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%I #21 Apr 28 2017 12:48:37
%S 0,0,1,2,2,3,1,2,3,2,3,2,1,2,1,2,2,2,3,5,3,2,3,4,3,4,3,4,8,3,1,2,3,2,
%T 3,6,2,2,3,1,3,4,3,2,3,2,3,3,3,4,6,7,3,4,4,4,5,4,5,4,4,5,1,2,5,2,3,5,
%U 4,2,3,6,3,4,6,7,7,2,3,2,3,4,8,3,3,4,7
%N a(n) = A240751(n) mod n, where A240751(n) = the smallest k such that in the prime power factorization of k! there exists at least one exponent n.
%H Robert G. Wilson v, <a href="/A284051/b284051.txt">Table of n, a(n) for n = 1..10000</a>
%F A240751(n) = n*A284050(n) + a(n). - _Antti Karttunen_, Mar 22 2017
%e A240751(5) = 12 so a(5) = 12 mod 5 == 2.
%t Table[k = 2; While[! MemberQ[FactorInteger[k!][[All, -1]], n], k++]; Mod[k, n], {n, 87}] (* _Michael De Vlieger_, Mar 24 2017 *)
%o (PARI) a(n) = A240751(n)%n \\ (For computation of A240751(n), see A240751)
%Y Cf. A240751, A284050.
%K nonn,easy
%O 1,4
%A _David A. Corneth_, Mar 19 2017
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