[seqfan] Re: a(n) divides the sum of the first a(n) terms of T.

[Max Alekseyev]

Franklin,

I still do not see why you claim that "otherwise a(k) would be n".
Consider the case when k appears among a(1), a(2), ..., a(k-1). Then
a(k) would not have a freedom to be n, since it would have to satisfy
the congruence:
a(1) + a(2) + ... + a(k) == 0 (mod k).

Regards,
Max


On Wed, Jun 11, 2014 at 11:03 PM, Frank Adams-Watters
<franktaw at netscape.net> wrote:
> Right. I needed to say k > n instead of >=.
>
>
> Franklin T. Adams-Watters
>
> -----Original Message-----
> From: Max Alekseyev <maxale at gmail.com>
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> Sent: Wed, Jun 11, 2014 9:58 pm
> Subject: [seqfan] Re: a(n) divides the sum of the first a(n) terms of T.
>
>
> On Wed, Jun 11, 2014 at 9:59 PM, Frank Adams-Watters
> <franktaw at netscape.net> wrote:
>>
>> Once you have a k such that k >= n and a(k) > n, n can no longer
>
> appear in
>>
>> the sequence; otherwise a(k) would be n.
>
>
> Take n = 2 and k = 2. We have a(k) = 3 > n, but n still appears in the
> sequence as a(3).
>
> Regards,
> Max
>
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