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

Eric Angelini Eric.Angelini at kntv.be
Thu May 20 13:48:22 CEST 2010

```[Franklin]:

> However, if we take n divides the sum of the first n terms of S, with
the weaker restriction that no term can have occurred before, we get
the very beautiful http://www.research.att.com/~njas/sequences/A019444.

... which is exactly what I had indicated in my very first post of monday:

>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

Best,
É.

-----Message d'origine-----
De : seqfan-bounces at list.seqfan.eu [mailto:seqfan-bounces at list.seqfan.eu] De la part de franktaw at netscape.net
Envoyé : jeudi 20 mai 2010 12:01
À : seqfan at list.seqfan.eu
Objet : [seqfan] Re: a(n) divides the sum of the first a(n) terms of S

However, if we take n divides the sum of the first n terms of S, with
the weaker restriction that no term can have occurred before, we get
the very beautiful http://www.research.att.com/~njas/sequences/A019444.

(I just submitted a formula for this:

Let s(n) = sum(k=1,n,a(k))/n = A019446(n). Then if s(n-1) does not
occur in a(1),...,a(n-1), a(n) = s(n) = s(n-1); otherwise, a(n) =
s(n-1) + n and s(n) = s(n-1) + 1.

and a corresponding PARI program.)

-----Original Message-----
From: franktaw at netscape.net

...

Of course n divides the sum of the first n terms of S is just the odd
numbers.