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.

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

Author

Luciano Ancora, Mar 14 2015

Keywords

• nonn
• easy

Comments

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

References

• 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.

Links

• 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).

Crossrefs

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

Programs

Mathematica

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]

Formula

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