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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 (list; graph; refs; listen; history; text; internal format)
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

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

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

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

Sequence in context: A102383 A070913 A266743 * A114280 A123564 A065882

Adjacent sequences:  A284047 A284048 A284049 * A284051 A284052 A284053

KEYWORD

nonn,easy

AUTHOR

David A. Corneth, Mar 19 2017

STATUS

approved

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Last modified March 19 00:03 EDT 2019. Contains 321305 sequences. (Running on oeis4.)