How can I find the generating functions for the following series coefficient vectors?

[Gerry]

Hi all,

how can i find the generating functions for the following ci vectors?

c2: [-2]
c3: [4, 2]
c4: [8, -16]
c5: [16, 80, 24]
c6: [-32, 320, -368]
c7: [64, 1120, 3136, 720]
c8: [128, -3584, 19712, -16896]
c9: [256, 10752, 102144, 209408, 40320]
c10: [-512, 30720, -462336, 1838080, -1297152]
c11: [1024, 84480, 1892352, 12869120, 21441024, 3628800]
c12: [2048, -225280, 7163904, -76595200, 245070848, -149944320]
c13: [4096, 585728, 25479168, 402980864, 2188865536, 3130103808,
479001600]
c14: [-8192, 1490944, -86102016, 1923530752, -16326582272,
44491259904, -24349317120]
c15: [16384, 3727360, 278806528, 8484270080, 105995681792,
487356047360, 618377527296, 87178291200]
..
..

Regards

Gerry

[Ray Vickson]

What do you mean by the generating function of a vector?

R.G. Vickson

[Gerry]

> R.G. Vickson- Hide quoted text -
>
> - Show quoted text -

Hi Ray,

notice that c[i,1] = 2^(i-1) and that every c[i] vector has Floor[(i
+1)/2] elements.
The last vector elements for odd i are (2i)! i.e. 1, 2, 24, 720,
40320,...
I would like to find a function to reproduce these coefficients.

Regards

Gerry

[Robert Israel]

Gerry <gerr...@gmail.com> writes:

> On Jan 18, 6:23=A0pm, Ray Vickson <RGVick...@shaw.ca> wrote:

> notice that c[i,1] =3D 2^(i-1) and that every c[i] vector has Floor[(i

> +1)/2] elements.
> The last vector elements for odd i are (2i)! i.e. 1, 2, 24, 720,
> 40320,...
> I would like to find a function to reproduce these coefficients.

Actually c[i,1] = -2^(i-1) if i==2 mod 4, and in general g[i,j] <= 0 if
i + 2 j is divisible by 4. But let's remove the distraction of the signs.

The generating function for |c[i,1]| is 1/(1 - 2 t)
The generating function for |c[i,2]| seems to be 2 t^3/(1 - 2 t)^4
The generating function for |c[i,3]| seems to be 8 (3+4 t) t^5/(1 -2 t)^7
The generating function for |c[i,4]| seems to be
16 (45 + 156 t + 68 t^2) t^7/(1 - 2 t)^10
The generating function for |c[i,5]| seems to be
128 (315 + 1944 t + 2304 t^2 + 496 t^3) t^9/(1 - 2 t)^13

So I guess the generating function for |c(i,n)| is of the form
p_n(t) t^(2 n - 1)/(1 - 2 t)^(3 n - 2)
where p_n is a polynomial of degree n-2 for n >= 2, but I don't
know how to generate p_n.
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

[Gerry]

On Jan 18, 9:05 pm, Robert Israel
<isr...@math.MyUniversitysInitials.ca> wrote:

> University of British Columbia            Vancouver, BC, Canada- Hide quoted text -

>
> - Show quoted text -

That's nice and you are right about the signs.

Now the terms for the function i'm looking for are though as follows :

For c(i,1)= 1/(1 - 2 t)

+16*t^4 = +16*Pi^(-5+n)*(-4+n)*(-3+n)*(-2+n)*(-1+n)*n
-32*t^5 = -32*Pi^(-6+n)*(-5+n)*(-4+n)*(-3+n)*(-2+n)*(-1+n)*n
+64*t^6 = +64*Pi^(-7+n)*(-6+n)*(-5+n)*(-4+n)*(-3+n)*(-2+n)*(-1+n)*n

For c(i,2)=2 t^3/(1 - 2 t)^4

80*t^5 = 80*Pi^(-3+n)*(-2+n)*(-1+n)*n
320*t^6 = 320*Pi^(-4+n)*(-3+n)*(-2+n)*(-1+n)*n
1120*t^7 = 1120*Pi^(-5+n)*(-4+n)*(-3+n)*(-2+n)*(-1+n)*n

So is there a solution such that the generating functions c(i,n), give
the RHS?

Also do you think p_n can be found if i generate more c vectors?
It will take me some more time to figure these out.

Regards

Gerry

[Gerry]

> Gerry- Hide quoted text -

>
> - Show quoted text -

Looking a little more at this i found the following vector going with
the polynomials p:

c = [6, 360, 45360, 5443200, 359251200, 5884534656000, 35307207936000,
144053408378880000, 1034591578977116160000,
3414152210624483328000000, 471153005066178699264000000];

p[1]=1
p[2]=5*y + 13
p[3]=35*y^2 + 273*y + 502
p[4]=175*y^3 + 2730*y^2 + 13589*y + 21306
p[5]=385*y^4 + 10010*y^3 + 94259*y^2 + 377938*y + 538008
p[6]=175175*y^5 + 6831825*y^4 + 103538435*y^3 + 758044287*y^2 +
2660892494*y + 3543275784
p[7]=25025*y^6 + 1366365*y^5 + 30325295*y^4 + 348795447*y^3 +
2181280088*y^2 + 6982138692*y + 8845239888
p[8]=2127125*y^7 + 154854700*y^6 + 4728173450*y^5 + 78251653480*y^4 +
755233160357*y^3 + 4229343941308*y^2 + 12638068099452*y +
15394725430128
p[9]=282907625*y^8 + 26480153700*y^7 + 1063755302610*y^6 +
23898135020760*y^5 + 327440340665529*y^4 + 2791475616916980*y^3 +
14390276321683660*y^2 + 40746063806217360*y + 48071685665851776
p[10]=15559919375*y^9 + 1820510566875*y^8 + 93047428724250*y^7 +
2721523258203750*y^6 + 50085306456159975*y^5 + 599729054869873467*y^4
+ 4655689995271406080*y^3 + 22489379545888409220*y^2 +
60956282318085298800*y + 70014256334003438208
p[11]=32534376875*y^10 + 4652415893125*y^9 + 294727990557000*y^8 +
10875107018425650*y^7 + 258355036674509775*y^6 +
4119694850769361941*y^5 + 44531362589168807110*y^4 +
321076797520661125460*y^3 + 1471165539194229003000*y^2 +
3844979393672142847584*y + 4316256737180311092480

Step 1:
========
p1[i] = 1/c[i] * p[i]

the coefficient of the highest monomial of p1[i] are 1/A047058
(Sextuple factorial numbers: 6^n*n!)

Step 2:
========
p2[i] = 6^i*i!/c[i]*p[i]);

the coefficient of the second highest monomial of p2[i] are A152741/5
(13 times triangular numbers 13n(n+1)/2)

Step 3:
=======

??? i don't know
does someone else know how to continue ?