Today is the $59^{th}$ day of the year.

$59$ is prime.

$59 = 1^2 + 3^2 + 7^2 = 3^2 + 5^2 + 5^2$

$(59, 1740, 1741)$ is a Pythagorean triple.

$59$ is a vile number, see A003159 and AviezriS.Fraenkel's original paper.

$59$ is a safe prime because $\frac {59 - 1}{2} = \frac {58}{2} = 29$ which is prime. $29$ is a Sophie Germain prime, see here and here.

## Thursday, 28 February 2013

## Wednesday, 27 February 2013

### WEDNESDAY, 27 FEBRUARY 2013

Today is the $58^{th}$ day of the year.

$58 = 2 \times 29$

$58 = 3^2 + 7 ^2$

$58 = 13 + 14 + 15 + 16$ which makes it a trapezoidal number, see here.

$58 = 2 + 3 + 5 + 7 + 11 + 13 + 17$ which means it is the sum of the first $7$ primes, see here.

$58$ is a hendecagonal number. A hendecagonal number is the equivalent of a triangular number but for a shape that has 11 sides. The first 11 of the 11-gonal or hendecagonal numbers is as follows:

$0, 1, 11, 30, 58, 95, 141, 196, 260, 333, 415$

This series can also be generated by writing the natural numbers in a triangular spiral and then reading off the line that starts 0, 1, 11,

..................36

................37..35

..............38..15..34

............39..16..14..33

..........40..17.. 3..13..32

........41..18.. 4.. 2..12..31

......42..19.. 5.. 0.. 1..11..30..58

....43..20.. 6.. 7.. 8.. 9..10..29..57

..44..21..22..23..24..25..26..27..28..56

45..46..47..48..49..50..51..52..53..54..55

$58 = 2 \times 29$

$58 = 3^2 + 7 ^2$

$58 = 13 + 14 + 15 + 16$ which makes it a trapezoidal number, see here.

$58 = 2 + 3 + 5 + 7 + 11 + 13 + 17$ which means it is the sum of the first $7$ primes, see here.

$58$ is a hendecagonal number. A hendecagonal number is the equivalent of a triangular number but for a shape that has 11 sides. The first 11 of the 11-gonal or hendecagonal numbers is as follows:

$0, 1, 11, 30, 58, 95, 141, 196, 260, 333, 415$

This series can also be generated by writing the natural numbers in a triangular spiral and then reading off the line that starts 0, 1, 11,

..................36

................37..35

..............38..15..34

............39..16..14..33

..........40..17.. 3..13..32

........41..18.. 4.. 2..12..31

......42..19.. 5.. 0.. 1..11..30..58

....43..20.. 6.. 7.. 8.. 9..10..29..57

..44..21..22..23..24..25..26..27..28..56

45..46..47..48..49..50..51..52..53..54..55

## Tuesday, 26 February 2013

### TUESDAY, 26 FEBRURAY 2013

Today is the $57^{th}$ day of the year.

$57 = 3 \times 19$

$57$ is the eighth member of the Tribonacci Numbers, see A000213. Tribonacci Numbers are a natural extension to the Fibonacci Numbers where, instead of the last two numbers being added together, the last three numbers are added together giving a definition of:

$a(n) = a(n - 1) + a(n - 2) + a(n - 3) with a(0) = a(1) = a(2) = 1$

This gives a sequence whose first seventeen numbers are:

$1, 1, 1, 3, 5, 9, 17, 31, 57, 105, 193, 355, 653, 1201, 2209, 4063, 7473$

As you will observe, this sequence gets large very quickly.

$57 = 3 \times 19$

$57$ is the eighth member of the Tribonacci Numbers, see A000213. Tribonacci Numbers are a natural extension to the Fibonacci Numbers where, instead of the last two numbers being added together, the last three numbers are added together giving a definition of:

$a(n) = a(n - 1) + a(n - 2) + a(n - 3) with a(0) = a(1) = a(2) = 1$

This gives a sequence whose first seventeen numbers are:

$1, 1, 1, 3, 5, 9, 17, 31, 57, 105, 193, 355, 653, 1201, 2209, 4063, 7473$

As you will observe, this sequence gets large very quickly.

## Monday, 25 February 2013

### MONDAY, 25 FEBRUARY 2013

Today is the $56^{th}$ day of the year.

$56 = 2^3 \times 7$

$56$ is the number of times that $11$ can be partitioned. A partition of a number is the number of different ways that the number can be written as a sum of integers where the order is not significant. Thus $11$ has the following partitions:

$56 = 2^3 \times 7$

$56$ is the number of times that $11$ can be partitioned. A partition of a number is the number of different ways that the number can be written as a sum of integers where the order is not significant. Thus $11$ has the following partitions:

- $11$
- $10 + 1$
- $9 + 2$
- $8 + 3$
- $7 + 4$
- $6 + 5$
- $9 + 1 + 1$
- $8 + 2 + 1$
- $7 + 3 + 1$
- $7 + 2 + 2$
- $6 + 4 + 1$
- $6 + 3 + 2$
- $5 + 5 + 1$
- $5 + 4 + 2$
- $5 + 3 + 3$
- $4 + 4 + 3$
- $8 + 1 + 1 + 1$
- $7 + 2 + 1 + 1$
- $6 + 3 + 1 + 1$
- $6 + 2 + 2 + 1$
- $5 + 4 + 1 + 1$
- $5 + 3 + 2 + 1$
- $5 + 2 + 2 + 2$
- $4 + 4 + 2 + 1$
- $4 + 3 + 3 + 1$
- $4 + 3 + 2 + 2$
- $3 + 3 + 3 + 2$
- $7 + 1 + 1 + 1 + 1$
- $6 + 2 + 1 + 1 + 1$
- $5 + 3 + 1 + 1 + 1$
- $5 + 2 + 2 + 1 + 1$
- $4 + 4 + 1 + 1 + 1$
- $4 + 3 + 2 + 1 + 1$
- $4 + 2 + 2 + 2 + 1$
- $3 + 3 + 3 + 1 + 1$
- $3 + 3 + 2 + 2 + 1$
- $3 + 2 + 2 + 2 + 2$
- $6 + 1 + 1 + 1 + 1 + 1$
- $5 + 2 + 1 + 1 + 1 + 1$
- $4 + 3 + 1 + 1 + 1 + 1$
- $4 + 2 + 2 + 1 + 1 + 1$
- $3 + 3 + 2 + 1 + 1 + 1$
- $3 + 2 + 2 + 2 + 1 + 1$
- $2 + 2 + 2 + 2 + 2 + 1$
- $5 + 1 + 1 + 1 + 1 + 1 + 1$
- $4 + 2 + 1 + 1 + 1 + 1 + 1$
- $3 + 3 + 1 + 1 + 1 + 1 + 1$
- $3 + 2 + 2 + 1 + 1 + 1 + 1$
- $2 + 2 + 2 + 2 + 1 + 1 + 1$
- $4 + 1 + 1 + 1 + 1 + 1 + 1 + 1$
- $3 + 2 + 1 + 1 + 1 + 1 + 1 + 1$
- $2 + 2 + 2 + 1 + 1 + 1 + 1 + 1$
- $3 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1$
- $2 + 2 + 1 + 1 + 1 + 1 + 1 + 1 + 1$
- $2 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1$
- $1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1$

## Sunday, 24 February 2013

### SUNDAY, 24 FEBRUARY 2013

Today is the $55^{th}$ day of the year.

$55 = 5 \times 11$

$55$ is a member of the Fibonacci Sequence, is a Triangular Number and is in Pascal's Triangle. Actually, the fact that it is a triangular number implies that it is in Pascal's triangle. This is because, from the third row of the triangle down, the third number in each row is the next member of the triangular number sequence, see here for an illustration.

Slightly less trivially, $55$ is a member of the Toothpick Sequence which is nicely illustrated here or there is an animated version here. To use this animation, leave everything as it is and just click the Next button. When you see the value 10 in the field labelled N: then there are 55 toothpicks in the diagram.

$55 = 5 \times 11$

$55$ is a member of the Fibonacci Sequence, is a Triangular Number and is in Pascal's Triangle. Actually, the fact that it is a triangular number implies that it is in Pascal's triangle. This is because, from the third row of the triangle down, the third number in each row is the next member of the triangular number sequence, see here for an illustration.

Slightly less trivially, $55$ is a member of the Toothpick Sequence which is nicely illustrated here or there is an animated version here. To use this animation, leave everything as it is and just click the Next button. When you see the value 10 in the field labelled N: then there are 55 toothpicks in the diagram.

## Saturday, 23 February 2013

### SATURDAY, 23 FEBRUARY 2013

Today is the $54^{th}$ day of the year.

$54 = 2 \times 3^3$

There are some prime numbers that are one greater than a square number. $2,917$ is one such number, see A002496. $2,917 = 2,916 + 1 = 54^2 + 1$, thus $54$ is one of the sequence of numbers $n$ such that $n^2 + 1$ is prime, see A005574. $n$ could be thought of as a Near Root Prime.

It is interesting to speculate on whether there is a number that is in this sequence, let's call it $p$, that when one is added to its square is also a member of this sequence i.e is there a number $p$ such that $p^2 + 1$ is a Near Root Prime. (cf Cunningham chains).

It is fairly easy to show that no such number exists. Assume that $p$ is a Near Root Prime then we know by definition that $p^2 + 1$ is prime. All primes greater than 2 are odd, therefore $p^2 + 1$ is odd. This means that $p^2$ is even which imples that $p$ is even assuming $p$ is greater than $1$. Inspection of A005574 bears this out.

If all the members of the Near Root Prime sequence are even , except the first member, then $p^2 + 1$ cannot be a Near Root Prime because it is odd.

$54 = 2 \times 3^3$

There are some prime numbers that are one greater than a square number. $2,917$ is one such number, see A002496. $2,917 = 2,916 + 1 = 54^2 + 1$, thus $54$ is one of the sequence of numbers $n$ such that $n^2 + 1$ is prime, see A005574. $n$ could be thought of as a Near Root Prime.

It is interesting to speculate on whether there is a number that is in this sequence, let's call it $p$, that when one is added to its square is also a member of this sequence i.e is there a number $p$ such that $p^2 + 1$ is a Near Root Prime. (cf Cunningham chains).

It is fairly easy to show that no such number exists. Assume that $p$ is a Near Root Prime then we know by definition that $p^2 + 1$ is prime. All primes greater than 2 are odd, therefore $p^2 + 1$ is odd. This means that $p^2$ is even which imples that $p$ is even assuming $p$ is greater than $1$. Inspection of A005574 bears this out.

If all the members of the Near Root Prime sequence are even , except the first member, then $p^2 + 1$ cannot be a Near Root Prime because it is odd.

## Friday, 22 February 2013

### FRIDAY, 22 FEBRUARY 2013

Today is the $53^{rd}$ day of the year.

$53$ is the $16^{th}$ prime number.

$53$ is also a Sophie Germain prime. Any prime, p, is a Sophie Germain prime if $2p + 1$ is also a prime. Since $107$ is a prime, the $28^{th}$, then $53$ is a Sophie Germain prime.

The first $23$ Sophie Germain primes are:

2, 3, 5, 11, 23, 29, 41, 53, 83, 89, 113, 131, 173, 179, 191, 233, 239, 251, 281, 293, 359, 419, 431, 443, 491, 509, 593, 641, 653, 659, 683, 719, 743, 761, 809, 911, 953

$53$ is the $16^{th}$ prime number.

$53$ is also a Sophie Germain prime. Any prime, p, is a Sophie Germain prime if $2p + 1$ is also a prime. Since $107$ is a prime, the $28^{th}$, then $53$ is a Sophie Germain prime.

The first $23$ Sophie Germain primes are:

2, 3, 5, 11, 23, 29, 41, 53, 83, 89, 113, 131, 173, 179, 191, 233, 239, 251, 281, 293, 359, 419, 431, 443, 491, 509, 593, 641, 653, 659, 683, 719, 743, 761, 809, 911, 953

### THURSDAY, 23 FEBRUARY 2013

Today is the $52^{nd}$ day of the year.

$52 = 2^2 \times 13$

$52$ is the $5^{th}$ Bell Number.

The $n^{th}$ Bell Number is the number of ways you can split up a set containing n elements. The first non-trivial example is when $n = 3$. Assume that we have a set that contains three elements 1, 2 and 3, then the following five partitions are the only distinct divisions of that set:

{1, 2, 3}

{{1} {2, 3}}

{{2} {1, 3}}

{{3} {1, 2}}

{{1} {2} {3}}

$52$ is the $5^{th}$ Bell Number and the $52$ distinct partitions are:

$52 = 2^2 \times 13$

$52$ is the $5^{th}$ Bell Number.

The $n^{th}$ Bell Number is the number of ways you can split up a set containing n elements. The first non-trivial example is when $n = 3$. Assume that we have a set that contains three elements 1, 2 and 3, then the following five partitions are the only distinct divisions of that set:

{1, 2, 3}

{{1} {2, 3}}

{{2} {1, 3}}

{{3} {1, 2}}

{{1} {2} {3}}

$52$ is the $5^{th}$ Bell Number and the $52$ distinct partitions are:

- {1, 2, 3, 4, 5}
- {{1}, {2, 3, 4, 5}}
- {{1}, {2, 3, 4, 5}}
- {{3}, {1, 2, 4, 5}}
- {{4}, {1, 2, 3, 5}}
- {{5}, {1, 2, 3, 4}}
- {{1, 2} {3, 4, 5}}
- {{2, 3} {1, 4, 5}}
- {{3, 4} {1, 2, 5}}
- {{4, 5} {1, 2, 3}}
- {{5, 1} {2, 3, 4}}
- {{1, 3} {2, 4, 5}}
- {{2, 4} {1, 3, 5}}
- {{3, 5} {1, 2, 4}}
- {{4, 1} {2, 3, 5}}
- {{5, 2} {1, 3, 4}}
- {{1} {2} {3, 4, 5}}
- {{1} {3} {2, 4, 5}}
- {{1} {4} {2, 3, 5}}
- {{1} {5} {2, 3, 4}}
- {{2} {3} {1, 4, 5}}
- {{2} {4} {1, 3, 5}}
- {{2} {5} {1, 2, 4}}
- {{3} {4} {1, 2, 5}}
- {{3} {5} {1, 2, 4}}
- {{4} {5} {1, 2, 3}}
- {{1} {2, 3} {4, 5}}
- {{1} {2, 4} {3, 5}}
- {{1} {2, 5} {3, 4}}
- {{2} {1, 3} {4, 5}}
- {{2} {1, 4} {3, 5}}
- {{2} {1, 5} {3, 4}}
- {{3} {1, 2} {4, 5}}
- {{3} {1, 4} {2, 5}}
- {{3} {1, 5} {2, 4}}
- {{4} {1, 2} {3, 5}}
- {{4} {1, 3} {2, 5}}
- {{4} {1, 5} {2, 3}}
- {{5} {1, 2} {3, 4}}
- {{5} {1, 3} {2, 4}}
- {{5} {1, 4} {2, 3}}
- {{1} {2} {3} {4, 5}}
- {{1} {2} {4} {3, 5}}
- {{1} {2} {5} {3, 4}}
- {{1} {3} {4} {2, 5}}
- {{1} {3} {5} {2, 4}}
- {{1} {4} {5} {2, 3}}
- {{2} {3} {4} {1, 5}}
- {{2} {3} {5} {1, 4}}
- {{2} {4} {5} {1, 3}}
- {{3} {4} {5} {1, 2}}
- {{1} {2} {3} {4} {5}}

## Wednesday, 20 February 2013

### WEDNESDAY, 20 FEBRUARY 2013

Today is the $51^{st}$ day of the year.

$51 = 3 \times 17$

$51$ is the sixth Motzkin Number. A Motzkin number is the number of different ways of drawing non-intersecting chords on a circle between n points on the circle's circumference. Here I will unashamedly copy the illustration from Wikipedia which shows that when $n = 4$ the Motzkin number is $9$ i.e. there are $9$ different ways to draw lines connecting $4$ points, not necessarily all of the four points, on the circumference of a circle without the lines intersecting.

The first 21 Motzkin numbers are:

1, 1, 2, 4, 9, 21, 51, 127, 323, 835, 2188, 5798, 15511, 41835, 113634, 310572, 853467, 2356779, 6536382, 18199284, 50852019, 142547559

Like the Fibonacci sequence, this sequence has a relationship that links the next number in the sequence to the previous two numbers. This relation is

$M_{n+1}=\frac{2n+3}{n+3}M_n+\frac{3n}{n+3}M_{n-1}$.

Thus

$M_6 = \frac{2.5+3}{5+3}M_5+\frac{3.5}{5+3}M_4$

knowing that $M_5 = 21$ and $M_4 = 9$ from the list above we get

$M_6 = \frac{13}{8}.21+\frac{15}{8}.9$

$M_6 = \frac {1} {8} (13. 21 + 15 . 9)$

$M_6 = \frac {1} {8} (273 + 135)$

$M_6 = \frac {1} {8} . 408$

Thus

$M_6 = 51$

$51 = 3 \times 17$

$51$ is the sixth Motzkin Number. A Motzkin number is the number of different ways of drawing non-intersecting chords on a circle between n points on the circle's circumference. Here I will unashamedly copy the illustration from Wikipedia which shows that when $n = 4$ the Motzkin number is $9$ i.e. there are $9$ different ways to draw lines connecting $4$ points, not necessarily all of the four points, on the circumference of a circle without the lines intersecting.

The first 21 Motzkin numbers are:

1, 1, 2, 4, 9, 21, 51, 127, 323, 835, 2188, 5798, 15511, 41835, 113634, 310572, 853467, 2356779, 6536382, 18199284, 50852019, 142547559

Like the Fibonacci sequence, this sequence has a relationship that links the next number in the sequence to the previous two numbers. This relation is

$M_{n+1}=\frac{2n+3}{n+3}M_n+\frac{3n}{n+3}M_{n-1}$.

Thus

$M_6 = \frac{2.5+3}{5+3}M_5+\frac{3.5}{5+3}M_4$

knowing that $M_5 = 21$ and $M_4 = 9$ from the list above we get

$M_6 = \frac{13}{8}.21+\frac{15}{8}.9$

$M_6 = \frac {1} {8} (13. 21 + 15 . 9)$

$M_6 = \frac {1} {8} (273 + 135)$

$M_6 = \frac {1} {8} . 408$

Thus

$M_6 = 51$

## Tuesday, 19 February 2013

### TUESDAY, 19 FEBRUARY 2013

Today is the $50^{th}$ day of the year.

$50$ is an odious number and a Harshad or Niven number.

An odious number is any number that has an odd number of ones in its binary expansion.

$50_{10} = 110010_{2}$ so it has three ones in its binary form and is therefore odious.

The first 31 odious numbers are:

1, 2, 4, 7, 8, 11, 13, 14, 16, 19, 21, 22, 25, 26, 28, 31, 32, 35, 37, 38, 41, 42, 44, 47, 49, 50, 52, 55, 56, 59, 61

Those numbers that are not odious are Evil Numbers.

The first 30 Evil Numbers are:

0, 3, 5, 6, 9, 10, 12, 15, 17, 18, 20, 23, 24, 27, 29, 30, 33, 34, 36, 39, 40, 43, 45, 46, 48, 51, 53, 54, 57, 58

A Harshad or Niven Number is any number that is divisible by the sum of its digits.

The sum of the digits of $50$ is $5 + 0 = 5$ and, clearly, $50$ is divisible by $5$.

The first 30 Harshad numbers are:

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 18, 20, 21, 24, 27, 30, 36, 40, 42, 45, 48, 50, 54, 60, 63, 70, 72, 80, 81

$50$ is an odious number and a Harshad or Niven number.

An odious number is any number that has an odd number of ones in its binary expansion.

$50_{10} = 110010_{2}$ so it has three ones in its binary form and is therefore odious.

The first 31 odious numbers are:

1, 2, 4, 7, 8, 11, 13, 14, 16, 19, 21, 22, 25, 26, 28, 31, 32, 35, 37, 38, 41, 42, 44, 47, 49, 50, 52, 55, 56, 59, 61

Those numbers that are not odious are Evil Numbers.

The first 30 Evil Numbers are:

0, 3, 5, 6, 9, 10, 12, 15, 17, 18, 20, 23, 24, 27, 29, 30, 33, 34, 36, 39, 40, 43, 45, 46, 48, 51, 53, 54, 57, 58

A Harshad or Niven Number is any number that is divisible by the sum of its digits.

The sum of the digits of $50$ is $5 + 0 = 5$ and, clearly, $50$ is divisible by $5$.

The first 30 Harshad numbers are:

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 18, 20, 21, 24, 27, 30, 36, 40, 42, 45, 48, 50, 54, 60, 63, 70, 72, 80, 81

## Monday, 18 February 2013

### MONDAY, 18 FEBRUARY 2013

Today is $49^{th}$ day of the year.

$49$ is a perfect square.

$49$ is also a lucky number.

The best description of lucky numbers that I have come across is from Ivars Peterson's MathTrak blog which I quote from below:

Hunting for prime numbers, those evenly divisible only by themselves and 1, requires a sieve to separate them from the rest. For example, the sieve of Eratosthenes, named for a Greek mathematician of the third century B.C., generates a list of prime numbers by the process of elimination.

To find all prime numbers less than, say, 100, the hunter writes down all the integers from 2 to 100 in order (1 doesn't count as a prime). First, 2 is circled, and all multiples of 2 (4, 6, 8, and so on) are struck from the list. That eliminates composite numbers that have 2 as a factor. The next unmarked number is 3. That number is circled, and all multiples of 3 are crossed out. The number 4 is already crossed out, and its multiples have also been eliminated. Five is the next unmarked integer. The procedure continues in this way until only prime numbers are left on the list. Though the sieving process is slow and tedious, it can be continued to infinity to identify every prime number.

Other types of sieves isolate different sequences of numbers. Around 1955, the mathematician Stanislaw Ulam (1909-1984) identified a particular sequence made up of what he called "lucky numbers," and mathematicians have been playing with them ever since.

Starting with a list of integers, including 1, the first step is to cross out every second number: 2, 4, 6, 8, and so on, leaving only the odd integers. The second integer not crossed out is 3. Cross out every third number not yet eliminated. This gets rid of 5, 11, 17, 23, and so on. The third surviving number from the left is 7; cross out every seventh integer not yet eliminated: 19, 39, ... Now, the fourth number from the beginning is 9. Cross out every ninth number not yet eliminated, starting with 27.

This particular sieving process yields certain numbers that permanently escape getting killed. That's why Ulam called them "lucky." See the table below for a list of lucky numbers less than 200.

1 3 7 9 13 15 21 25 31 33 37 43 49 51 63 67 69 73 75 79 87 93 99 105 111 115 127 129 133 135 141 151 159 163 169 171 189 193 195

These lucky numbers should not be confused with Euler's Lucky Numbers, see http://oeis.org/A014556.

For those that are interested, the $66$ days of this year that are related to a lucky number are:

$49$ is a perfect square.

$49$ is also a lucky number.

The best description of lucky numbers that I have come across is from Ivars Peterson's MathTrak blog which I quote from below:

Hunting for prime numbers, those evenly divisible only by themselves and 1, requires a sieve to separate them from the rest. For example, the sieve of Eratosthenes, named for a Greek mathematician of the third century B.C., generates a list of prime numbers by the process of elimination.

To find all prime numbers less than, say, 100, the hunter writes down all the integers from 2 to 100 in order (1 doesn't count as a prime). First, 2 is circled, and all multiples of 2 (4, 6, 8, and so on) are struck from the list. That eliminates composite numbers that have 2 as a factor. The next unmarked number is 3. That number is circled, and all multiples of 3 are crossed out. The number 4 is already crossed out, and its multiples have also been eliminated. Five is the next unmarked integer. The procedure continues in this way until only prime numbers are left on the list. Though the sieving process is slow and tedious, it can be continued to infinity to identify every prime number.

Other types of sieves isolate different sequences of numbers. Around 1955, the mathematician Stanislaw Ulam (1909-1984) identified a particular sequence made up of what he called "lucky numbers," and mathematicians have been playing with them ever since.

Starting with a list of integers, including 1, the first step is to cross out every second number: 2, 4, 6, 8, and so on, leaving only the odd integers. The second integer not crossed out is 3. Cross out every third number not yet eliminated. This gets rid of 5, 11, 17, 23, and so on. The third surviving number from the left is 7; cross out every seventh integer not yet eliminated: 19, 39, ... Now, the fourth number from the beginning is 9. Cross out every ninth number not yet eliminated, starting with 27.

This particular sieving process yields certain numbers that permanently escape getting killed. That's why Ulam called them "lucky." See the table below for a list of lucky numbers less than 200.

1 3 7 9 13 15 21 25 31 33 37 43 49 51 63 67 69 73 75 79 87 93 99 105 111 115 127 129 133 135 141 151 159 163 169 171 189 193 195

These lucky numbers should not be confused with Euler's Lucky Numbers, see http://oeis.org/A014556.

For those that are interested, the $66$ days of this year that are related to a lucky number are:

Lucky Number | Date |

1 | Tuesday 1 January |

3 | Thursday 3 January |

7 | Monday 7 January |

9 | Wednesday 9 January |

13 | Sunday 13 January |

15 | Tuesday 15 January |

21 | Monday 21 January |

25 | Friday 25 January |

31 | Thursday 31 January |

33 | Saturday 2 February |

37 | Wednesday 6 February |

43 | Tuesday 12 February |

49 | Monday 18 February |

51 | Wednesday 20 February |

63 | Monday 4 March |

67 | Friday 8 March |

69 | Sunday 10 March |

73 | Thursday 14 March |

75 | Saturday 16 March |

79 | Wednesday 20 March |

87 | Thursday 28 March |

93 | Wednesday 3 April |

99 | Tuesday 9 April |

105 | Monday 15 April |

111 | Sunday 21 April |

115 | Thursday 25 April |

127 | Tuesday 7 May |

129 | Thursday 9 May |

133 | Monday 13 May |

135 | Wednesday 15 May |

141 | Tuesday 21 May |

151 | Friday 31 May |

159 | Saturday 8 June |

163 | Wednesday 12 June |

169 | Tuesday 18 June |

171 | Thursday 20 June |

189 | Monday 8 July |

193 | Friday 12 July |

195 | Sunday 14 July |

201 | Saturday 20 July |

205 | Wednesday 24 July |

211 | Tuesday 30 July |

219 | Wednesday 7 August |

223 | Sunday 11 August |

231 | Monday 19 August |

235 | Friday 23 August |

237 | Sunday 25 August |

241 | Thursday 29 August |

259 | Monday 16 September |

261 | Wednesday 18 September |

267 | Tuesday 24 September |

273 | Monday 30 September |

283 | Thursday 10 October |

285 | Saturday 12 October |

289 | Wednesday 16 October |

297 | Thursday 24 October |

303 | Wednesday 30 October |

307 | Sunday 3 November |

319 | Friday 15 November |

321 | Sunday 17 November |

327 | Saturday 23 November |

331 | Wednesday 27 November |

339 | Thursday 5 December |

349 | Sunday 15 December |

357 | Monday 23 December |

361 | Friday 27 December |

## Sunday, 17 February 2013

### SUNDAY, 17 FEBRUARY 2013

Today is $48^{th}$ day of the year.

$48$ is the double factorial of $6$, written $6!!$

The value of the double factorial of a number $n$ is defined as follows:

If $n = -1$ or $n = 0$ then $n!! = 1$ otherwise $n!! = n.(n - 2)!!$

Given $n = 6$ then $6!! = 6.4!! = 6.4.2!! = 6.4.2.0!! = 6.4.2.1 = 48$

However, if $n = 5$ then $5!! = 5.3!! = 5.3.1!! = 5.3.1 = 15$

Thus, $6!!.5!! = 6.4.2.1.5.3.1 = 6.5.4.3.2.1 = 6!$

or, equivalently:

$$ n!!.(n - 1)!! = n! $$

$48$ is the double factorial of $6$, written $6!!$

The value of the double factorial of a number $n$ is defined as follows:

If $n = -1$ or $n = 0$ then $n!! = 1$ otherwise $n!! = n.(n - 2)!!$

Given $n = 6$ then $6!! = 6.4!! = 6.4.2!! = 6.4.2.0!! = 6.4.2.1 = 48$

However, if $n = 5$ then $5!! = 5.3!! = 5.3.1!! = 5.3.1 = 15$

Thus, $6!!.5!! = 6.4.2.1.5.3.1 = 6.5.4.3.2.1 = 6!$

or, equivalently:

$$ n!!.(n - 1)!! = n! $$

## Saturday, 16 February 2013

### SATURDAY, 16 FEBRUARY 2013

Today is $47^{th}$ day of the year.

47 is a prime number.

47 is the $8^{th}$ Lucas Number, see http://oeis.org/A000032

The Lucas Number are like the Fibonacci numbers in that the $n^{th}$ number is the sum of the previous two numbers, i.e. the $(n-1)^{th}$ and the $(n-2)^{th}$. In the case of the Fibonacci numbers the first two numbers are 0 and 1 giving the sequence

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, ...

whereas the Lucas Numbers start with 2 and 1 giving the following sequence:

2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, 843, ...

You may be able spot an interesting link between these two series; the $n^{th}$ Lucas Number is equal to the sum of the $(n-1)^{th}$ and the $(n+1)^{th}$ Fibonacci Number i.e the eighth Lucas Number, 47 , is equal to the seventh Fibonacci Number, 13, plus the ninth Fibonacci Number, 34.

We could write this as:

$$L_{n}=F_{n-1}+F_{n+1}$$

Using the same nomenclature, Wikipedia informs us that there are a number of other identities:

\begin{align}L_{m+n} = L_{m+1}F_{n}+L_mF_{n-1}\end{align}

\begin{align}L_n^2 = 5 F_n^2 + 4 (-1)^n\end{align}

\begin{align}F_{2n} = L_n F_n\end{align}

\begin{align}F_n = {L_{n-1}+L_{n+1} \over 5}\end{align}

Equation 1

With $m=3$ and $n=5$ then $L_8 = L_4.F_5 + L_3.F_4 = 7.5 + 4.3 = 35 + 12 = 47$

Equation 2

With $n = 8$ then $L_8^2 = 5.F_8^2 + 4.(-1)^8 = 5.{21}^2 + 4 = 5.441 + 4 = 2,209 = 47^2$

Equation 3

With $n = 8$ then $F_{16} = L_8.F_8 = 47.21 = 987$

Equation 4

With $n = 7$ then $F_7 = {L_6 + L_8 \over 5} = {18 + 47 \over 5} = {65 \over 5} = 13$

47 is a prime number.

47 is the $8^{th}$ Lucas Number, see http://oeis.org/A000032

The Lucas Number are like the Fibonacci numbers in that the $n^{th}$ number is the sum of the previous two numbers, i.e. the $(n-1)^{th}$ and the $(n-2)^{th}$. In the case of the Fibonacci numbers the first two numbers are 0 and 1 giving the sequence

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, ...

whereas the Lucas Numbers start with 2 and 1 giving the following sequence:

2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, 843, ...

You may be able spot an interesting link between these two series; the $n^{th}$ Lucas Number is equal to the sum of the $(n-1)^{th}$ and the $(n+1)^{th}$ Fibonacci Number i.e the eighth Lucas Number, 47 , is equal to the seventh Fibonacci Number, 13, plus the ninth Fibonacci Number, 34.

We could write this as:

$$L_{n}=F_{n-1}+F_{n+1}$$

Using the same nomenclature, Wikipedia informs us that there are a number of other identities:

\begin{align}L_{m+n} = L_{m+1}F_{n}+L_mF_{n-1}\end{align}

\begin{align}L_n^2 = 5 F_n^2 + 4 (-1)^n\end{align}

\begin{align}F_{2n} = L_n F_n\end{align}

\begin{align}F_n = {L_{n-1}+L_{n+1} \over 5}\end{align}

Equation 1

With $m=3$ and $n=5$ then $L_8 = L_4.F_5 + L_3.F_4 = 7.5 + 4.3 = 35 + 12 = 47$

Equation 2

With $n = 8$ then $L_8^2 = 5.F_8^2 + 4.(-1)^8 = 5.{21}^2 + 4 = 5.441 + 4 = 2,209 = 47^2$

Equation 3

With $n = 8$ then $F_{16} = L_8.F_8 = 47.21 = 987$

Equation 4

With $n = 7$ then $F_7 = {L_6 + L_8 \over 5} = {18 + 47 \over 5} = {65 \over 5} = 13$

## Friday, 15 February 2013

### FRIDAY, 15 FEBRUARY 2013

Today is 46th day of the year.

Take one of the pizzas mentioned in yesterdays post and make nine straight cuts with a pizza cutter. What is the maximum number of pieces of pizza (not necessarily the same size) that can be created with this approach? Not surprisingly it 46. An illustration of the first few members of this sequence, the Lazy Caterer's Sequence, can be found at http://oeis.org/A000124/a000124.gif. It looks like this:

Take one of the pizzas mentioned in yesterdays post and make nine straight cuts with a pizza cutter. What is the maximum number of pieces of pizza (not necessarily the same size) that can be created with this approach? Not surprisingly it 46. An illustration of the first few members of this sequence, the Lazy Caterer's Sequence, can be found at http://oeis.org/A000124/a000124.gif. It looks like this:

## Thursday, 14 February 2013

### THURSDAY, 14 FEBRUARY 2013

Today is the 45th day of the year.

If you had a choice of two pizza toppings from a selection of 10 then there would be 45 different combinations from which to choose.

Let us assume that the available toppings are Pepperoni, Cheese, Sausage, Mushrooms, Pineapple, Bacon, Ham, Shrimp, Onions and Green Peppers, then the daily toppings could have been:

If you had a choice of two pizza toppings from a selection of 10 then there would be 45 different combinations from which to choose.

Let us assume that the available toppings are Pepperoni, Cheese, Sausage, Mushrooms, Pineapple, Bacon, Ham, Shrimp, Onions and Green Peppers, then the daily toppings could have been:

Date |
Topping 1 |
Topping 2 |

Tuesday 01 January | Pepperoni | Cheese |

Wednesday 02 January | Pepperoni | Sausage |

Thursday 03 January | Pepperoni | Mushrooms |

Friday 04 January | Pepperoni | Pineapple |

Saturday 05 January | Pepperoni | Bacon |

Sunday 06 January | Pepperoni | Ham |

Monday 07 January | Pepperoni | Shrimp |

Tuesday 08 January | Pepperoni | Onions |

Wednesday 09 January | Pepperoni | Green Peppers |

Thursday 10 January | Cheese | Sausage |

Friday 11 January | Cheese | Mushrooms |

Saturday 12 January | Cheese | Pineapple |

Sunday 13 January | Cheese | Bacon |

Monday 14 January | Cheese | Ham |

Tuesday 15 January | Cheese | Shrimp |

Wednesday 16 January | Cheese | Onions |

Thursday 17 January | Cheese | Green Peppers |

Friday 18 January | Sausage | Mushrooms |

Saturday 19 January | Sausage | Pineapple |

Sunday 20 January | Sausage | Bacon |

Monday 21 January | Sausage | Ham |

Tuesday 22 January | Sausage | Shrimp |

Wednesday 23 January | Sausage | Onions |

Thursday 24 January | Sausage | Green Peppers |

Friday 25 January | Mushrooms | Pineapple |

Saturday 26 January | Mushrooms | Bacon |

Sunday 27 January | Mushrooms | Ham |

Monday 28 January | Mushrooms | Shrimp |

Tuesday 29 January | Mushrooms | Onions |

Wednesday 30 January | Mushrooms | Green Peppers |

Thursday 31 January | Pineapple | Bacon |

Friday 01 February | Pineapple | Ham |

Saturday 02 February | Pineapple | Shrimp |

Sunday 03 February | Pineapple | Onions |

Monday 04 February | Pineapple | Green Peppers |

Tuesday 05 February | Bacon | Ham |

Wednesday 06 February | Bacon | Shrimp |

Thursday 07 February | Bacon | Onions |

Friday 08 February | Bacon | Green Peppers |

Saturday 09 February | Ham | Shrimp |

Sunday 10 February | Ham | Onions |

Monday 11 February | Ham | Green Peppers |

Tuesday 12 February | Shrimp | Onions |

Wednesday 13 February | Shrimp | Green Peppers |

Thursday 14 February | Onions | Green Peppers |

## Wednesday, 13 February 2013

### WEDNESDAY, 13 FEBRUARY 2013

Today is $44^{th}$ day of the year.

$$44^{16} + 1 = 197,352,587,024,076,973,231,046,657$$

$197,352,587,024,076,973,231,046,657$ is a prime number.

$$44^{16} + 1 = 197,352,587,024,076,973,231,046,657$$

$197,352,587,024,076,973,231,046,657$ is a prime number.

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