[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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