A002022 -id:A002022 - cp's OEIS frontend

A002021 Pile of coconuts problem: (n-1)*(n^n - 1), n even; n^n - n + 1, n odd.

1, 3, 25, 765, 3121, 233275, 823537, 117440505, 387420481, 89999999991, 285311670601, 98077104930805, 302875106592241, 144456088732254195, 437893890380859361, 276701161105643274225, 827240261886336764161, 668888937280041138782191, 1978419655660313589123961

Offset: 1

N. J. A. Sloane

This is a generalization (from n = 5) of Ben Ames Williams's published problem. For a given n, the problem is effectively as follows. A successful monkey-share process removes 1 coconut for a monkey followed by an exact share of 1/n from the coconut pile. Determine the least initial number of coconuts for a monkey-share to succeed n times, leaving a multiple of n to be shared equally at the end. The problem in the D'Agostino link is slightly different, requiring a coconut for the monkey in the final division. - Peter Munn, Jun 14 2023

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 1..100
Anonymous, The Monkey and the Coconuts (with FormulaOne program)
Santo D'Agostino, "The Coconut Problem"; Updated With Solution, May 2011.
Mark Richardson, A Needlessly Complicated and Unhelpful Solution to Ben Ames Williams' Famous Coconuts Problem, The Winnower, Authorea (2016) Vol. 3.
R. S. Underwood and Robert E. Moritz, Problem 3242, Amer. Math. Monthly, 35 (1928), 47-48.
Robert G. Wilson v, Letter to N. J. A. Sloane, Oct. 1993

Cf. A002022, A006091.

seq(`if`(n::even, (n-1)*(n^n - 1),n^n-n+1),n=1..30); # Robert Israel, Aug 26 2016

Table[If[EvenQ[n],(n-1)(n^n-1),n^n-n+1],{n,30}] (* Harvey P. Dale, Apr 21 2012 *)

def a(n): return (n-1)*(n**n - 1) if n%2 == 0 else n**n - n + 1
print([a(n) for n in range(1, 20)]) # Michael S. Branicky, Feb 07 2022

E.g.f.: (1-x)*exp(x)-(W(x)+2)*(2*W(x)+1)/(2*(1+W(x))^3)-W(-x)/(2*(1+W(-x))^3) where W is the Lambert W function. - Robert Israel, Aug 26 2016

a(n) = 1-n-(-n)^n+(1+(-1)^n)*n^(n+1)/2. - Wesley Ivan Hurt, Nov 09 2023

More terms from Harvey P. Dale, Apr 21 2012

A006091 a(n) = n^n - n + 1.

Original entry on oeis.org

1, 3, 25, 253, 3121, 46651, 823537, 16777209, 387420481, 9999999991, 285311670601, 8916100448245, 302875106592241, 11112006825558003, 437893890380859361, 18446744073709551601, 827240261886336764161, 39346408075296537575407, 1978419655660313589123961

Offset: 1

N. J. A. Sloane

Related to famous "coconuts" problem - cf. A002021, A002022.

Archimedeans Problems Drive, Eureka, 41 (1981), 7.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Archimedeans Problems Drive, Problems for 1980, Eureka, 41 (1981), 6-7. (Annotated scanned copy)
J. Burkardt, The Coconut puzzle (version 3)
Santo D'Agostino, "The Coconut Problem"; Updated With Solution, May 2011.
Mark Richardson, A Needlessly Complicated and Unhelpful Solution to Ben Ames Williams' Famous Coconuts Problem, The Winnower, Authorea (2016) Vol. 3.

Cf. A014293.

[n^n - n + 1: n in [1..20]]; // Vincenzo Librandi, Aug 23 2011

Table[n^n-n+1,{n,20}] (* Harvey P. Dale, Jun 09 2011 *)

A006091(n)=n^n-n+1 \\ M. F. Hasler, Oct 30 2017

E.g.f.: 1/(1 + LambertW(-x)) + exp(x)*(1 - x) - 2. - Ilya Gutkovskiy, Oct 30 2017

A254029 Positive solutions of Monkey and Coconuts Problem for the classic case (5 sailors, 1 coconut to the monkey): a(n) = 15625*n - 4 for n >= 1.

Original entry on oeis.org

15621, 31246, 46871, 62496, 78121, 93746, 109371, 124996, 140621, 156246, 171871, 187496, 203121, 218746, 234371, 249996, 265621, 281246, 296871, 312496, 328121, 343746, 359371, 374996, 390621, 406246, 421871, 437496, 453121, 468746

Offset: 1

Luciano Ancora, Mar 14 2015

The sequence lists the numbers of coconuts originally collected on a pile. This is the case s=5, c=1 in the general formula b(n) = n*s^(s+1) - c*(s-1).

{a(n) = 5^6*n - 4}{n>=1} gives the positive solutions to the following problem: co(k) = (4/5)*(co(k-1) - 1), for k >= 0, with co(0) = a, and the requirement c0(5) - 1 == 0 (mod 5). This gives co(5) - 1 = (2^10*a - 7*3^3*61)/5^5, with a = a(n), n >= 1. See a formula below. - Richard S. Fischer and _Wolfdieter Lang, Jun 01 2023

Charles S. Ogilvy and John T. Anderson, Excursions in Number Theory, Oxford University Press, 1966, pages 52-54.
Miodrag S. Petković, "The sailors, the coconuts, and the monkey", Famous Puzzles of Great Mathematicians, Amer. Math. Soc.(AMS), 2009, pages 52-56.

Luciano Ancora, Table of n, a(n) for n = 1..1000
Umberto Cerruti, Marinai e noci di cocco, Divertiamoci con la Matematica (in Italian)
Santo D'Agostino, "The Coconut Problem"; Updated With Solution, May 2011.
Eric Weisstein's World of Mathematics, Monkey and Coconut Problem
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A002021, A002022, A006091, A014293, A085283, A085606, A158421.

s = 5; c = 1; Table[n s^(s + 1) - c (s - 1), {n, 1, 30}] (* or *)
CoefficientList[Series[(15621 + 4 x)/(-1 + x)^2, {x, 0, 29}], x]

G.f.: x*(15621 + 4*x)/(1 - x)^2.
a(n) = 2*a(n-1) - a(n-2) = a(n-1) + 15625, with a(0) = -4 and a(-1) = -(4 + 5^6). a(n) = 5^6*n - 4.
a(n) = (15*c(n) + 11) + 265*(c(n) + 1)/2^10, with c(n) = A158421(n) = 2^10*n - 1, for n >= 1. - Richard S. Fischer and Wolfdieter Lang, Jun 01 2023

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A002021 Pile of coconuts problem: (n-1)*(n^n - 1), n even; n^n - n + 1, n odd.

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A006091 a(n) = n^n - n + 1.

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A254029 Positive solutions of Monkey and Coconuts Problem for the classic case (5 sailors, 1 coconut to the monkey): a(n) = 15625*n - 4 for n >= 1.

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