[19] [4,2,2,2,2,2,2,1,1,1,1,1,1,1]$
$[20] [4,2,2,2,2,2,2,2,1,1,1,1,1]$
$[21] [4,2,2,2,2,2,2,2,2,1,1,1]$
$[22] [4,2,2,2,2,2,2,2,2,2,1]$
$[23] [4,4,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1]$
$[24] [4,4,2,1,1,1,1,1,1,1,1,1,1,1,1,1]$
$[25] [4,4,2,2,1,1,1,1,1,1,1,1,1,1,1]$
$[26] [4,4,2,2,2,1,1,1,1,1,1,1,1,1]$
$[27] [4,4,2,2,2,2,1,1,1,1,1,1,1]$
$[28] [4,4,2,2,2,2,2,1,1,1,1,1]$
$[29] [4,4,2,2,2,2,2,2,1,1,1]$
$[30] [4,4,2,2,2,2,2,2,2,1]$
$[31] [4,4,4,1,1,1,1,1,1,1,1,1,1,1]$
$[32] [4,4,4,2,1,1,1,1,1,1,1,1,1]$
$[33] [4,4,4,2,2,1,1,1,1,1,1,1]$
$[34] [4,4,4,2,2,2,1,1,1,1,1]$
$[35] [4,4,4,2,2,2,2,1,1,1]$
$[36] [4,4,4,2,2,2,2,2,1]$
$[37] [4,4,4,4,1,1,1,1,1,1,1]$
$[38] [4,4,4,4,2,1,1,1,1,1]$
$[39] [4,4,4,4,2,2,1,1,1]$
$[40] [4,4,4,4,2,2,2,1]$
$[41] [4,4,4,4,4,1,1,1]$
$[42] [4,4,4,4,4,2,1]$
$[43] [8,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1]$
$[44] [8,2,1,1,1,1,1,1,1,1,1,1,1,1,1]$
$[45] [8,2,2,1,1,1,1,1,1,1,1,1,1,1]$
$[46] [8,2,2,2,1,1,1,1,1,1,1,1,1]$
$[47] [8,2,2,2,2,1,1,1,1,1,1,1]$
$[48] [8,2,2,2,2,2,1,1,1,1,1]$
$[49] [8,2,2,2,2,2,2,1,1,1]$
$[50] [8,2,2,2,2,2,2,2,1]$
$[51] [8,4,1,1,1,1,1,1,1,1,1,1,1]$
$[52] [8,4,2,1,1,1,1,1,1,1,1,1]$
$[53] [8,4,2,2,1,1,1,1,1,1,1]$
$[54] [8,4,2,2,2,1,1,1,1,1]$
$[55] [8,4,2,2,2,2,1,1,1]$
$[56] [8,4,2,2,2,2,2,1]$
$[57] [8,4,4,1,1,1,1,1,1,1]$
$[58] [8,4,4,2,1,1,1,1,1]$
$[59] [8,4,4,2,2,1,1,1]$
$[60] [8,4,4,2,2,2,1]$
$[61] [8,4,4,4,1,1,1]$
$[62] [8,4,4,4,2,1]$
$[63] [8,8,1,1,1,1,1,1,1]$
$[64] [8,8,2,1,1,1,1,1]$
$[65] [8,8,2,2,1,1,1]$
$[66] [8,8,2,2,2,1]$
$[67] [8,8,4,1,1,1]$
$[68] [8,8,4,2,1]$
$[69] [16,1,1,1,1,1,1,1]$
$[70] [16,2,1,1,1,1,1]$
$[71] [16,2,2,1,1,1]$
$[72] [16,2,2,2,1]$
$[73] [16,4,1,1,1]$
$[74] [16,4,2,1]$

## Thursday, 14 March 2013

### THURSDAY, 14 MARCH 2013

Today is the $73^{rd}$ day of the year.

$73$ is prime and an Emirp (see Tuesday 12 March).

$73_{10} = 1001001_2$ which means that $73$ is palindromic in binary.

There are two primitive Pythagorean triples containing $73$
$(55, 48, 73)$
$(73, 2664, 2665)$

The date today is $14^{th}$ March. If one writes the date in either the ISO format (see XKCD 1179) of $2013.3.14$ or the American format of $3.14.2013$ then it contains the sub-string $3.14$ which is the beginning of the ratio we know as $\pi$. Thus, today is known by many as Pi Day and it is used as an excuse to try and get people interested in mathematics. There is even a website.

Since $\pi$ continues infinitely without repetition or pattern then, it is proposed, any sequence of digits turns up sooner or later. There is a $\pi$ search website here that allows one to see if a particular digit sequence occurs within the $20,000,000$ digits. Out of interest the date in English format, $14022013$, occurs at position $40,231,854$.
$0$ first occurs at position $32$
$1$ first occurs at position $1$
$2$ first occurs at position $6$
$3$ first occurs at position $9$
$4$ first occurs at position $2$
$5$ first occurs at position $4$
$6$ first occurs at position $7$
$7$ first occurs at position $13$
$8$ first occurs at position $11$
$9$ first occurs at position $5$
$10$ first occurs at position $49$
...
$73$ first occurs at position $299$.
$2$ occurs for the eighth time at position $73$
The sequence of positions of $n$ within the expansion of $\pi$ which starts $32, 1, 6, 9, 2, 4, 7, 13, 11, 5, 49$ is sequence A014777 in OEIS (and $014777$ occurs at position $2,025,745$).
My birthday in $ddmmyyyy$ format occurs at position $198,662,921$.

$73$ is the fourth member of sequence A023001,  whose first nine members are:
$0, 1, 9, 73, 585, 4681, 37449, 299593, 2396745$
The definition of this sequence is $a(n) = \frac {8^n - 1}{7}$. What is fascinating about this is the implication that every power of $8$ less one is divisible by seven. Can anyone find a proof for that?

## Wednesday, 13 March 2013

### WEDNESDAY, 13 MARCH 2013

Today is the $72^{nd}$ day of the year.

$72 = 13 + 17 + 19 + 23$ which means it is the sum of four consecutive primes, see A034963.

$72$ is a member of the following primitive Pythagorean triples:

$(65, 72, 97 )$
$(1295, 72, 1297)$

$72 = 2^3 \times 3^2$

The factorisation of $72$ above has a nice symmetry to it. It is of the form $n^{n+1} \times (n+1)^n$.
The first seven numbers of such a sequence are as follows:
$0^1 \times 1^0 = 0$
$1^2 \times 2^1 = 2$
$2^3 \times 3^2 = 72$
$3^4 \times 4^3 = 5184$
$4^5 \times 5^4 = 640000$
$5^6 \times 6^5 = 121500000$
$6^7 \times 7^6 = 32934190464$
Surely nobody has thought of putting that into the On-Line Encyclopedia of Integer Sequences?

Of course they have, it is A051443.

Imagine a ruler that instead of having marks at regular intervals only had them at irregular intervals but in a specific irregular way. One specific irregular way would be to have the marks at integer intervals but in such a way that no two pairs of marks are the same distance apart. An example of a ruler of this type would be one which had marks at $2, 3$ and $6$. The distances between the pairs are all different:
$3 - 2 = 1$
$6 - 2 = 4$
$6 - 3 = 3$
A ruler like this is called a Golomb Ruler for Solomon W. Golomb, see A003022.
Now consider a ruler with marks at $0, 1, 4$ and $6$. the distances between all the pairs of marks on this ruler are
$1 - 0 = 1$
$4 - 0 = 4$
$6 - 0 = 6$
$4 - 1 = 3$
$6 - 1 = 5$
$6 - 4 = 2$
Inspection shows that all possible distances up to the length of the ruler can be measured with this arrangement. A ruler like this is called a perfect Golomb ruler.
The number of marks on a Golomb ruler is referred to as its Order and the largest number is referred to as its Length$^*$. Which leads us to the definition of an Optimal Golomb ruler:
A Golomb Ruler is Optimal if there exists no shorter Golomb ruler of the same Order.
A003022 is a list of length of the optimal rulers for each value of $n$. From the relevant page in Wikipedia we can see that the Optimal ruler for eleven marks is $72$ units and has marks at either:
$0, 1, 4, 13, 28, 33, 47, 54, 64, 70, 72$
or
$0, 1, 9, 19, 24, 31, 52, 56, 58, 69, 72$

Consider calculating the distances between all these marks for the first arrangement.

 0 1 4 13 28 33 47 54 64 70 72 0 1 4 13 28 33 47 54 64 70 72 1 3 12 27 32 46 53 63 69 71 4 9 24 29 43 50 60 66 68 13 15 20 34 41 51 57 59 28 5 19 26 36 42 44 33 14 21 31 37 39 47 7 17 23 25 54 10 16 18 64 6 8 70 2 72

A quick count shows that there are $10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 55$ values. We know that they are all different because this is the definition of a Golomb ruler. However, $55 < 72$ so there cannot be all possible distances between $1$ and $72$ represented on this ruler. We can conclude, therefore, that this is not a perfect Golomb ruler.

$^*$ Assuming it starts at zero. More technically the length is the difference between the largest and smallest number.

## Tuesday, 12 March 2013

### TUESDAY, 12 MARCH 2013

Today is the $71^{st}$ day of the year.

$71$ is prime.

$71$ is also an Emirp. An Emirp is a prime number whose digits, when reveresed, also form a prime number. In this case, of course, it is $17$. See A006567.

If one takes any four consecutive numbers, multiplies them together and adds one then the resulting number is a perfect square. Here are the first eleven of these calculations:
$(1 \times 2 \times 3 \times 4) + 1 = 25 = 5^2$
$(2 \times 3 \times 4 \times 5) + 1 = 121 = 11^2$
$(3 \times 4 \times 5 \times 6) + 1 = 361 = 19^2$
$(4 \times 5 \times 6 \times 7) + 1 = 841 = 29^2$
$(5 \times 6 \times 7 \times 8) + 1 = 1681 = 41^2$
$(6 \times 7 \times 8 \times 9) + 1 = 3025 = 55^2$
$(7 \times 8 \times 9 \times 10) + 1 = 5041 = 71^2$
$(8 \times 9 \times 10 \times 11) + 1 = 7921 = 89^2$
$(9 \times 10 \times 11 \times 12) + 1 = 11881 = 109^2$
$(10 \times 11 \times 12 \times 13) + 1 = 17161 = 131^2$
$(11 \times 12 \times 13 \times 14) + 1 = 24025 = 155^2$
$(12 \times 13 \times 14 \times 15) + 1 = 32761 = 181^2$
The sequence of roots of these calculations is $5, 11, 19, 29, 41, 55, 71, 89, ...$
Not suprisingly this sequence is a sequence at the On-Line Encyclopedia of Integer Sequences, it is A028387.
The sequence has a formula of $n + (n + 1)^2$. As can be observered, $71$ is the seventh member of the sequence and $71 = 7 + 8^2$

## Monday, 11 March 2013

### MONDAY, 11 MARCH 2013

Today is the $70^{th}$ day of the year.

$70 = 2 \times 5 \times 7$

Consider a Fibonacci-like sequence where $a(0) = 0$ and $a(1) = 1$ and the recurrence equation is $a(n) = 2a(n - 1) + a(n - 2)$ then the first twelve members of the sequence are:
$0, 1, 2, 5, 12, 29, 70, 169, 408, 985, 2378, 5741$
which includes today's number, $70$, at the sixth position. These are known as the Pell Numbers, see A000129.

All of these Fibonacci-like sequences can be generalised as
$a(0) = p, a(1) = q, a(n) = s.a(n - 1) + r.a(n - 2)$.
This is known as the Horodam Sequence, see Wolfram Mathworld.
In the case of the Pell Numbers the four constants, $(p, q, r, s)$, are $(0, 1, 1, 2)$

Consider a triangle of numbers constructed so that the first two lines are the same as Pascal's Triangle. All subsequent lines start and end with the number one, with all intervening numbers being the sum of the triangle of three numbers above them, giving the following:

1
1.... 1
1.... 3.... 1
1.... 5.... 5.... 1
1.... 7....13.... 7.... 1
1.... 9....25....25.... 9.... 1
1....11....41....63....41....11.... 1

If each line is summed then we get:

1                   =   1
1  +  1                =   2
1  +  3  +  1             =   5
1  +  5  +  5  +  1          =  12
1  +  7  + 13  +  7  +  1       =  29
1  +  9  + 25  + 25  +  9  +  1    =  70
1  + 11  + 41  + 63  + 41  + 11  +  1 = 169

Inspection shows that the results of these sums are the same as the Pell Numbers (search for Patrick Costello in A000129).

## Sunday, 10 March 2013

### SUNDAY, 10 MARCH 2013

Today is the $69^{th}$ day of the year

$69 = 3 \times 23$

$69$ is a member of the following two Pythagorean triples $(69, 260, 269)$, $(69, 2380, 2381)$.

$69$ is a Lucky Number, see A000959.
Lucky numbers are defined by a sieve like process. Starting with a list of the Natural Numbers$^*$ delete every every second number noting that $2$ is the next number in the list after the number one. We now have a list that looks like $1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21...$ or all the odd numbers.
We continue the process by deleting every $x^{th}$ number where $x$ is the next number in the list that we haven't previously used which means that we use $3$ next i.e. delete every third number in the list to give us $1, 3, 7, 9, 13, 15, 19, 21...$. The next number remaining in the list that we haven't previously used is $7$. Deleting every seventh number leaves the list as $1, 3, 7, 9, 13, 15, 21, ...$.
The  next number remaining in the list that we haven't previously used is $9$ so we delete every ninth number , then every thirteenth, then every fifteenth and so on. The numbers that remain are the Lucky Numbers.
Although the whole sequence can never be determined it is easy, though laborious, to determine whether a number is a member of the sequence by following the above process until either the number in question is removed from the list or the next round of deletions starts past the current position of the number in question in the list.
As far as I can ascertain there is no simple test for whether a number is Lucky.

$^*$ For this purpose the Natural Numbers are considered to be all the positive integers without zero

## Saturday, 9 March 2013

### SATURDAY, 9 MARCH 2013

Today is the $68^{th}$ day of the year.

$68 = 2^2 * 17$

$68 = 31 + 37$ which makes it the sum of two successive primes, see A001043.

Since $6^2 + 8^2 = 36 + 64 = 100$ and $1^2 + 0^2 + 0^2 = 1 + 0 + 0 = 1$ then $68$ is a Happy Number, see A007770.

$68$ is the tenth member of the Tribonacci Numbers, see A001590 and Tuesday, 26 February.

$68$ is the number of digits in the number $52!$, see A034886.

Taking the average of the prime factors of $68$ gives $\frac {2 + 2 + 17}{3} = \frac {21}{3} = 7$. $7$ is prime so $68$ is the $16^{th}$ member of the sequence of non-prime numbers where the arithmetic mean of the factors is prime, see A134344.

## Friday, 8 March 2013

### FRIDAY, 8 MARCH 2013

Today is the $67^{th}$ day of the year.

$67$ is a prime number and is the nineteenth in the sequence of prime numbers, see A000040.

$67$ is a toothpick number, see A139250 and a nice illustration here.

Consider the powers of $2$ modulo 67.
$2^0 = 1 = 1$
$2^{1} = 2 * 2^{0} = 2 * 1 = 2$
$2^{2} = 2 * 2^{1} = 2 * 2 = 4$
$2^{3} = 2 * 2^{2} = 2 * 4 = 8$
$2^{4} = 2 * 2^{3} = 2 * 8 = 16$
$2^{5} = 2 * 2^{4} = 2 * 16 = 32$
$2^{6} = 2 * 2^{5} = 2 * 32 = 64$
$2^{7} = 2 * 2^{6} = 2 * 64 = 61$
$2^{8} = 2 * 2^{7} = 2 * 61 = 55$
$2^{9} = 2 * 2^{8} = 2 * 55 = 43$
$2^{10} = 2 * 2^{9} = 2 * 43 = 19$
$2^{11} = 2 * 2^{10} = 2 * 19 = 38$
$2^{12} = 2 * 2^{11} = 2 * 38 = 9$
$2^{13} = 2 * 2^{12} = 2 * 9 = 18$
$2^{14} = 2 * 2^{13} = 2 * 18 = 36$
$2^{15} = 2 * 2^{14} = 2 * 36 = 5$
$2^{16} = 2 * 2^{15} = 2 * 5 = 10$
$2^{17} = 2 * 2^{16} = 2 * 10 = 20$
$2^{18} = 2 * 2^{17} = 2 * 20 = 40$
$2^{19} = 2 * 2^{18} = 2 * 40 = 13$
$2^{20} = 2 * 2^{19} = 2 * 13 = 26$
$2^{21} = 2 * 2^{20} = 2 * 26 = 52$
$2^{22} = 2 * 2^{21} = 2 * 52 = 37$
$2^{23} = 2 * 2^{22} = 2 * 37 = 7$
$2^{24} = 2 * 2^{23} = 2 * 7 = 14$
$2^{25} = 2 * 2^{24} = 2 * 14 = 28$
$2^{26} = 2 * 2^{25} = 2 * 28 = 56$
$2^{27} = 2 * 2^{26} = 2 * 56 = 45$
$2^{28} = 2 * 2^{27} = 2 * 45 = 23$
$2^{29} = 2 * 2^{28} = 2 * 23 = 46$
$2^{30} = 2 * 2^{29} = 2 * 46 = 25$
$2^{31} = 2 * 2^{30} = 2 * 25 = 50$
$2^{32} = 2 * 2^{31} = 2 * 50 = 33$
$2^{33} = 2 * 2^{32} = 2 * 33 = 66$
$2^{34} = 2 * 2^{33} = 2 * 66 = 65$
$2^{35} = 2 * 2^{34} = 2 * 65 = 63$
$2^{36} = 2 * 2^{35} = 2 * 63 = 59$
$2^{37} = 2 * 2^{36} = 2 * 59 = 51$
$2^{38} = 2 * 2^{37} = 2 * 51 = 35$
$2^{39} = 2 * 2^{38} = 2 * 35 = 3$
$2^{40} = 2 * 2^{39} = 2 * 3 = 6$
$2^{41} = 2 * 2^{40} = 2 * 6 = 12$
$2^{42} = 2 * 2^{41} = 2 * 12 = 24$
$2^{43} = 2 * 2^{42} = 2 * 24 = 48$
$2^{44} = 2 * 2^{43} = 2 * 48 = 29$
$2^{45} = 2 * 2^{44} = 2 * 29 = 58$
$2^{46} = 2 * 2^{45} = 2 * 58 = 49$
$2^{47} = 2 * 2^{46} = 2 * 49 = 31$
$2^{48} = 2 * 2^{47} = 2 * 31 = 62$
$2^{49} = 2 * 2^{48} = 2 * 62 = 57$
$2^{50} = 2 * 2^{49} = 2 * 57 = 47$
$2^{51} = 2 * 2^{50} = 2 * 47 = 27$
$2^{52} = 2 * 2^{51} = 2 * 27 = 54$
$2^{53} = 2 * 2^{52} = 2 * 54 = 41$
$2^{54} = 2 * 2^{53} = 2 * 41 = 15$
$2^{55} = 2 * 2^{54} = 2 * 15 = 30$
$2^{56} = 2 * 2^{55} = 2 * 30 = 60$
$2^{57} = 2 * 2^{56} = 2 * 60 = 53$
$2^{58} = 2 * 2^{57} = 2 * 53 = 39$
$2^{59} = 2 * 2^{58} = 2 * 39 = 11$
$2^{60} = 2 * 2^{59} = 2 * 11 = 22$
$2^{61} = 2 * 2^{60} = 2 * 22 = 44$
$2^{62} = 2 * 2^{61} = 2 * 44 = 21$
$2^{63} = 2 * 2^{62} = 2 * 21 = 42$
$2^{64} = 2 * 2^{63} = 2 * 42 = 17$
$2^{65} = 2 * 2^{64} = 2 * 17 = 34$

Every number between $1$ and $66$ appears just once. There are no repetitions. This makes $2$ a primitive root of, the prime, $67$. See A001122.

This is the above list in modulo $67$ order.

$2^{0} = 1$
$2^{1} = 2 * 2^{0} = 2 * 1 = 2$
$2^{39} = 2 * 2^{38} = 2 * 35 = 3$
$2^{2} = 2 * 2^{1} = 2 * 2 = 4$
$2^{15} = 2 * 2^{14} = 2 * 36 = 5$
$2^{40} = 2 * 2^{39} = 2 * 3 = 6$
$2^{23} = 2 * 2^{22} = 2 * 37 = 7$
$2^{3} = 2 * 2^{2} = 2 * 4 = 8$
$2^{12} = 2 * 2^{11} = 2 * 38 = 9$
$2^{16} = 2 * 2^{15} = 2 * 5 = 10$
$2^{59} = 2 * 2^{58} = 2 * 39 = 11$
$2^{41} = 2 * 2^{40} = 2 * 6 = 12$
$2^{19} = 2 * 2^{18} = 2 * 40 = 13$
$2^{24} = 2 * 2^{23} = 2 * 7 = 14$
$2^{54} = 2 * 2^{53} = 2 * 41 = 15$
$2^{4} = 2 * 2^{3} = 2 * 8 = 16$
$2^{64} = 2 * 2^{63} = 2 * 42 = 17$
$2^{13} = 2 * 2^{12} = 2 * 9 = 18$
$2^{10} = 2 * 2^{9} = 2 * 43 = 19$
$2^{17} = 2 * 2^{16} = 2 * 10 = 20$
$2^{62} = 2 * 2^{61} = 2 * 44 = 21$
$2^{60} = 2 * 2^{59} = 2 * 11 = 22$
$2^{28} = 2 * 2^{27} = 2 * 45 = 23$
$2^{42} = 2 * 2^{41} = 2 * 12 = 24$
$2^{30} = 2 * 2^{29} = 2 * 46 = 25$
$2^{20} = 2 * 2^{19} = 2 * 13 = 26$
$2^{51} = 2 * 2^{50} = 2 * 47 = 27$
$2^{25} = 2 * 2^{24} = 2 * 14 = 28$
$2^{44} = 2 * 2^{43} = 2 * 48 = 29$
$2^{55} = 2 * 2^{54} = 2 * 15 = 30$
$2^{47} = 2 * 2^{46} = 2 * 49 = 31$
$2^{5} = 2 * 2^{4} = 2 * 16 = 32$
$2^{32} = 2 * 2^{31} = 2 * 50 = 33$
$2^{65} = 2 * 2^{64} = 2 * 17 = 34$
$2^{38} = 2 * 2^{37} = 2 * 51 = 35$
$2^{14} = 2 * 2^{13} = 2 * 18 = 36$
$2^{22} = 2 * 2^{21} = 2 * 52 = 37$
$2^{11} = 2 * 2^{10} = 2 * 19 = 38$
$2^{58} = 2 * 2^{57} = 2 * 53 = 39$
$2^{18} = 2 * 2^{17} = 2 * 20 = 40$
$2^{53} = 2 * 2^{52} = 2 * 54 = 41$
$2^{63} = 2 * 2^{62} = 2 * 21 = 42$
$2^{9} = 2 * 2^{8} = 2 * 55 = 43$
$2^{61} = 2 * 2^{60} = 2 * 22 = 44$
$2^{27} = 2 * 2^{26} = 2 * 56 = 45$
$2^{29} = 2 * 2^{28} = 2 * 23 = 46$
$2^{50} = 2 * 2^{49} = 2 * 57 = 47$
$2^{43} = 2 * 2^{42} = 2 * 24 = 48$
$2^{46} = 2 * 2^{45} = 2 * 58 = 49$
$2^{31} = 2 * 2^{30} = 2 * 25 = 50$
$2^{37} = 2 * 2^{36} = 2 * 59 = 51$
$2^{21} = 2 * 2^{20} = 2 * 26 = 52$
$2^{57} = 2 * 2^{56} = 2 * 60 = 53$
$2^{52} = 2 * 2^{51} = 2 * 27 = 54$
$2^{8} = 2 * 2^{7} = 2 * 61 = 55$
$2^{26} = 2 * 2^{25} = 2 * 28 = 56$
$2^{49} = 2 * 2^{48} = 2 * 62 = 57$
$2^{45} = 2 * 2^{44} = 2 * 29 = 58$
$2^{36} = 2 * 2^{35} = 2 * 63 = 59$
$2^{56} = 2 * 2^{55} = 2 * 30 = 60$
$2^{7} = 2 * 2^{6} = 2 * 64 = 61$
$2^{48} = 2 * 2^{47} = 2 * 31 = 62$
$2^{35} = 2 * 2^{34} = 2 * 65 = 63$
$2^{6} = 2 * 2^{5} = 2 * 32 = 64$
$2^{34} = 2 * 2^{33} = 2 * 66 = 65$
$2^{33} = 2 * 2^{32} = 2 * 33 = 66$