A006091 -id:A006091 - 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

A002022 In the pile of coconuts problem, the number of coconuts that remain to be shared equally at the end of the process.

Original entry on oeis.org

0, 6, 240, 1020, 78120, 279930, 40353600, 134217720, 31381059600, 99999999990, 34522712143920, 106993205379060, 51185893014090744, 155568095557812210, 98526125335693359360, 295147905179352825840, 239072435685151324847136

Offset: 2

N. J. A. Sloane

See A002021 for further description of the problem.

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 = 2..100
Anonymous, The Monkey and the Coconuts (with FormulaOne program)
Santo D'Agostino, "The Coconut Problem"; Updated With Solution, May 2011.
R. S. Underwood and Robert E. Moritz, Problem 3242, Amer. Math. Monthly, 35 (1928), 47-48.

Cf. A002021, A006091.

f := proc(n) if n mod 2 = 1 then RETURN((n-1)^n-(n-1)) else RETURN((n-1)^(n+1)-(n-1)) fi; end:

Rest[Table[If[OddQ[n],(n-1)^n-(n-1),(n-1)^(n+1)-(n-1)],{n,30}]] (* Harvey P. Dale, Oct 21 2011 *)

Formula and more terms from James Sellers, Feb 10 2000

Detail added to the name by Peter Munn, Jun 16 2023

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

A193871 Square array T(n,k) = k^n - k + 1 read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 7, 7, 1, 1, 15, 25, 13, 1, 1, 31, 79, 61, 21, 1, 1, 63, 241, 253, 121, 31, 1, 1, 127, 727, 1021, 621, 211, 43, 1, 1, 255, 2185, 4093, 3121, 1291, 337, 57, 1, 1, 511, 6559, 16381, 15621, 7771, 2395, 505, 73, 1, 1, 1023, 19681, 65533, 78121, 46651, 16801, 4089, 721, 91, 1

Offset: 1

Omar E. Pol, Aug 21 2011

The columns give 1^n-0, 2^n-1, 3^n-2, 4^n-3, 5^n-4, etc.

The main diagonal gives A006091, which is a sequence related to the famous "coconuts" problem.

			Array begins:
  1,   1,    1,     1,     1,    1,    1,   1,   1,   1
  1,   3,    7,    13,    21,   31,   43,  57,  73
  1,   7,   25,    61,   121,  211,  337, 505
  1,  15,   79,   253,   621, 1291, 2395
  1,  31,  241,  1021,  3121, 7771
  1,  63,  727,  4093, 15621
  1, 127, 2185, 16381
  1, 255, 6559
  1, 511
  1

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

Row 1: A000012. Rows 2,3: A002061, A061600 but both without repetitions.
Columns 1..10: A000012, positives A000225, A058481, A141725, A164785, A164784, A164783, A164786, A177095, A170955.
Cf. A276135.

Table[k^# - k + 1 &[n - k + 1], {n, 11}, {k, n}] // Flatten (* Michael De Vlieger, Nov 16 2016 *)

cp's OEIS Frontend

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

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Maple

Mathematica

Python

Formula

Extensions

A002022 In the pile of coconuts problem, the number of coconuts that remain to be shared equally at the end of the process.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Maple

Mathematica

Extensions

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.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Mathematica

Formula

A193871 Square array T(n,k) = k^n - k + 1 read by antidiagonals.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica