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a(n) = (4^(valuation(n, 4) + 1) - 1) / 3.
3

%I #27 Nov 27 2022 02:09:33

%S 1,1,1,5,1,1,1,5,1,1,1,5,1,1,1,21,1,1,1,5,1,1,1,5,1,1,1,5,1,1,1,21,1,

%T 1,1,5,1,1,1,5,1,1,1,5,1,1,1,21,1,1,1,5,1,1,1,5,1,1,1,5,1,1,1,85,1,1,

%U 1,5,1,1,1,5,1,1,1,5,1,1,1,21,1,1,1,5,1,1,1,5,1,1,1,5,1,1,1,21,1,1,1,5

%N a(n) = (4^(valuation(n, 4) + 1) - 1) / 3.

%C Sum of powers of 4 dividing n.

%H Amiram Eldar, <a href="/A323921/b323921.txt">Table of n, a(n) for n = 1..10000</a>

%F G.f.: Sum_{k>=0} 4^k * x^(4^k) / (1 - x^(4^k)).

%F L.g.f.: -log(Product_{k>=0} (1 - x^(4^k))).

%F Dirichlet g.f.: zeta(s) / (1 - 4^(1 - s)).

%F From _Amiram Eldar_, Nov 27 2022: (Start)

%F Multiplicative with a(2^e) = (4^floor((e+2)/2)-1)/3, and a(p^e) = 1 for p != 2.

%F Sum_{k=1..n} a(k) ~ n*log_4(n) + (1/2 + (gamma - 1)/log(4))*n, where gamma is Euler's constant (A001620). (End)

%t Table[(4^(IntegerExponent[n, 4] + 1) - 1)/3, {n, 1, 100}]

%t nmax = 100; CoefficientList[Series[Sum[4^k x^(4^k)/(1 - x^(4^k)), {k, 0, Floor[Log[4, nmax]] + 1}], {x, 0, nmax}], x] // Rest

%o (Python)

%o def A323921(n): return ((1<<((~n&n-1).bit_length()&-2)+2)-1)//3 # _Chai Wah Wu_, Jul 09 2022

%o (PARI) a(n) = (4^(valuation(n, 4) + 1) - 1) / 3; \\ _Michel Marcus_, Jul 09 2022

%Y Cf. A001620, A038712, A080278, A088842, A234957, A235127, A339747, A339748.

%K nonn,mult

%O 1,4

%A _Ilya Gutkovskiy_, Dec 15 2020