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A254029
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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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4
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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
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OFFSET
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1,1
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COMMENTS
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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
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REFERENCES
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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.
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LINKS
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FORMULA
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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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MATHEMATICA
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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]
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PROG
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(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)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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