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

Eric Angelini Eric.Angelini at kntv.be
Wed May 19 01:13:44 CEST 2010

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Yes, Maximilian, my typos, sorry -- you
have it right, many thanks !
And sorry again for the late answer to
problem -- I was out of town.
Best,
E.

-------- Message d'origine--------
De: seqfan-bounces at list.seqfan.eu de la part de Maximilian Hasler
Date: mar. 18/05/2010 18:44
À: Sequence Fanatics Discussion list
Objet : [seqfan] Re: a(n) divides the sum of the first a(n) terms of S

>> S = 1,3,5,6,10,11,12,13,14,15,20,22,24,26,28,29,30,31,32,48,49,65,
>>...
>> 10 d.e. 1+3+5+6+10+11+12=48 (i.seg. of 10 terms)
>
> 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...

(sorry, now (I think) I understand and everything is consistent,
only in the example the sum should go up to the 10th term,
...+15  = 90 (== 0 mod 10), + 20 = 110 (==0 mod 11),
+22 = 132 (== 0 mod 12), + 24 = 156 (==0 mod 13),
+26 = 182 (== 0 mod 14), + 28 = 210 (==0 mod 15),
+ 29 +...+ 48 = 380 (==0 mod 20)

then ... + 49 + 65 = 494 (== 10 mod 22) ?
Maybe another typo ?
actually it seems that between indices a(n) and a(n+1), elements
increase with step of 1,
and then the a(n+1)-th element makes a jump sufficiently large to
satisfy the criterion.
Did I get it right, this time ?
Then it is easy to write the associated code....

{c=#a=[1,3];while(#a<100,
a=concat( a,a[#a] + if( #a+1<a[c],1,
(a[c]-sum(i=1,#a,a[i],a[#a])%a[c++-1])  )));a}

[1, 3, 5, 6, 10, 11, 12, 13, 14, 15, 20, 22, 24, 26, 28, 29, 30, 31,
32, 48, 49, 55, 56, 60, 61, 67, 68, 72, 89, 93, 97, 101, 102, 103,
104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 141,
161, 162, 163, 164, 165, 166, 175, 188, 189, 190, 191, 222, 269, 270,
271, 272, 273, 274, 313, 378, 379, 380, 381, 452, 453, 454, 455, 456,
457, 458, 459, 460, 461, 462, 463, 464, 465, 466, 467, 468, 470, 471,
472, 473, 553, 554, 555, 556, 602, 603, 604, 605]

So the 65 should be 55... :-)

Maximilian

>
> 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
>>
>> Best,
>> É.
>

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