A284117 Sum of proper prime power divisors of n.
0, 0, 0, 4, 0, 0, 0, 12, 9, 0, 0, 4, 0, 0, 0, 28, 0, 9, 0, 4, 0, 0, 0, 12, 25, 0, 36, 4, 0, 0, 0, 60, 0, 0, 0, 13, 0, 0, 0, 12, 0, 0, 0, 4, 9, 0, 0, 28, 49, 25, 0, 4, 0, 36, 0, 12, 0, 0, 0, 4, 0, 0, 9, 124, 0, 0, 0, 4, 0, 0, 0, 21, 0, 0, 25, 4, 0, 0, 0, 28, 117, 0, 0, 4, 0, 0, 0, 12, 0, 9, 0, 4, 0, 0, 0, 60, 0, 49, 9, 29
Offset: 1
Examples
a(8) = 12 because 12 has 6 divisors {1, 2, 3, 4, 6, 12} among which 2 are proper prime powers {4, 8} therefore 4 + 8 = 12.
Links
Programs
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Maple
f:= n -> add(t[1]*(t[1]^t[2]-t[1])/(t[1]-1), t=ifactors(n)[2]): map(f, [$1..100]); # Robert Israel, Mar 31 2017
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Mathematica
nmax = 100; Rest[CoefficientList[Series[Sum[Boole[PrimePowerQ[k] && PrimeOmega[k] > 1] k x^k/(1 - x^k), {k, 1, nmax}], {x, 0, nmax}], x]] Table[Total[Select[Divisors[n], PrimePowerQ[#1] && PrimeOmega[#1] > 1 &]], {n, 100}] f[p_, e_] := (p^(e + 1) - 1)/(p - 1) - p - 1; a[1] = 0; a[n_] := Plus @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Jul 24 2024 *) -
PARI
concat([0, 0, 0], Vec(sum(k=1, 100, (isprimepower(k) && bigomega(k)>1) * k * x^k/(1 - x^k)) + O(x^101))) \\ Indranil Ghosh, Mar 21 2017
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PARI
a(n) = sumdiv(n, d, d*(isprimepower(d) && !isprime(d))); \\ Michel Marcus, Apr 01 2017
Formula
G.f.: Sum_{p prime, k>=2} p^k*x^(p^k)/(1 - x^(p^k)).
a(n) = Sum_{d|n, d = p^k, p prime, k >= 2} d.
a(n) = 0 if n is a squarefree (A005117).
Additive with a(p^e) = (p^(e+1)-1)/(p-1) - p - 1. - Amiram Eldar, Jul 24 2024