%I #20 Aug 23 2021 06:01:24
%S 0,0,1,1,0,2,1,2,0,1,0,3,2,2,1,1,1,3,0,1,2,1,0,4,2,1,0,2,1,4,2,2,0,2,
%T 0,2,2,3,1,3,1,2,1,3,1,1,0,5,2,0,0,2,2,2,1,3,2,0,1,5,3,3,1,1,2,2,0,2,
%U 1,3,0,4,2,3,0,2,1,2,1,4,2,0,0,6,1,1,0
%N Let m! have prime factorization Product (p_j^e_j); a(n) = number of distinct prime factors p_j such that e_j = n has no solution for any m!.
%C If n belongs to A048247 then a(n) is equal to zero.
%C For a given prime p and n satisfying p^k + p^(k-1) + ... + 1 <= n < p^(k+1) + ... + 1 for some k, let r_k = n mod (p^k + p^(k-1) + ... + 1), r_(k-1) = r_k mod (p^(k-1) + ... + 1), and so on down to r_1 = r_2 mod (p + 1). Then, p^n appears in a factorial m! iff none of the r_i is congruent to -1. - _Charlie Neder_, Nov 03 2018
%H Jinyuan Wang, <a href="/A226460/b226460.txt">Table of n, a(n) for n = 0..1000</a>
%e For n = 11, there are three distinct prime factors (3, 5, 11) in factorization of m!.
%e 3^10 divides 26! ( 26! is not divisible by 3^11).
%e 3^13 divides 27!.
%e 5^10 divides 49! ( 49! is not divisible by 5^11).
%e 5^12 divides 50!.
%e 11^10 divides 120! ( 120! is not divisible by 11^11).
%e 11^12 divides 121!.
%e The exponent of three distinct prime factors never becomes equal to 11. (It searches for all the exponent of prime factorization of factorials [A000142].)
%e Therefore a(11)=3.
%o (PARI) is(k, p) = my(c, s); while(s<k, c++; s+=1+valuation(c, p)); s>k;
%o a(n) = sum(p=2, n, isprime(p)&&is(n, p)); \\ _Jinyuan Wang_, Aug 22 2021
%Y Cf. A000142, A048247, A115627.
%K nonn,easy
%O 0,6
%A _Naohiro Nomoto_, Jun 08 2013