[seqfan] Re: Unknown sequence related to Bernoulli-numbers/zeta() at negative arguments / solution
Gottfried Helms
helms at uni-kassel.de
Sat Jun 21 12:47:28 CEST 2014
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
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