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A304037
If n = Product (p_j^k_j) then a(n) = Sum (pi(p_j)^k_j), where pi() = A000720.
3
0, 1, 2, 1, 3, 3, 4, 1, 4, 4, 5, 3, 6, 5, 5, 1, 7, 5, 8, 4, 6, 6, 9, 3, 9, 7, 8, 5, 10, 6, 11, 1, 7, 8, 7, 5, 12, 9, 8, 4, 13, 7, 14, 6, 7, 10, 15, 3, 16, 10, 9, 7, 16, 9, 8, 5, 10, 11, 17, 6, 18, 12, 8, 1, 9, 8, 19, 8, 11, 8, 20, 5, 21, 13, 11, 9, 9, 9, 22, 4, 16, 14, 23, 7, 10, 15, 12, 6
OFFSET
1,3
FORMULA
If gcd(u,v) = 1 then a(u*v) = a(u) + a(v).
a(p^k) = A000720(p)^k where p is a prime.
a(A002110(m)^k) = 1^k + 2^k + ... + m^k.
As an example:
a(A000040(k)) = k.
a(A006450(k)) = A000040(k).
a(A038580(k)) = A006450(k).
a(A001248(k)) = a(A011757(k)) = A000290(k).
a(A030078(k)) = a(A055875(k)) = A000578(k).
a(A002110(k)) = a(A011756(k)) = A000217(k).
a(A061742(k)) = A000330(k).
a(A115964(k)) = A000537(k).
a(A080696(k)) = A007504(k).
a(A076954(k)) = A001923(k).
EXAMPLE
a(72) = 5 because 72 = 2^3*3^2 = prime(1)^3*prime(2)^2 and 1^3 + 2^2 = 5.
MATHEMATICA
a[n_] := Plus @@ (PrimePi[#[[1]]]^#[[2]]& /@ FactorInteger[n]); a[1] = 0; Table[a[n], {n, 1, 88}]
KEYWORD
nonn
AUTHOR
Ilya Gutkovskiy, May 05 2018
STATUS
approved