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

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

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.

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

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.

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
$\frac {p^n - 1}{p - 1} \in \mathbb{N}  \forall n, p \in \mathbb{N}$