Coin partitions
2015-09-08
Problem 078: Coin partitions
Let p(n)
represent the number of different ways in which n coins can be separated into piles.
For example, five coins can be separated into piles in exactly seven different ways, so p(5)=7
.
OOOOO
OOOO O
OOO OO
OOO O O
OO OO O
OO O O O
O O O O O
Find the least value of n for which p(n) is divisible by one million.
Solution:
vXX OOO
# ... #
. . . .
. . .
. . . .
# ... #
>$.@
v+1p+1/g01\+1%g01:g04_^#:%g02g05$$ <
>"}}@"**20p"~|"+10p111p1>:40p050p0>:60p::2/1+:*3*\:2/1+\2%2*1-*+2/:40g`#^_4 v
^+1p05-\g05*-1*2%2/2g06g+1/g01\+1%g01:-\g0<
# ... #
. . . .
. . .
. . . .
# ... #
>$.@
v+1p+1/g01\+1%g01:g04_^#:%g02g05$$ <
>"}}@"**20p"~|"+10p111p1>:40p050p0>:60p::2/1+:*3*\:2/1+\2%2*1-*+2/:40g`#^_4 v
^+1p05-\g05*-1*2%2/2g06g+1/g01\+1%g01:-\g0<
Explanation:
Again the algorithm is from MathBlog. I can't really say that I understand the algorithm fully (and the MathBlog guy says he neither).
But for the best explanation you better read his article.
Interpreter steps: | 1 191 633 332 |
Execution time (BefunExec): | 2min 50s (6.97 MHz) |
Program size: | 251 x 256 |
Solution: | 55374 |
Solved at: | 2015-09-08 |