A336649 Sum of divisors of A336651(n) (odd part of n divided by its largest squarefree divisor).
1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 6, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 8, 6, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 6, 1, 1, 1, 1, 1, 40, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 8, 4, 6, 1, 1, 1, 1, 1
Offset: 1
Links
- Antti Karttunen, Table of n, a(n) for n = 1..65537
Programs
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Mathematica
f[2, e_] := 1; f[p_, e_] := (p^e - 1)/(p-1); a[1] = 1; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Dec 07 2020 *)
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PARI
A336649(n) = { my(f=factor(n)); prod(i=1, #f~, if((2==f[i,1])||(1==f[i,2]),1,(((f[i,1]^(f[i,2]))-1) / (f[i,1]-1)))); };
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PARI
A000265(n) = (n>>valuation(n,2)); A335341(n) = if(1==n,n,sigma(n/factorback(factorint(n)[, 1]))); A336649(n) = A335341(A000265(n));
Formula
Multiplicative with a(2^e) = 1, a(p^1) = 1 and a(p^e) = (p^e - 1)/(p-1) if e > 1.
Dirichlet g.f.: zeta(s-1) * zeta(s) * (1 - 1/(1-2^s+2^(2*s-1))) * Product_{p prime} (1 - 1/p^(s-1) + 1/p^(2*s-1)). - Amiram Eldar, Dec 18 2023