OFFSET
1,4
COMMENTS
In Ben Ames Williams's coconuts problem, a pile of coconuts remains the next day that is divisible by n sailors. Integers in the sequence multiplied by (n^2)-n determine the size of the divisible pile.
LINKS
Mark Richardson, Table of n, a(n) for n = 1..101
M. B. Richardson, A Needlessly Complicated and Unhelpful Solution to Ben Ames Williams' Coconuts Problem, The Winnower, 3 (2016), e147175.52128. doi: 10.15200/winn.147175.52128
FORMULA
a(n) = ((n-1)^(n-1) - 1)/n, if n is odd.
a(n) = ((n-1)^n - 1)/n, if n is even.
a(n) = ((n-1)^(n - 1/2 + (-1)^n/2) - 1)/n = (n^A052928(n) - 1)/n. - Omar E. Pol, Aug 24 2016
MAPLE
seq(((n-1)^(n-1/2+(-1)^n/2)-1)/n, n=1..30); # Robert Israel, Aug 26 2016
MATHEMATICA
Join[{0}, Table[((n - 1)^(n - 1/2 + (-1)^n/2) - 1)/n, {n, 2, 30}]] (* Bruno Berselli, Aug 26 2016 *)
PROG
(PARI) a(n) = ((n-1)^(n-1/2+(-1)^n/2)-1)/n \\ Felix Fröhlich, Aug 26 2016
(Magma) [((n-1)^(n-1 div 2 +(-1)^n div 2)-1) div n: n in [1..25]]; // Vincenzo Librandi, May 15 2019
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Mark Richardson, Aug 21 2016
STATUS
approved