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%I #14 Nov 27 2022 02:10:01
%S 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,
%T 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,
%U 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
%N a(n) = (5^(valuation(n, 5) + 1) - 1) / 4.
%C Sum of powers of 5 dividing n.
%C Denominator of the quotient sigma(5*n) / sigma(n).
%H Amiram Eldar, <a href="/A339747/b339747.txt">Table of n, a(n) for n = 1..10000</a>
%F G.f.: Sum_{k>=0} 5^k * x^(5^k) / (1 - x^(5^k)).
%F L.g.f.: -log(Product_{k>=0} (1 - x^(5^k))).
%F Dirichlet g.f.: zeta(s) / (1 - 5^(1 - s)).
%F a(n) = sigma(n)/(sigma(5*n) - 5*sigma(n)), where sigma(n) = A000203(n). - _Peter Bala_, Jun 10 2022
%F From _Amiram Eldar_, Nov 27 2022: (Start)
%F Multiplicative with a(5^e) = (5^(e+1)-1)/4, and a(p^e) = 1 for p != 5.
%F 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)
%t Table[(5^(IntegerExponent[n, 5] + 1) - 1)/4, {n, 1, 100}]
%t 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
%o (PARI) a(n) = (5^(valuation(n, 5) + 1) - 1)/4; \\ _Amiram Eldar_, Nov 27 2022
%Y Cf. A000203, A001620, A038712, A060904, A080278, A088842, A112765, A323921, A339748.
%K nonn,mult
%O 1,5
%A _Ilya Gutkovskiy_, Dec 15 2020
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