Thursday 28 February 2013

THURSDAY, 28 FEBRUARY 2013

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.



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

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.



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:

  1. $11$
  2. $10 + 1$
  3. $9 + 2$
  4. $8 + 3$
  5. $7 + 4$
  6. $6 + 5$
  7. $9 + 1 + 1$
  8. $8 + 2 + 1$
  9. $7 + 3 + 1$
  10. $7 + 2 + 2$
  11. $6 + 4 + 1$
  12. $6 + 3 + 2$
  13. $5 + 5 + 1$
  14. $5 + 4 + 2$
  15. $5 + 3 + 3$
  16. $4 + 4 + 3$
  17. $8 + 1 + 1 + 1$
  18. $7 + 2 + 1 + 1$
  19. $6 + 3 + 1 + 1$
  20. $6 + 2 + 2 + 1$
  21. $5 + 4 + 1 + 1$
  22. $5 + 3 + 2 + 1$
  23. $5 + 2 + 2 + 2$
  24. $4 + 4 + 2 + 1$
  25. $4 + 3 + 3 + 1$
  26. $4 + 3 + 2 + 2$
  27. $3 + 3 + 3 + 2$
  28. $7 + 1 + 1 + 1 + 1$
  29. $6 + 2 + 1 + 1 + 1$
  30. $5 + 3 + 1 + 1 + 1$
  31. $5 + 2 + 2 + 1 + 1$
  32. $4 + 4 + 1 + 1 + 1$
  33. $4 + 3 + 2 + 1 + 1$
  34. $4 + 2 + 2 + 2 + 1$
  35. $3 + 3 + 3 + 1 + 1$
  36. $3 + 3 + 2 + 2 + 1$
  37. $3 + 2 + 2 + 2 + 2$
  38. $6 + 1 + 1 + 1 + 1 + 1$
  39. $5 + 2 + 1 + 1 + 1 + 1$
  40. $4 + 3 + 1 + 1 + 1 + 1$
  41. $4 + 2 + 2 + 1 + 1 + 1$
  42. $3 + 3 + 2 + 1 + 1 + 1$
  43. $3 + 2 + 2 + 2 + 1 + 1$
  44. $2 + 2 + 2 + 2 + 2 + 1$
  45. $5 + 1 + 1 + 1 + 1 + 1 + 1$
  46. $4 + 2 + 1 + 1 + 1 + 1 + 1$
  47. $3 + 3 + 1 + 1 + 1 + 1 + 1$
  48. $3 + 2 + 2 + 1 + 1 + 1 + 1$
  49. $2 + 2 + 2 + 2 + 1 + 1 + 1$
  50. $4 + 1 + 1 + 1 + 1 + 1 + 1 + 1$
  51. $3 + 2 + 1 + 1 + 1 + 1 + 1 + 1$
  52. $2 + 2 + 2 + 1 + 1 + 1 + 1 + 1$
  53. $3 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1$
  54. $2 + 2 + 1 + 1 + 1 + 1 + 1 + 1 + 1$
  55. $2 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1$
  56. $1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1$

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.

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.

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

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:
  1. {1, 2, 3, 4, 5}
  2. {{1}, {2, 3, 4, 5}}
  3. {{1}, {2, 3, 4, 5}}
  4. {{3}, {1, 2, 4, 5}}
  5. {{4}, {1, 2, 3, 5}}
  6. {{5}, {1, 2, 3, 4}}
  7. {{1, 2} {3, 4, 5}}
  8. {{2, 3} {1, 4, 5}}
  9. {{3, 4} {1, 2, 5}}
  10. {{4, 5} {1, 2, 3}}
  11. {{5, 1} {2, 3, 4}}
  12. {{1, 3} {2, 4, 5}}
  13. {{2, 4} {1, 3, 5}}
  14. {{3, 5} {1, 2, 4}}
  15. {{4, 1} {2, 3, 5}}
  16. {{5, 2} {1, 3, 4}}
  17. {{1} {2} {3, 4, 5}}
  18. {{1} {3} {2, 4, 5}}
  19. {{1} {4} {2, 3, 5}}
  20. {{1} {5} {2, 3, 4}}
  21. {{2} {3} {1, 4, 5}}
  22. {{2} {4} {1, 3, 5}}
  23. {{2} {5} {1, 2, 4}}
  24. {{3} {4} {1, 2, 5}}
  25. {{3} {5} {1, 2, 4}}
  26. {{4} {5} {1, 2, 3}}
  27. {{1} {2, 3} {4, 5}}
  28. {{1} {2, 4} {3, 5}}
  29. {{1} {2, 5} {3, 4}}
  30. {{2} {1, 3} {4, 5}}
  31. {{2} {1, 4} {3, 5}}
  32. {{2} {1, 5} {3, 4}}
  33. {{3} {1, 2} {4, 5}}
  34. {{3} {1, 4} {2, 5}}
  35. {{3} {1, 5} {2, 4}}
  36. {{4} {1, 2} {3, 5}}
  37. {{4} {1, 3} {2, 5}}
  38. {{4} {1, 5} {2, 3}}
  39. {{5} {1, 2} {3, 4}}
  40. {{5} {1, 3} {2, 4}}
  41. {{5} {1, 4} {2, 3}}
  42. {{1} {2} {3} {4, 5}}
  43. {{1} {2} {4} {3, 5}}
  44. {{1} {2} {5} {3, 4}}
  45. {{1} {3} {4} {2, 5}}
  46. {{1} {3} {5} {2, 4}}
  47. {{1} {4} {5} {2, 3}}
  48. {{2} {3} {4} {1, 5}}
  49. {{2} {3} {5} {1, 4}}
  50. {{2} {4} {5} {1, 3}}
  51. {{3} {4} {5} {1, 2}}
  52. {{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
$M_6 = 51$

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

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:


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:
$$ n!!.(n -  1)!! = n! $$

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 $F_7 = {L_6 + L_8 \over 5} = {18 + 47 \over 5} = {65 \over 5} = 13$

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:


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:

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.