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

Eric Angelini Eric.Angelini at kntv.be
Thu May 20 10:15:25 CEST 2010

```Nice !

------ Message d'origine--------
De: seqfan-bounces at list.seqfan.eu de la part de franktaw at netscape.net
Date: jeu. 20/05/2010 09:34
À: seqfan at list.seqfan.eu
Objet : [seqfan] Re: a(n) divides the sum of the first a(n) terms of S

One can also look at n divides the sum of the first a(n) terms of S,
which gives us:

1, 3, 4, 7, 8, 9, 12, 16, 18, 19, 20, 23, 24, 25, 26, 33, 34, 42, 46,
48, 49, 50, 59, 61, 63, 65, 66, 67, 68, 69, 70, 71, 72, 78, 79, 80, 81,
82, 83, 84, 85, 98, 99, 100, 101, 115, 116, 131, 133, 155, 156, 157,
158, 159, 160, 161, 162, 163, 169, 170, 189, 190, 208, 209, 231, 242,
244, 246, 248, 250, 252, 254, 255, 256, 257, 258, 259, 263, 296, 298,
300, 302, 304, 306, 308, 309, 310, 311, 312, 313, 314, 315, 316, 317,
318, 319, 320, 358, 360, 404

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

-----Original Message-----
From: Eric Angelini <Eric.Angelini at kntv.be>

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)

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