%I #32 Jun 25 2024 11:42:25

%S 1,3,1,6,1,3,1,10,1,3,1,6,1,3,1,15,1,3,1,6,1,3,1,10,1,3,1,6,1,3,1,21,

%T 1,3,1,6,1,3,1,10,1,3,1,6,1,3,1,15,1,3,1,6,1,3,1,10,1,3,1,6,1,3,1,28,

%U 1,3,1,6,1,3,1,10,1,3,1,6,1,3,1,15,1,3,1,6,1,3,1,10,1,3,1,6,1

%N a(n) = A000217(A001511(n)), where A001511 is one more than the 2-adic valuation of n, and A000217(n) is the n-th triangular number, binomial(n+1, 2).

%C Row sums of A115363. In general, the row sums of ((1,x) - m(x,x^2))^(-2) are obtained by following the ruler function A001511(n) by the solution of the recurrence a(1)=1, a(n) = n*m^(n-1) + a(n-1), n > 1.

%C The Stephan formula says this is the Dirichlet convolution of A000012 with A104117. - _R. J. Mathar_, Feb 07 2011

%H Antti Karttunen, <a href="/A115364/b115364.txt">Table of n, a(n) for n = 1..16383</a>

%F a(n) = binomial(A007814(n)+2, 2) = binomial(A001511(n)+1, 2).

%F Dirichlet g.f.: zeta(s)*(2^s/(2^s-1))^2. - _Ralf Stephan_, Jun 17 2007

%F Multiplicative with a(2^k) = A000217(k+1), a(p^k) = 1 for odd primes p. - _Antti Karttunen_, Nov 02 2018

%F O.g.f.: Sum_{k >= 1} k*x^(2^(k-1))/(1 - x^(2^(k-1))). More generally, if f(n) is an arithmetic function and g(n) := Sum_{k = 1..n} f(k), then Sum_{k >= 1} f(k)*x^(2^(k-1))/(1 - x^(2^(k-1))) = Sum_{n >= 1} g(A001511(n))*x^n. This is the case f(n) = n. - _Peter Bala_, Mar 26 2019

%F Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 4. - _Amiram Eldar_, Oct 22 2022

%F More precise asymptotics: Sum_{k=1..n} a(k) ~ 4*n - log(n)*(log(n) + 2*log(4*Pi))/(4*log(2)^2). - _Vaclav Kotesovec_, Jun 25 2024

%t Array[PolygonalNumber[IntegerExponent[#, 2] + 1] &, 93] (* _Michael De Vlieger_, Nov 02 2018 *)

%o (PARI) A115364(n) = binomial(valuation(n,2)+2,2); \\ _Antti Karttunen_, Nov 02 2018

%o (Python)

%o def A115364(n): return (m:=((~n & n-1).bit_length()+1))*(m+1)>>1 # _Chai Wah Wu_, Jul 02 2022

%Y Cf. A000217, A001511, A007814, A104117, A115363.

%Y Cf. also A094290.

%K easy,mult,nonn

%O 1,2

%A _Paul Barry_, Jan 21 2006

%E Formula corrected and the name changed by _Antti Karttunen_, Nov 02 2018