

A158366


Least k such that n! divides (n+k)!/(n+1)!.


0



1, 2, 3, 4, 5, 4, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 26, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 40, 43, 44, 43, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 64, 67, 68, 69, 70, 71, 72
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OFFSET

1,2


COMMENTS

Motivated by 6!=10!/7!.
It appears that for most n, a(n) = n. The sequence of n for which a(n) != n begins: 6, 28, 42, 45, 66, 77, 91, 110, 126 ... that is probably A120624.  Michel Marcus, Aug 21 2013


LINKS



PROG

(PARI) for (n=1, 100, k=1; while((n+k)!/(n+1)!%n!, k++); print1(k, ", "))


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



