

A255933


a(n) is the largest integer m such that s/(m!1) is integer, where s is the sum of all previous terms; a(1)=1.


0



1, 2, 2, 3, 2, 3, 2, 3, 2, 3, 4, 2, 2, 2, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 4, 5, 2, 2, 2, 3, 2, 3, 4, 2, 2, 2, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 4, 2, 2, 5, 2, 3, 2, 3, 4, 2, 2, 2, 2, 3, 2
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OFFSET

1,2


COMMENTS

For all n>1, a(n) exists and is at least 2, since 2 gives a denominator (2!1) = 1, thus an integer.
The sequence of partial sums is: 1,3,5,8,10,13,15,18,20,23,27,29,31,33,35,38,...
The record values occur at n=1,2,4,11,49,286,1997,...


LINKS

Table of n, a(n) for n=1..107.


MAPLE

a(5) = 2 since (1+2+2+3)/(n!1) = 8/(2!1) = 8, an integer.
a(6) = 3 since (1+2+2+3+2)/(n!1) = 10/(3!1) = 2, an integer.


PROG

(PARI) lista(nn) = {v = [1]; s = 1; print1(s, ", "); for (n=2, nn, k = 2; while(k!1 <= s, k++); until (type(s/(k!1)) == "t_INT", k); s += k; print1(k, ", "); v = concat(v, k); ); } \\ Michel Marcus, Mar 11 2015


CROSSREFS

Sequence in context: A163178 A219545 A029374 * A319396 A049237 A162361
Adjacent sequences: A255930 A255931 A255932 * A255934 A255935 A255936


KEYWORD

nonn


AUTHOR

Neri Gionata, Mar 11 2015


STATUS

approved



