[seqfan] Re: Unknown sequence related to Bernoulli-numbers/zeta() at negative arguments / solution

[Gottfried Helms]

Am 21.06.2014 08:16 schrieb Paul D Hanna:
> Hi Gottfried,
>     You were correct in that it a0() and a1() are related by convolution.
> Define
>
>  A1 = 1/4*x^2/2! + 1/8*x^3/3! + 1/48*x^4/4! - 1/48*x^5/5! - 1/96*x^6/6!
> + 1/72*x^7/7! + 101/8640*x^8/8! - 3/160*x^9/9! - 13/576*x^10/10! + 1/24*x^11/11!
> + 7999/120960*x^12/12! - 691/5040*x^13/13! - 2357/8640*x^14/14! + 5/8*x^15/15!
> + 52037/34560*x^16/16! +...
>
> then sqrt(2*A1) = log( (exp(x)-1)/x )
>
>    = 1/2*x + 1/12*x^2/2! - 1/120*x^4/4! + 1/252*x^6/6! - 1/240*x^8/8!
>       + 1/132*x^10/10! +...
>  .
Hi Seqfans, hi Paul -

 your idea was perfect. It is immediately generalizable to all sequences a_k(n).
 I'll show how this can be done:

 Let M be the (transposed) Carleman-matrix (of theoretically infinite size)
 for the function

   f(x) = log( x/(exp(x)-1))    //note: to adapt signs I took
                                // the reciprocal in the log-function

 then with the (formal) Vandermondevector (of infinite size)

    V(x)=[1, x, x^2, x^3, x^4,...]

 we can write (by the general construction of a Carleman-matrix)
 the dot-product:

     V(x) * M = [1,  f(x),  f(x)^2, f(x)^3, ... ]
              = V( f(x) )

 Then with the scaling by factorials
    G = diagonal(0!,1!,2!,3!,4!,...) and its inverse
    g = G^-1 = diagonal(1/0!,1/1!,1/2!,1/3!,1/4!,...)

 we get the similarity-scaled version:

   A =  G * M * G^-1

                     (which is sometimes called the "Bell-matrix of f(x)" ).

 The columns of A give now the terms of all a_k(n). For a better
 systematic notation I should now renumber my sequences such that now

    a_0(n) = [1,0,0,...]
    a_1(n) = [0,-1/2,-1/12,...]      ( =  a0(n+1) the previous notation)
    a_2(n) = [0,0,1/4,1/12,...]      ( =  a1(n+1) the previous notation)
    ...

The matrix A defines the sequences of any order by its columns:

  a_0(n)  a_1(n)    a_2(n)     a_3(n)     a_4(n)    a_5(n)      a_6(n      a_7(n) ...
---------------------------------------------------------------------------------
A=
      1       .         .          .          .         .            .          .   ...
      0    -1/2         .          .          .         .            .          .   ...
      0   -1/12       1/4          .          .         .            .          .
      0       0       1/8       -1/8          .         .            .          .
      0   1/120      1/48       -1/8       1/16         .            .          .
      0       0     -1/48      -5/96       5/48     -1/32            .          .
      0  -1/252     -1/96     13/576       5/64     -5/64         1/64          .
      0       0      1/72      7/192    -7/1152   -35/384        7/128     -1/128
      0   1/240  101/8640      -1/64  -469/6912    -7/288       35/384     -7/192
      0       0    -3/160  -101/1920     -5/384  133/1536        7/120    -21/256
      0  -1/132   -13/576    47/2304     67/576  745/9216    -245/3072  -133/1536
      0       0      1/24   143/1152   275/4608  -121/768  -3179/18432  847/18432
     ...    ...       ...       ...         ...       ...          ...        ...

(A quick check with the new sequence a3(n) and a4(n) shows pretty good
 results with the further coefficients of linear-regressions on my data
 and this suggests strongly that this is in fact the sought general solution)

 Gottfried Helms