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A325704
If n = prime(i_1)^j_1 * ... * prime(i_k)^j_k, then a(n) is the numerator of the reciprocal factorial sum j_1/i_1! + ... + j_k/i_k!.
4
0, 1, 1, 2, 1, 3, 1, 3, 1, 7, 1, 5, 1, 25, 2, 4, 1, 2, 1, 13, 13, 121, 1, 7, 1, 721, 3, 49, 1, 5, 1, 5, 61, 5041, 5, 3, 1, 40321, 361, 19, 1, 37, 1, 241, 7, 362881, 1, 9, 1, 4, 2521, 1441, 1, 5, 7, 73, 20161, 3628801, 1, 8, 1, 39916801, 25, 6, 121, 181, 1
OFFSET
1,4
COMMENTS
Alternatively, if n = prime(i_1) * ... * prime(i_k), then a(n) is the numerator of 1/i_1! + ... + 1/i_k!.
FORMULA
a(n) = A318573(A325709(n)).
MATHEMATICA
Table[Total[Cases[If[n==1, {}, FactorInteger[n]], {p_, k_}:>k/PrimePi[p]!]], {n, 100}]//Numerator
PROG
(PARI) A325704(n) = { my(f=factor(n)); numerator(sum(i=1, #f~, f[i, 2]/(primepi(f[i, 1])!))); }; \\ Antti Karttunen, Nov 17 2019
KEYWORD
nonn,frac
AUTHOR
Gus Wiseman, May 18 2019
STATUS
approved