a(n) = sigma_2(a(n-1)): 10|a(n>3) ?

[Richard Guy]

This sequence seems not to be in OEIS.  Here's a
hand calculation of the next term

$(943950)=$(2.3.5^2.7.29.31)=5.10.651.50.842.962=1318281510000

It would certainly be remarkable if there were a subsequent term
not divisible by  2  and  5.   R.

On Wed, 14 May 2008, Max Alekseyev wrote:

> Zak,
>
> I think, it's a tough question.
> An affirmative answer to your question would imply, in particular,
> that in this sequence there are no squares (since a term following a
> square in this sequence is odd). But existence of squares in this
> sequence is a hard question by itself, I believe.
>
> Regards,
> Max
>
> On Wed, May 14, 2008 at 12:09 AM, zak seidov <zakseidov at yahoo.com> wrote:
>> Rule:
>> a(n>1)=sigma_2(a(n-1)) with a(1)=2.
>> SEQ starts:
>> 2,5,26,850,943950.
>> My Q:
>> Are all next terms divisible by 10?
>> thanks, zak
>>
>>
>>
>>
>>
>
>
>




Hi,

* zak seidov <zakseidov at yahoo.com> [May 14. 2008 09:04]:
> Dear SF gurus,
> to your kind consideration,
> thanks, zak
> 

> %N A1 Related to sum of two squares


Hmmm, _very_ vague indeed...


> %S A1 
> 1,2,1,4,2,9,1,38,67,344,2202,13139,1637801,4372282,31146832303,53840975396
> 
> %C A1 Puzzle? Worth submitting?

Is it a puzzle or isn't it?  Do you know?


>  Hint1: sequence starts with {1,2,1,4}. Then each time
> two more terms are added.

Friggin BY WHAT F%$&%$ing rule ???


>  Hint2: similar sequence starting with {1,1,1,2};
> {1,1,1,2,1,3,1,7,4,22,6,158,500,3500,58000,556000,75000000,1975000000

(HintF++: and I am plagued by ideas as least as you are).



> %A A1 Zak Seidov (zakseidov(AT)yahoo.com), May 14 2008


Worth submitting?  Yes, but only if you solve this one:


recurrence with coefficients over the most simple ring containing
1+sqrt(a+b*sqrt(c)) where a is a sum of three squares, b is the least
positive number that can be written in two different ways as the sum
of two cubes u^3+v^3 where both of u and v are positive, and c equals
floor(Re(C)) where C is the smallest (wrt. L2-norm) complex number

best regards,   jj