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

Maximilian Hasler maximilian.hasler at gmail.com
Tue May 18 18:08:50 CEST 2010

```On Mon, May 17, 2010 at 7:42 PM, Eric Angelini <Eric.Angelini at kntv.be> wrote:
>
> 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)

I'm not sure I understand correctly.
Up to the "6",
"(i.seg. of N terms)" seems to mean the N-th partial sum a(1)+...+a(N)
but 10 does not divide 48, nor are there 10 terms in the sum...

So is the intended definition
"a(n) = the least possible integer > a(n-1) such that a(k) divides
a(1)+...+a(a(k)) for all k" ?

(I write it in that complicated way because the sum is not yet known
when a(n) is computed,
but maybe that's not necessary.)

Maximilian

> 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.

```