[seqfan] a(n) divides the sum of the first a(n) terms of S

[Eric Angelini]

Hello SeqFans,

S = 1,3,5,6,10,11,12,13,14,15,20,22,24,26,28,29,30,31,32,48,49,65,...

1 divides exactly 1 (initial segment of 1 term)
3  d.e. 1+3+5=9 (i.seg. of 3 terms)
5  d.e. 1+3+5+6+10=25 (i.seg. of 5 terms)
6  d.e. 1+3+5+6+10+11=36 (i.seg. of 6 terms)
10 d.e. 1+3+5+6+10+11+12=48 (i.seg. of 10 terms)
etc.

This evoques:
http://www.research.att.com/~njas/sequences/A019444

"a_1, a_2, ..., is a permutation of the positive integers such that
 the average of each initial segment is an integer, using the greedy
 algorithm to define a_n."

... but S, here, is monotonically increasing and we always use the
smallest available integer not leading to a contradiction.

Best,
É.