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a(n) = (6^(valuation(n, 6) + 1) - 1) / 5.
3

%I #9 Dec 12 2022 01:41:29

%S 1,1,1,1,1,7,1,1,1,1,1,7,1,1,1,1,1,7,1,1,1,1,1,7,1,1,1,1,1,7,1,1,1,1,

%T 1,43,1,1,1,1,1,7,1,1,1,1,1,7,1,1,1,1,1,7,1,1,1,1,1,7,1,1,1,1,1,7,1,1,

%U 1,1,1,43,1,1,1,1,1,7,1,1,1,1,1,7,1,1,1,1,1,7,1,1,1,1,1,7,1,1,1,1

%N a(n) = (6^(valuation(n, 6) + 1) - 1) / 5.

%C Sum of powers of 6 dividing n.

%H Amiram Eldar, <a href="/A339748/b339748.txt">Table of n, a(n) for n = 1..10000</a>

%F G.f.: Sum_{k>=0} 6^k * x^(6^k) / (1 - x^(6^k)).

%F L.g.f.: -log(Product_{k>=0} (1 - x^(6^k))).

%F Dirichlet g.f.: zeta(s) / (1 - 6^(1 - s)).

%t Table[(6^(IntegerExponent[n, 6] + 1) - 1)/5, {n, 1, 100}]

%t nmax = 100; CoefficientList[Series[Sum[6^k x^(6^k)/(1 - x^(6^k)), {k, 0, Floor[Log[6, nmax]] + 1}], {x, 0, nmax}], x] // Rest

%Y Cf. A038712, A080278, A088842, A122841, A234959, A323921, A339747.

%K nonn

%O 1,6

%A _Ilya Gutkovskiy_, Dec 15 2020