A284233 Sum of odd prime power divisors of n (not including 1).
0, 0, 3, 0, 5, 3, 7, 0, 12, 5, 11, 3, 13, 7, 8, 0, 17, 12, 19, 5, 10, 11, 23, 3, 30, 13, 39, 7, 29, 8, 31, 0, 14, 17, 12, 12, 37, 19, 16, 5, 41, 10, 43, 11, 17, 23, 47, 3, 56, 30, 20, 13, 53, 39, 16, 7, 22, 29, 59, 8, 61, 31, 19, 0, 18, 14, 67, 17, 26, 12, 71, 12, 73, 37, 33, 19, 18, 16, 79, 5
Offset: 1
Examples
a(15) = 8 because 15 has 4 divisors {1, 3, 5, 15} among which 2 are odd prime powers {3, 5} therefore 3 + 5 = 8.
Links
- Eric Weisstein's World of Mathematics, Prime Power.
- Index entries for sequences related to sums of divisors.
Crossrefs
Programs
-
Mathematica
nmax = 80; Rest[CoefficientList[Series[Sum[Boole[PrimePowerQ[k] && Mod[k, 2] == 1] k x^k/(1 - x^k), {k, 1, nmax}], {x, 0, nmax}], x]] Table[Total[Select[Divisors[n], PrimePowerQ[#] && Mod[#, 2] == 1 &]], {n, 80}] f[p_, e_] := (p^(e + 1) - 1)/(p - 1) - 1; f[2, e_] := 0; a[1] = 0; a[n_] := Plus @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Jul 24 2024 *)
Formula
a(n) = Sum_{d|n, d = p^k, p prime, p > 2, k > 0} d.
a(p^k) = p*(p^k - 1)/(p - 1) for p is a prime > 2.
a(2^k*p) = p for p is a prime > 2.
a(2^k) = 0.
Additive with a(2^e) = 0, and a(p^e) = (p^(e+1)-1)/(p-1) - 1 for an odd prime p. - Amiram Eldar, Jul 24 2024