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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. 4
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 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). 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 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] PROG (Python 3.x) seq=[] for x in range (1, 1000000): total_c, i = x, 1 while i < 6: if (total_c)%5 == 1: total_c = total_c - (total_c)//5 -1 if i == 5: #print (x, total_c) break i += 1 if total_c%5 == 1: seq.append(x) # Glen Gilchrist, Jan 28 2023 CROSSREFS Cf. A002021, A002022, A006091, A014293, A085283, A085606, A158421. Sequence in context: A205654 A206183 A205365 * A129152 A223185 A097219 Adjacent sequences: A254026 A254027 A254028 * A254030 A254031 A254032 KEYWORD nonn,easy AUTHOR Luciano Ancora, Mar 14 2015 STATUS approved

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Last modified December 2 02:40 EST 2023. Contains 367505 sequences. (Running on oeis4.)