login
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.
3
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
1,1
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
LINKS
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
EXAMPLE
For n = 5, p = a(n) + 1 = 3 is the prime such that A240751(5)! = 12! is the least factorial that has exponent 5.
MATHEMATICA
Table[k = 2; While[! MemberQ[FactorInteger[k!][[All, -1]], n], k++]; Floor[k/n], {n, 87}] (* Michael De Vlieger, Mar 24 2017 *)
PROG
(PARI) a(n) = A240751(n)\n \\ (for computation of A240751(n), see A240751)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
David A. Corneth, Mar 19 2017
STATUS
approved