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A061770 Numbers m = a(n) > a(n-1) such that there exists a smallest integer k > 1 such that k!/(k+1)^m is an integer. 2

%I #10 Apr 17 2015 09:59:27

%S 0,1,2,5,7,8,9,10,11,14,17,19,21,28,35,44,58,88,95,103,110,178,179,

%T 185,208,222,287,313,334,358,371,419,479,502,558,629,670,718,838,1006,

%U 1118,1259,1438

%N Numbers m = a(n) > a(n-1) such that there exists a smallest integer k > 1 such that k!/(k+1)^m is an integer.

%C Original name: The least exponent m = a(n) > a(n-1) for which k is the first number where k!/(k+1)^m is an integer.

%e a(5) = 8 because the first integer k > 1 such that (k+1)^8 divides k! is k = 39, which is larger than the first integer k > 1 such that (k+1)^7 divides k! (k = 35).

%e 6 is not in the sequence because the first integer k > 1 such that (k+1)^6 divides k! is k = 23, which is equal to the first integer k > 1 such that (k+1)^5 divides k!.

%t l = 0; Do[k = Max[l - 1, 1]; While[ !IntegerQ[ k! / (k + 1)^n], k++ ]; If[ k > l, l = k; Print[n] ], {n, 0, 1500} ]

%o (PARI) b(n)=k=2;while(k!%(k+1)^n,k++);k

%o print1(0,", ");for(n=1,100,if(b(n)>b(n-1),print1(n,", "))) \\ _Derek Orr_, Apr 16 2015

%Y Locations of records in A061768.

%K nonn

%O 0,3

%A _Robert G. Wilson v_, Jun 21 2001

%E Name and example edited by _Derek Orr_, Apr 16 2015

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Last modified March 29 06:15 EDT 2024. Contains 371265 sequences. (Running on oeis4.)