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A061770
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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.
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2
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0, 1, 2, 5, 7, 8, 9, 10, 11, 14, 17, 19, 21, 28, 35, 44, 58, 88, 95, 103, 110, 178, 179, 185, 208, 222, 287, 313, 334, 358, 371, 419, 479, 502, 558, 629, 670, 718, 838, 1006, 1118, 1259, 1438
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OFFSET
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0,3
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COMMENTS
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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.
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LINKS
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EXAMPLE
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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).
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!.
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MATHEMATICA
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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} ]
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PROG
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(PARI) b(n)=k=2; while(k!%(k+1)^n, k++); k
print1(0, ", "); for(n=1, 100, if(b(n)>b(n-1), print1(n, ", "))) \\ Derek Orr, Apr 16 2015
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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Name and example edited by Derek Orr, Apr 16 2015
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STATUS
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approved
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