[seqfan] Re: decode sequence

Max Alekseyev maxale at gmail.com
Fri Sep 4 19:00:13 CEST 2015


Hi Gottfried,

I've just realized this question of you from 2007 (quoted below) is related
to your recent MO question (and my answer):
http://mathoverflow.net/questions/208873/is-there-a-systematic-relation-between-the-generating-functions-for-the-rows-vs/

Also, the sequence b(n) you mentioned is now http://oeis.org/A180609 (added
by Paul D. Hanna in 2010).

Regards,
Max


On Tue, Aug 21, 2007 at 4:20 AM, Gottfried Helms <
Annette.Warlich at t-online.de> wrote:

> Am 21.08.2007 01:54 schrieb Max Alekseyev:
> > Gottfried,
> >
> > I have not got your question.
> > Do you ask help in identifying your sequences? If so, did you try to
> > feed them to Superseeker?
> >
> > In general, it is hard to guess the nature of a new sequence just from
> > numerical values. It would make much sense if you provided detailed
> > description of how to compute the numerical values of your sequence.
> > The current description ("context") is rather obscure.
> >
> > Regards,
> > Max
>
> Max -
>
>  thanks for your comments.
>  The sequence comes up in the following situation in a variant
>  of the tetration/powertower/hyperexpoentiation - context.
>
>  Assume the triangular matrix of stirling-numbers 2'nd kind
> S2 =
>   1  .   .   .   .  .
>   0  1   .   .   .  .
>   0  1   1   .   .  .
>   0  1   3   1   .  .
>   0  1   7   6   1  .
>   0  1  15  25  10  1
>
> do a similarity-scaling by diagonal factorials F=diag(0!,1!,2!,3!,...)
> f=F^-1
>
> fS2F = f * S2 * F =
>   1      .     .    .  .  .
>   0      1     .    .  .  .
>   0    1/2     1    .  .  .
>   0    1/6     1    1  .  .
>   0   1/24  7/12  3/2  1  .
>   0  1/120   1/4  5/4  2  1
>
> Then , with a vector with consecutive powers of x,
>
>      V(x) = colvector(1,x,x^2,x^3,x^4,...)
>
> actually with its transpose to a rowvector denoted by ~,
> we get, assuming infinite dimension
>
>   V(x)~ * fS2F = V(y)~
>                = V(e^x -1)~
>                = rowvector(1, e^x -1, (e^x -1)^2, ...)
>
> again a vector V() containing consecutive powers.
> The interesting scalar result is in the second column of
> the result vector; we may say with fS2F we have an operator
> which implements
>
>   f: x  -> e^x - 1
>
> Thus this process may be iterated. The second iteration is
> now, letting y = e^x-1
>
>   V(y)~ * fS2F = V(z)~
>                = V( e^y - 1)~
>                = V( e^(e^x -1) - 1)~
>                = rowvector(1, e^(e^x -1) - 1, (e^(e^x -1) - 1)^2, (e^(e^x
> -1) - 1)^3,...)
>
> and so on.
>
>
>
> Iteration can now be expressed in two ways:
>
> a) the scalar notation
>   f:  x -> e^x -1
>   f°1 := f
>   f°2 := 2'nd iteration
>   f°b := b-fold iteration
>
> b) matrix-notation
>
>   f°b equivalences   V(x)~ * fS2F^b = V(z)~
>
> where z is then the above described powertower of height b.
>
> ----------------------------------
>
> The benefit of the matrix-notation for this variant of the tetration-
> problem is now, that we can use the matrix-logarithm and
> ~ exponential to compute even fractional, real or complex
> powers of fS2F, and can thus define a fractional, real or even
> complex-fold iteration on f.
>
> The complex powers of fS2F can be determined by simply
> multiplicating the matrix-logarithm times b and exponentiate again
>
>   fS2F^b = MExp( b * MLog(fS2F) )
>
> Thus, to compute f°b for a given vector V(x), it suffices now to
> compute the powerseries constructed by the vector-multiplication
>
>     z = V(x)~ * fS2F^b [,1]        where [,1] denotes the 2'nd column
>       = sum{k=0..inf} x^k * s_k    where s_k denotes the k'th entry
>                                    of that column
>
> To get things simpler again it would suffice to compute s_k depending
> on b.
> This is indeed possible and also gives only finite polynomials in b,
> since fS2F is triangular and has a unit-diagonal, so the computation
> of the Mlog and MExp employs only nilpotent matrices, where the
> involved series-computations terminate after finite number of terms if
> dimension of fS2F is finite.
>
> What we get for s_k are polynomials in b of order k-1.
> The first few are
>
>                                        0 = s_0
>                                        1 = s_1
>                                    1/2*b = s_2
>                           1/4*b^2-1/12*b = s_3
>                  1/8*b^3-5/48*b^2+1/48*b = s_4
>     1/16*b^4-13/144*b^3+1/24*b^2-1/180*b = s_5
>     ...
>
> Here my question begins. To avoid the computation of the
> matrix-logarithm and exponentiation it would suffice to know
> the coefficients of b for each s_k; and my hope is to find
> a simpler description for them than by expliciting the
> log/exp-application referring all the entries of fS2F.
>
> The coefficients of b form a triangle of pol-coeffs, call it PC:
> PC=
>   0        .         .        .         .     .
>   1        .         .        .         .     .
>   0      1/2         .        .         .     .
>   0    -1/12       1/4        .         .     .
>   0     1/48     -5/48      1/8         .     .
>   0   -1/180      1/24  -13/144      1/16     .
>   0  11/8640  -91/5760  89/1728  -77/1152  1/32
>  ....
>
> ( b^0   b^1      b^2       b^3      b^4      b^5   the columns are
>                                                    related to the powers
>                                                    of b)
> where formally
>
>    PC * V(b) = colvector( s_0, s_1, s_2,....)
>
> and the second column contains the sequence in question.
>
> I already found, that the subsequent columns can be described
> by simple scalings of the second column, so the only difficult
> aspect in the matrix is the sequence in the second column, which goes
> (well, actually n begins at 2 here, I messed it in my previous post)
>
> a(n)= 1/2  -1/12  1/48  -1/180  11/8640  -1/6720  -11/241920  29/1451520
>       493/43545600  -2711/239500800  -6203/3592512000  2636317/373621248000
>      -10597579/10461394944000  -439018457/78460462080000 ...
>
> and may be rescaled to integers by factorials
>
> c(n) = sequence ( F^2 * PC [,1])
>
>             where F^2 = diag( 0!^2 , 1!^2, 2!^2 , ... )
>
>      = 2     -3    12     -80     660     -3780      -73920     2630880
>        149083200    -18035740800       -396166256640  273606679906560
>       -7698990233975040      -9568219721630169600   ...
>
>
> (I've just started superseeker while I'm writing this, completely
> forgotten!)
>
> Gottfried
>
>
>
>


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