A131709

[Max Alekseyev]

On 10/31/07, koh <zbi74583 at boat.zero.ad.jp> wrote:

> > The total number of partitions (i.e., a(4) in A131709) is
> > 4 + 4 + 32 + 4 + 61 = 105
> >
>     I think that the last term is 60 , so it bedcomes 104.

Agree.

>     My result corresponded with your classification is as follows.
>     a(3) = 3^4 + 2*(3^2-1) - 1 + 2+ 3 + 3
>            = {5)+21} + {3)-16} - {I} + {2)-2} + {4)-1} +  {1)-1}
>     Where,"I" means the " impossible case".
>     I didn't counted it.
>
>     I have awoke that I counted twice the case of  two 4-cycle.
>     So, the second term must be 2*(3^2-1)  = 16
>     Hence a(3) = 104
>
>     Both results are the same.

Yes, a(3)=104 and I suggest the following recurrent formula:
a(n+1) = 10*(a(n)-a(n-1)) + 4
giving
4, 14, 104, 904, 8004, ...

[...]

>     a(4)
>     = 3^6 + 2*3^4 - 2*3^2 + 2*3^2 - 1 + 2*1 + 3*3^2 + 2*1 + 2*3 + 3
>     = 729 + 162 + 1 + 27 + 11
>     = 930
>
>     It is not correct.
>     I counted the cycles twice in 2nd and 3rd and 4th term and mistook the sign of 6th term.
>
>     The total number is the following.
>     = 3^6 + 2*(3^4-2) - 2*(3^2-1) + 2*(3^2-1)  - 1 -2*1 + 3*3^2 + 2*1 + 2*3 + 3
>     = 922
>
>     Corresponding cycles :
>         nc      14c              14c              16c            16c  24c    24c         18c     16c+14c   1.10c
>         where "nc" means No cycle , "14c" means One 4 cyclth.

I counted yet another way and still got a(4)=904 that agrees with the
recurrence above.
Please double check your computation of a(4).

Regards,
Max