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A339747
a(n) = (5^(valuation(n, 5) + 1) - 1) / 4.
3
1, 1, 1, 1, 6, 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, 1, 1, 1, 1, 31, 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, 1, 1, 1, 1, 31, 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, 1, 1, 1, 1, 31, 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, 1, 1, 1, 1, 31
OFFSET
1,5
COMMENTS
Sum of powers of 5 dividing n.
Denominator of the quotient sigma(5*n) / sigma(n).
LINKS
FORMULA
G.f.: Sum_{k>=0} 5^k * x^(5^k) / (1 - x^(5^k)).
L.g.f.: -log(Product_{k>=0} (1 - x^(5^k))).
Dirichlet g.f.: zeta(s) / (1 - 5^(1 - s)).
a(n) = sigma(n)/(sigma(5*n) - 5*sigma(n)), where sigma(n) = A000203(n). - Peter Bala, Jun 10 2022
From Amiram Eldar, Nov 27 2022: (Start)
Multiplicative with a(5^e) = (5^(e+1)-1)/4, and a(p^e) = 1 for p != 5.
Sum_{k=1..n} a(k) ~ n*log_5(n) + (1/2 + (gamma - 1)/log(5))*n, where gamma is Euler's constant (A001620). (End)
MATHEMATICA
Table[(5^(IntegerExponent[n, 5] + 1) - 1)/4, {n, 1, 100}]
nmax = 100; CoefficientList[Series[Sum[5^k x^(5^k)/(1 - x^(5^k)), {k, 0, Floor[Log[5, nmax]] + 1}], {x, 0, nmax}], x] // Rest
PROG
(PARI) a(n) = (5^(valuation(n, 5) + 1) - 1)/4; \\ Amiram Eldar, Nov 27 2022
KEYWORD
nonn,mult
AUTHOR
Ilya Gutkovskiy, Dec 15 2020
STATUS
approved