A115364 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).
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, 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, 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
Offset: 1
Links
- Antti Karttunen, Table of n, a(n) for n = 1..16383
Programs
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Mathematica
Array[PolygonalNumber[IntegerExponent[#, 2] + 1] &, 93] (* Michael De Vlieger, Nov 02 2018 *)
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PARI
A115364(n) = binomial(valuation(n,2)+2,2); \\ Antti Karttunen, Nov 02 2018
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Python
def A115364(n): return (m:=((~n & n-1).bit_length()+1))*(m+1)>>1 # Chai Wah Wu, Jul 02 2022
Formula
Dirichlet g.f.: zeta(s)*(2^s/(2^s-1))^2. - Ralf Stephan, Jun 17 2007
Multiplicative with a(2^k) = A000217(k+1), a(p^k) = 1 for odd primes p. - Antti Karttunen, Nov 02 2018
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
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 4. - Amiram Eldar, Oct 22 2022
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
Extensions
Formula corrected and the name changed by Antti Karttunen, Nov 02 2018
Comments