A284050 a(n) = floor(A240751(n) / n), where A240751(n) = the smallest k such that in the prime power factorization of k! there exists at least one exponent n.

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

Offset: 1

Author

David A. Corneth, Mar 19 2017

Keywords

• nonn
• easy

Comments

For n > 2, p = a(n) + 1 is the prime that has exponent n in A240751(n)! (see A240751 for an outline of a proof).

First occurrence of p-1: 1, 2, 12, 29, 186, 2865, 3265, 379852, 7172525, ..., (A240764). - Robert G. Wilson v, Apr 15 2017. Comment changed by David A. Corneth, Apr 15 2017

Examples

			For n = 5, p = a(n) + 1 = 3 is the prime such that A240751(5)! = 12! is the least factorial that has exponent 5.
		

Links

• Robert G. Wilson v, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A240751, A240755, A240764, A284051, A157928.

Programs

Mathematica

Table[k = 2; While[! MemberQ[FactorInteger[k!][[All, -1]], n], k++]; Floor[k/n], {n, 87}] (* Michael De Vlieger, Mar 24 2017 *)

PARI

a(n) = A240751(n)\n \\ (for computation of A240751(n), see A240751)

Formula

A240751(n) = n*a(n) + A284051(n). - Antti Karttunen, Mar 22 2017

a(n) = A240755(n) - 1 for n > 2 and a(n) = A240755(n) for n < 3. I.e., A240755(n) - A157928(n+1). - David A. Corneth, Mar 27 2017