A381958 Numerator of the sum of the reciprocals of the indices of distinct prime factors of n.

0, 1, 1, 1, 1, 3, 1, 1, 1, 4, 1, 3, 1, 5, 5, 1, 1, 3, 1, 4, 3, 6, 1, 3, 1, 7, 1, 5, 1, 11, 1, 1, 7, 8, 7, 3, 1, 9, 2, 4, 1, 7, 1, 6, 5, 10, 1, 3, 1, 4, 9, 7, 1, 3, 8, 5, 5, 11, 1, 11, 1, 12, 3, 1, 1, 17, 1, 8, 11, 19, 1, 3, 1, 13, 5, 9, 9, 5, 1, 4, 1, 14, 1, 7, 10, 15, 3, 6, 1, 11, 5, 10, 13, 16, 11

Offset: 1

Author

Ilya Gutkovskiy, Mar 11 2025

Keywords

• nonn
• frac

Examples

			0, 1, 1/2, 1, 1/3, 3/2, 1/4, 1, 1/2, 4/3, 1/5, 3/2, 1/6, 5/4, 5/6, 1, 1/7, 3/2, 1/8, 4/3, ...
		

Links

• Index entries for sequences computed from indices in prime factorization

Crossrefs

Cf. A000720, A028235, A028236, A066328, A083345, A318573, A379141, A381959 (denominators).

Programs

Mathematica

Join[{0}, Table[Plus @@ (1/PrimePi[#[[1]]] & /@ FactorInteger[n]), {n, 2, 95}] // Numerator]
nmax = 95; CoefficientList[Series[Sum[x^Prime[k]/(k (1 - x^Prime[k])), {k, 1, nmax}], {x, 0, nmax}], x] // Numerator // Rest

PARI

a(n) = my(f=factor(n)); numerator(sum(k=1, #f~, 1/primepi(f[k,1]))); \\ Michel Marcus, Mar 11 2025

Formula

If n = Product (p_j^k_j) then a(n) = numerator of Sum (1/pi(p_j)).

G.f. for fractions: Sum_{k>=1} x^prime(k) / (k*(1 - x^prime(k))).