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A276135
Ben Ames Williams's Monkey and Coconuts Problem.
2
0, 0, 1, 20, 51, 2604, 6665, 720600, 1864135, 348678440, 909090909, 261535698060, 685853880635, 281241170407092, 740800455037201, 410525522232055664, 1085102592571150095, 781282469559318055056, 2070863582910344082917, 1879498672877297909667780, 4993219047619047619047619, 5577014881186619679500164220
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
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
Sequence in context: A049390 A220040 A128905 * A350115 A211143 A183047
KEYWORD
nonn,easy
AUTHOR
Mark Richardson, Aug 21 2016
STATUS
approved