A330706 Numbers m such that the prime factorization of m! contains no composite exponents.

1, 2, 3, 4, 5, 8, 14

Offset: 1

Author

Devansh Singh, Mar 29 2020

Keywords

• nonn
• fini
• full

Comments

This sequence is finite and a(7) = 14 is the last term. Nagura (see references) proves that for n >= 25, there is always a prime between n and 1.2*n. Hence, for any prime p > 25, there is always a number m between 4p and 4.8*p, and so floor(m/p) = 4. Since by assumption p > 4, floor(floor(m/p)/p) = 0 and so m! is divisible by p^4 but not p^5. It remains to check the primes up to 25 individually. - Charles R Greathouse IV, Apr 14 2020

Examples

			4 is a term since 4! = (2^3)*(3^1) and the multiplicity of 2 is 3 which is prime and the multiplicity of 3 is 1.
		

Links

• J. Nagura, On the interval containing at least one prime number, Proc. Japan Acad., 28 (1952), 177-181.

Crossrefs

Cf. A000142, A115627.

Programs

Mathematica

Select[Range[100], !AnyTrue[FactorInteger[#!][[;;,2]], CompositeQ] &] (* Amiram Eldar, Mar 29 2020 *)

PARI

ok(n)={my(f=factor(n!)[,2]); for(i=1, #f, if(f[i]<>1 && !isprime(f[i]), return(0))); 1}
{select(ok, [1..100])} \\ Andrew Howroyd, Mar 29 2020

PARI

f(m,p)=my(s); while(m\=p, s+=m); s;
is(n)=forprime(p=2,n\4+1, if(!isprime(f(n,p)), return(0))); 1;
select(is,[1..25]) \\ Charles R Greathouse IV, Apr 14 2020