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A348841
Number of primes with even exponents >= 2 in the prime power factorization of n!, for n >= 1.
3
0, 0, 0, 0, 0, 2, 2, 1, 1, 3, 3, 2, 2, 2, 2, 2, 2, 3, 3, 4, 2, 2, 2, 4, 4, 4, 3, 4, 4, 5, 5, 4, 2, 4, 4, 4, 4, 4, 4, 4, 4, 3, 3, 4, 5, 7, 7, 8, 8, 7, 5, 6, 6, 8, 6, 4, 2, 4, 4, 6, 6, 6, 7, 7, 5, 6, 6, 7, 7, 6, 6, 7, 7, 7, 6, 7, 7, 10, 10, 9
OFFSET
1,6
COMMENTS
The restriction to positive exponents in the prime factor factorization is used to avoid the ambiguity due to p^0 = 1 for any prime. Then a(n) = A000720(n) - A055460(n), for n >= 1.
LINKS
FORMULA
a(n) = A000720(n) - A055460(n), for n >= 1.
a(n) = A162641(A000142(n)). - Michel Marcus, Nov 03 2021
EXAMPLE
n = 12: 12! = 479001600 = 2^10 * 5^2 * 3^5 * 7^1 * 11^1, hence a(12) = 2, A055460(12) = 3 and A000720(12) = 5. This latter equation holds because 2, 3, 5, 7, 11 are the primes not exceeding 12.
MATHEMATICA
Table[Length@Select[FactorInteger[n!], EvenQ@Last@#&], {n, 80}] (* Giorgos Kalogeropoulos, Nov 02 2021 *)
PROG
(PARI) a(n) = my(f=factor(n!)); #select(x->(! (x%2)), f[, 2]); \\ Michel Marcus, Nov 03 2021
(PARI) a(n) = my(res = 0); forprime(p = 2, n\2, res+=(val(n, p)%2==0)); res
val(n, p) = my(r=0); while(n, r+=n\=p); r \\ David A. Corneth, Nov 03 2021
CROSSREFS
KEYWORD
nonn
AUTHOR
Wolfdieter Lang, Nov 02 2021
STATUS
approved