A129628 Inverse Moebius transform of A001511.
1, 3, 2, 6, 2, 6, 2, 10, 3, 6, 2, 12, 2, 6, 4, 15, 2, 9, 2, 12, 4, 6, 2, 20, 3, 6, 4, 12, 2, 12, 2, 21, 4, 6, 4, 18, 2, 6, 4, 20, 2, 12, 2, 12, 6, 6, 2, 30, 3, 9, 4, 12, 2, 12, 4, 20, 4, 6, 2, 24, 2, 6, 6, 28, 4, 12, 2, 12, 4, 12, 2, 30, 2, 6, 6, 12, 4, 12, 2, 30, 5, 6, 2, 24, 4, 6, 4, 20
Offset: 1
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Andrew Howroyd)
Crossrefs
Programs
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Maple
seq(add(padic[ordp](2*d, 2), d in numtheory[divisors](n)), n=1..100); # Ridouane Oudra, Sep 30 2024
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Mathematica
f[p_, e_] := If[p==2, (e+1)*(e+2)/2, e+1]; a[1] = 1; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Sep 30 2020 *)
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PARI
a(n)={sumdiv(n, d, 1 + valuation(d, 2))} \\ Andrew Howroyd, Aug 04 2018
Formula
Dirichlet g.f.: zeta(s)^2 * 2^s/(2^s-1). - Ralf Stephan, Jun 17 2007, corrected by Vaclav Kotesovec, Feb 02 2019
a(n) = Sum_{d|n} A001511(d). - Andrew Howroyd, Aug 04 2018
Sum_{k=1..n} a(k) ~ 2*n * (2*gamma - 1 + log(n/2)), where gamma is the Euler-Mascheroni constant A001620. - Vaclav Kotesovec, Feb 02 2019
Multiplicative with a(2^e) = (e+1)*(e+2)/2, and a(p^e) = e+1 for p > 2. - Amiram Eldar, Sep 30 2020
From Ridouane Oudra, Sep 30 2024: (Start)
a(n) = Sum_{i=0..A007814(n)} tau(n/2^i).
a(n) = Sum_{d|2*n} A007814(d).
Comments