Triangle
Gottfried Helms
helms at uni-kassel.de
Tue Apr 15 18:09:58 CEST 2008
Am 15.04.2008 15:55 schrieb franktaw at netscape.net:
> If you are going to post something like this, please let us know
> how you encountered it. Did you come up with some rules to
> generate it? Did you find it in a publication? Give us whatever
> information you have.
>
> Franklin T. Adams-Watters
Franklin -
this (and similar matrices) occur in a much complicated
structure, for which I try to find a simpler generation
rule.
Context:
If you consider the function
U_t(x) = t^x - 1
and the iterate
U_t(x,h) = U_t(t^x - 1,h-1)
then U_t(x,h) is a powerseries in x, where each term
is also a function of t and h.
So
U_t(x,h) = a1(t,h)*x/1! + a2(t,h)*x^2/2! + a3(t,h)*x^3/3! + ...
The coeffient-functions a_k() can be expressed as matrices, row-
resp. column-scaled by t resp t^h.
It is possible to get all coefficients a_k by symbolic
eigendecomposition of the involved matrix-operator; however
this is extremely memory-consuming. To get only 64 terms
for U_t(x,h) you need about 256 Mb memory, with more than geometric
growth-rate of memory depending on number of required terms.
I could identify the matrix-coefficents for the first
few a_k (say k=1..32), but there is not hope to be
able to improve this much. However, the matrices look
a bit systematic, related to the Stirling-numbers, and
also seem to follow some general rules so that is hope,
one could determine each coefficent *individually*.
This could be possible by solving for the unknown coefficients
in the a_k()-matrices by known initial values and some hypothetical
conditions.
In other words: this would allow to determine, say, a_96() just by
computation of its matrixcoefficients - without the symbolic
computation of eigensystem of the 96x96-matrix-operator for the
U_t()-function (which should be impossible due to time and memory
requirement).
A simple recursion scheme would then be a extremely useful
tool for the number-theoretical analysis of the U_t(x,h)-function
Well, I felt that the matrix, which I posted before, might have
come across someone and possibly a link to a related discussion could
be given, so I just posted the simple example.
-------------------------------------
Here is even more context:
The general problem is to solve for non-square matrices, which
contain coefficients to determine the powerseries-coefficients a_k.
Let's call these matrices M_1,M_2,M_3 for a_1,a_2,... respectively.
If I don't use the eigensystem-solver, the entries in M-matrices
are unknwon except some known "initial" values (first,last row).
According to a hypothesis of mine they satisfy also enough conditions,
that the unknown entries may be determined uniquely for each M_
individually.
Example, nearly the most simple one.
M_3 = RS= D_3 =
| -1 7 -12 6 | | 0 | | -1 |
| -b0 b1 -b2 b3 | | ? | | 1 |
| -c0 c1 -c2 c3 | | ? | | 1 |
| -6 11 -6 1 | | 0 | | 0 |
| 0 0 0 0 | | 0 | | -1 |
| 0 0 0 0 | | 0 | | -1 |
| 0 0 0 0 | | 0 | | 1 |
The unknown entries in M_3 are sought, such that the condition,
that their rowsums agree to some compositions of shifts of D_3,
is satisfied.
For this case, we see immediately, that the row-sums equal zero,
since we have the two rows of known init-values, and they provide
zeros. So using
D_3 * k0 = RS
is only possible for k0 = 0, and RS is then completely zero,
giving first relations between b and c-unknowns.
The next condition involves a vertical shift of columns in M_3:
M_3 = RS= D_3 =
| -1 | |-1 | | -1 |
| -b0 7 | | ? | | 1 |
| -c0 b1 -12 | | ? | | 1 |
| -6 c1 -b2 6 | | ? | | 0 |
| 0 11 -c2 b3 | | ? | | -1 |
| 0 0 -6 c3 | | ? | | -1 |
| 0 0 0 1 | | 1 | | 1 |
Again, the RS must match an integer composition of D_3,
k0 * D_3 = RS
and k0 = 1 in this case, and we have some more equations
to relate the unknowns - but that might not be enough
conditions.
Next is equivalent.
M_3 = RS= D_3 =
| -1 | |-1 | | -1 |
| -b0 | | ? | | 1 -1 |
| -c0 7 | | ? | | 1 1 -1 |
| -6 b1 | | ? | | 0 1 1 -1 |
| 0 c1 -12 | | ? | | -1 0 1 1 |
| 0 11 -b2 | | ? | | -1 -1 0 1 |
| 0 0 -c2 6 | | ? | | 1 -1 -1 0 |
| 0 0 -6 b3 | | ? | | 1 -1 -1 |
| 0 0 0 c3 | | ? | | 1 -1 |
| 0 0 0 1 | | 1 | | 1 |
Here we encounter compositions of shifted D_3 involved;
we seek a columnvector K=[k0,k1,k2,k3] , which allows
RS = D_3 * K
Obviously k0=1 and k3=1, but k1 and k2 are now two more
unknowns.
If we still don't have enough equations to determine
all the unknowns, this scheme can be continued.
--------------------------------------------------
But this is a very complicated process, epecially if the
M_matrices are bigger.
So I tried to find an alternative: in this case whether I
can determine the needed K-vectors apriori.
The K-vectors for consecutive M_2,M_3,M_4,,,, ,all with
the same shift, combined give some matrices, and
the posted matrix is just one example.
If I would be able to determine the K-matrices, by some
handy recursive generation rule, then also the solution
for the M-matrices would be immediate and I could generate
many more terms of the powerseries for the U_t()-function
for further investigation. (well, it is no problem to
get them numerically, however for a deeper consideration
of convergence and such I need a symbolic description in
terms of t and h)
Gottfried Helms
>
> -----Original Message-----
> From: Gottfried Helms <helms at uni-kassel.de>
>
> Did someone come across this triangle?
>
> 1
> 1 1 1
> 1 3 4 3 1
> 1 7 13 19 13 6 1
> 1 15 40 85 96 75 35 10 1
> 1 31 121 335 560 616 471 240 80 15 1
> ...
>
>
> (not found in OEIS)
>
> Apparently related to Stirling numbers (1'st or 2'nd kind)
> 1
> 1 1
> 1 3 1
> 1 7 6 1
> ...
>
> Gottfried Helms
>
>
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