A183097 a(n) = sum of powerful divisors d (including 1) of n.
1, 1, 1, 5, 1, 1, 1, 13, 10, 1, 1, 5, 1, 1, 1, 29, 1, 10, 1, 5, 1, 1, 1, 13, 26, 1, 37, 5, 1, 1, 1, 61, 1, 1, 1, 50, 1, 1, 1, 13, 1, 1, 1, 5, 10, 1, 1, 29, 50, 26, 1, 5, 1, 37, 1, 13, 1, 1, 1, 5, 1, 1, 10, 125, 1, 1, 1, 5, 1, 1, 1, 130, 1, 1, 26, 5, 1, 1, 1, 29, 118, 1, 1, 5, 1, 1, 1, 13, 1, 10, 1, 5, 1, 1, 1, 61, 1, 50, 10, 130
Offset: 1
Examples
For n = 12, set of such divisors is {1, 4}; a(12) = 1+4 = 5.
Links
Programs
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Maple
A183097 := proc(n) local a,pe,p,e ; a := 1; for pe in ifactors(n)[2] do p := op(1,pe) ; e := op(2,pe) ; if e > 1 then a := a* ( (p^(e+1)-1)/(p-1)-p) ; end if; end do: a ; end proc: # R. J. Mathar, Jun 02 2020
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Mathematica
fun[p_,e_] := (p^(e+1)-1)/(p-1) - p; a[1] = 1; a[n_] := Times @@ (fun @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, May 14 2019 *)
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PARI
A183097(n) = sumdiv(n, d, ispowerful(d)*d); \\ Antti Karttunen, Oct 07 2017
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PARI
a(n) = {my(f = factor(n)); prod(i = 1, #f~, (f[i,1]^(f[i,2]+1)-1) / (f[i,1]-1) - f[i,1]);} \\ Amiram Eldar, Dec 24 2023
Formula
a(1) = 1, a(p) = 1, a(pq) = 1, a(pq...z) = 1, a(p^k) = ((p^(k+1)-1) / (p-1))-p, for p, q = primes, k = natural numbers, pq...z = product of k (k > 2) distinct primes p, q, ..., z.
From Amiram Eldar, Dec 24 2023: (Start)
Dirichlet g.f.: zeta(s) * zeta(2*s-2) * zeta(3*s-3) / zeta(6*s-6).
Sum_{k=1..n} a(k) ~ c * n^(3/2), where c = zeta(3/2)^2/(3*zeta(3)) = 1.892451... . (End)
Comments