A226460 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!.

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, 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, 1, 3, 0, 4, 2, 3, 0, 2, 1, 2, 1, 4, 2, 0, 0, 6, 1, 1, 0

Offset: 0

Author

Naohiro Nomoto, Jun 08 2013

Keywords

• nonn
• easy

Comments

If n belongs to A048247 then a(n) is equal to zero.

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

Examples

			For n = 11, there are three distinct prime factors (3, 5, 11) in factorization of m!.
3^10 divides 26! ( 26! is not divisible by 3^11).
3^13 divides 27!.
5^10 divides 49! ( 49! is not divisible by 5^11).
5^12 divides 50!.
11^10 divides 120! ( 120! is not divisible by 11^11).
11^12 divides 121!.
The exponent of three distinct prime factors never becomes equal to 11. (It searches for all the exponent of prime factorization of factorials [A000142].)
Therefore a(11)=3.
		

Links

• Jinyuan Wang, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A000142, A048247, A115627.

Programs

PARI

is(k, p) = my(c, s); while(sk;
a(n) = sum(p=2, n, isprime(p)&&is(n, p)); \\ Jinyuan Wang, Aug 22 2021