decode sequence
Gottfried Helms
Annette.Warlich at t-online.de
Tue Aug 21 10:20:25 CEST 2007
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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