Let p(n) represent the number of different ways in which n coins can be separated into piles. For example, five coins can separated into piles in exactly seven different ways, so p(5)=7.

Analysis

This problem is really asking to find the first term in the sequence of integer partitions that’s divisible by 1,000,000.

A partition of an integer, n, is one way of describing how many ways the sum of positive integers, ≤ n, can be added together to equal n, regardless of order. The function p(n) is used to denote the number of partitions for n. Below we show our 5 “coins” as addends to evaluate 7 partitions, that is p(5)=7.

We have the following generating function which uses our pentagonal numbers as exponents:

Solution

Runs < 10 seconds in Perl.

($p[0], $n, $i, $m) = (1, 0, 1, 1e6);
 
for $i (1..250) {             #generate pentagonal numbers for generating function
  $p = $i*(3 * $i - 1)/2;
  push @k, $p, $p+$i;
}
 
while ($p[$n++]) {         #expand generating function to calculate p(n)
  my ($p, $i) = (0, 0);
  $p += ($i%4>1 ? -1 : 1 ) * $p[$n - $k[$i++]] while ($k[$i] <= $n); 
  $p[$n] = $p % $m; 
} 
print "Answer to PE78 = ",$n-1;

Comments

The full answer is:
363253009254357859308323315773967616467
158361736338932270710864607092686080534
895417314045435376684389911706807452721
591544937406153858232021581676352762505
545553421158554245989201590354130448112
450821973350979535709118842524107301749
07784762924663654000000

Pentagonal numbers are generated by the formula, P_n=n(3n−1)/2. The first ten pentagonal numbers are:

1, 5, 12, 22, 35, 51, 70, 92, 117, 145, …

It can be seen that P₄ + P₇ = 22 + 70 = 92 = P₈. However, their difference, 70 − 22 = 48, is not pentagonal.

Find the pair of pentagonal numbers, P_j and P_k, for which their sum and difference is pentagonal and D = |P_k − P_j| is minimised; what is the value of D?

Analysis

How many pentagonal numbers are required to search through to solve the problem? We wrote the following program and started with 10,000 pentagonal numbers and, after finding the correct answer, adjusted the requirement down to 2,400. When the problem asks for D = |P_k − P_j| to be minimized it simply means the smallest pair which, by our method, is the first occurrence found.

A set (hashable) is used to both hold the pentagonal numbers and to check if a number is pentagonal. All that’s left is to permute the pairs and check that the difference and sum are pentagonal numbers.
Runs < 2 seconds in Python.

pent = set( n * (3*n - 1) / 2 for n in range(2,2400) ) 
 
def pe44():
  for pj in pent: 
    for pk in pent: 
      if pj - pk in pent and pj + pk in pent: 
        return pj - pk
 
print "Answer to PE44 = ",pe44()

So, a more challenging program might be to solve the problem without having to guess at how many pentagonal numbers are required. That would require a function to determine if a number is pentagonal. (When the equation (√(1 + 24 · n) + 1) / 6 evaluates to an integer, n, is a pentagonal number.)

It turned out to be an interesting discovery and took longer to find an answer. But, as luck would have it, was very helpful for solving problem 45.

Solution

Runs < 6 seconds in Python.

from math import sqrt
 
def is_pentagonal(n):
  p = ( sqrt(1 + 24*n) + 1 ) / 6
  return p==int(p)
 
def pe44():
  i=0;
  pent = [1]
  while True:
    i+=1
    pent.append( i * (3*i - 1) / 2 )
    for j in range(2, len(pent)-1):
      if is_pentagonal(pent[i]-pent[j]) and is_pentagonal(pent[i]+pent[j]):
        return pent[i]-pent[j]
 
print "Answer to PE44 = ",pe44()

Comments

See Problem 45.

Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:

It can be verified that T₂₈₅ = P₁₆₅ = H₁₄₃ = 40755.

Find the next triangle number that is also pentagonal and hexagonal.

Analysis

Note that Hexagonal numbers are a subset of Triangle numbers so we only determine the first occurrence of H_i = P_j to find our answer. We can safely start our search from 166 as eluded to in the problem’s description.

From our experience with problem 44 we were able to make our search trivial without the use of arrays.

Solution

Runs < 1 second in Python.

from math import sqrt
 
def is_pentagonal(n):
  p = ( sqrt(1 + 24*n) + 1 ) / 6
  return p==int(p)
 
n = 165 
while True:
  n += 1 
  hex = n * (2 * n-1)
  if is_pentagonal(hex): break 
 
print "Answer to PE45 = ",hex

Comments

See Problem 44.
The next number in the series is 57722156241751, calculated in about 15 seconds.