A255933 a(n) is the largest integer m such that s/(m!-1) is an integer, where s is the sum of all previous terms; a(1)=1.

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

Offset: 1

Author

Neri Gionata, Mar 11 2015

Keywords

• nonn

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,...

Programs

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.

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