A286875 If n = Product (p_j^k_j) then a(n) = Sum (k_j >= 2, p_j^k_j).
0, 0, 0, 4, 0, 0, 0, 8, 9, 0, 0, 4, 0, 0, 0, 16, 0, 9, 0, 4, 0, 0, 0, 8, 25, 0, 27, 4, 0, 0, 0, 32, 0, 0, 0, 13, 0, 0, 0, 8, 0, 0, 0, 4, 9, 0, 0, 16, 49, 25, 0, 4, 0, 27, 0, 8, 0, 0, 0, 4, 0, 0, 9, 64, 0, 0, 0, 4, 0, 0, 0, 17, 0, 0, 25, 4, 0, 0, 0, 16, 81, 0, 0, 4, 0, 0, 0, 8, 0, 9, 0, 4, 0, 0, 0, 32, 0, 49, 9, 29, 0, 0, 0, 8, 0, 0, 0, 31
Offset: 1
Examples
a(360) = a(2^3*3^2*5) = 2^3 + 3^2 = 17.
Links
Crossrefs
Programs
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Mathematica
Table[DivisorSum[n, # &, CoprimeQ[#, n/#] && PrimePowerQ[#] && !PrimeQ[#] &], {n, 108}] f[p_, e_] := If[e == 1, 0, p^e]; a[1] = 0; a[n_] := Plus @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Jul 24 2024 *)
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PARI
A286875(n) = { my(f=factor(n)); for (i=1, #f~, if(f[i, 2] < 2, f[i, 1] = 0)); vecsum(vector(#f~,i,f[i,1]^f[i,2])); }; \\ Antti Karttunen, Oct 07 2017
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Python
from sympy import primefactors, isprime, gcd, divisors def a(n): return sum(d for d in divisors(n) if gcd(d, n//d)==1 and len(primefactors(d))==1 and not isprime(d)) print([a(n) for n in range(1, 109)]) # Indranil Ghosh, Aug 02 2017
Formula
a(n) = Sum_{d|n, d = p^k, p prime, k >= 2, gcd(d, n/d) = 1} d.
a(A005117(k)) = 0.
Additive with a(p^e) = p^e if e >= 2, and 0 otherwise. - Amiram Eldar, Jul 24 2024
Comments