

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
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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*(s1).
{a(n) = 5^6*n  4}_{n>=1} gives the positive solutions to the following problem: co(k) = (4/5)*(co(k1)  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 5254.
Miodrag S. Petković, "The sailors, the coconuts, and the monkey", Famous Puzzles of Great Mathematicians, Amer. Math. Soc.(AMS), 2009, pages 5256.


LINKS



FORMULA

G.f.: x*(15621 + 4*x)/(1  x)^2.
a(n) = 2*a(n1)  a(n2) = a(n1) + 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)


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



