Could any (or all) of Mathematica, Maxima, Axiom derive and verify/validate recurrence formula derived from gf ?

Wed May 14 21:45:18 CEST 2008

Here's result of my (interactive) session 
with Mathematica for A024352 
%N A024352 Numbers which are the difference of two
squares
%F Recurrency a[n]=a[n-1]+a[n-3]-a[n-4],a[1]==3,
a[2]==5, a[3]==7, a[4]==8
%F  a(n)= (1/9)*(15 + 15*n + 3*Cos[(2*n*Pi)/3] - 
      Sqrt[3]*Sin[(2*n*Pi)/3])}}
%C First 200 terms
{3,5,7,8,10,12,13,15,17,18,20,22,23,25,27,28,30,32,33,35,37,38,40,42,43,45,47,48,50,52,53,55,57,58,60,62,63,65,67,68,70,72,73,75,77,78,80,82,83,85,87,88,90,92,93,95,97,98,100,102,103,105,107,108,110,112,113,115,117,118,120,122,123,125,127,128,130,132,133,135,137,138,140,142,143,145,147,148,150,152,153,155,157,158,160,162,163,165,167,168,170,172,173,175,177,178,180,182,183,185,187,188,190,192,193,195,197,198,200,202,203,205,207,208,210,212,213,215,217,218,220,222,223,225,227,228,230,232,233,235,237,238,240,242,243,245,247,248,250,252,253,255,257,258,260,262,263,265,267,268,270,272,273,275,277,278,280,282,283,285,287,288,290,292,293,295,297,298,300,302,303,305,307,308,310,312,313,315,317,318,320,322,323,325,327,328,330,332,333,335}
%t Mathematica Table[(1/9)*(15 + 15*n +
3*Cos[(2*n*Pi)/3] - 
      Sqrt[3]*Sin[(2*n*Pi)/3]),{n,200}]//Simplify

enjoy, zak

--- Alexander Povolotsky <apovolot at gmail.com> wrote:

> Dear - Zak, Thanks!
> 
> %F A024352 gp > ggf([3, 5, 7, 8, 9, 11, 12, 13, 15,
> 16, 17, 19, 20,
> 21, 23, 24, 25,
> 27, 28, 29, 31, 3, 33,35, 36, 37, 39, 40, 41, 43,
> 44, 45, 47, 48, 49,
> 51, 52, 53, 55, 56, 57, 59, 60, 61, 63, 64, 6, 67,
> 68, 69, 71, 72, 73,
> 75, 76, 77, 79, 80, 81, 83, 84, 85, 87, 88, 89, 91,
> 92, 93])
> = (-x^4 - 2*x^3 + 2*x^2 + 2*x + 3)/(x^4 - x^3 - x +
> 1)
> 
> %F A132395 ggf([1, 23, 1981, 270623, 38808661,
> 5586675623,
> 804461802541, 115842285207023, 1681286711863621,
> 2402105260570936823,
> 345903157236903231901, 49810054638975637017023,
> 717264767977969018307381])
> = (-132*x + 1)/(1584*x^2 - 155*x + 1)
> 
> %F A136211 ggf([1, 4, 5, 19, 24, 91, 115, 436, 551,
> 2089, 2640, 10009,
> 12649, 47956])
> = (-x^3 + 4*x + 1)/(x^4 - 5*x^2 + 1)
> 
> %F A101202 ggf([142857, 285714, 428571, 571428,
> 714285, 857142,
> 999999, 1142856, 1285713, 1428570, 1571427, 1714284,
> 1857141, 1999998,
> 2142855, 2285712, 2428569, 2571426, 2714283,
> 2857140, 2999997,
> 3142854, 3285711, 3428568, 3571425])
> = 142857/(x^2 - 2*x + 1)
> 
> %F A005131 ggf([1, 0, 1, 1, 2, 1, 1, 4, 1, 1, 6, 1,
> 1, 8, 1, 1, 10, 1,
> 1, 12, 1, 1, 14,
> 1, 1, 16, 1, 1, 18, 1, 1, 20, 1, 1, 22, 1, 1, 24, 1,
> 1, 26, 1, 1, 28,
> 1, 1, 30, 1, 1, 32,
> 1, 1, 34, 1, 1, 36, 1, 1, 38, 1, 1, 40, 1, 1, 42])
> = (-x^5 + 2*x^4 - x^3 + x^2 + 1)/(x^6 - 2*x^3 + 1)
> 
> %F A011934 ggf([0, 1, 7, 20, 44, 81, 135, 208, 304,
> 425, 575, 756,
> 972, 1225, 1519, 1856, 2240, 2673, 3159, 3700, 4300,
> 4961, 5687, 6480,
> 7344, 8281, 9295, 10388, 11564, 12825, 14175, 15616,
> 17152, 18785,
> 20519, 22356, 24300, 26353, 28519, 30800, 33200])
> = (x^3 + 4*x^2 + x)/(x^5 - 3*x^4 + 2*x^3 + 2*x^2 -
> 3*x + 1)
> 
> Sorry for sending more than one ;-)
> 
> Thanks again,
> Best Regards,
> Alex
> --------------------------------------------------
> On 5/14/08, zak seidov <zakseidov at yahoo.com> wrote:
> > Alexander,
> > send me your GF,
> > and i'll try,
> > zak
> > PS with  Mathematica
> > --- Alexander Povolotsky <apovolot at gmail.com>
> wrote:
> >
> > > Hi,
> > >
> > > I am addressing this email to the subset of
> people
> > > on SeqFan list,
> > > who, as I have spotted, are using some of those
> > > programs.
> > >
> > > Could any (or all) of Mathematica, Maxima, Axiom
> > > derive and
> > > verify/validate recurrence formula derived from
> the
> > > gf ?
> > >
> > > Thanks,
> > > Regards,
> > > Alex
> > > ================================================
> > > On 5/13/08, Ralf Stephan <ralf at ark.in-berlin.de>
> > > wrote:
> > >  You wrote
> > >
> > > > Are there any PARI/GP scripts to resolve given
> > > generating formula into:
> > > > a) recurrent formula
> > > > b) closed form formula
> > >
> > > Alex, I don't know of such scripts but the
> > > recurrence can be easily
> > > read off the g.f denominator:
> > > And as to b) no there are no such scripts
> because
> > > Pari isn't
> > > suited to symbolic computation. I'm sure there
> are
> > > scripts/functions
> > > for/in Mathematica, Maxima, Axiom.
> > >
> > >  Regards,
> > >  ralf
> > >
> >
> >
> >
> >
> >
> 

Nice ;-)

On 14/5/08 18:21, "Joerg Arndt" <arndt at jjj.de> wrote:

> 
> Worth submitting?  Yes, but only if you solve this one:
> 
> seq:  5656, 677655, 2323, 1, 1, 45, 56, 67, 8, 3, 787, 77, 812, 2, 2, 34, ...
> 
> Hint: seq starts with 5656 and is related to a 3rd order linear
> recurrence with coefficients over the most simple ring containing
> 1+sqrt(a+b*sqrt(c)) where a is a sum of three squares, b is the least
> positive number that can be written in two different ways as the sum
> of two cubes u^3+v^3 where both of u and v are positive, and c equals
> floor(Re(C)) where C is the smallest (wrt. L2-norm) complex number
> such that zeta(1/2+eps+I*C) == 0 where I==sqrt(-1) and -1/2<eps<+1/2.
> 

ja> From seqfan-owner at ext.jussieu.fr  Wed May 14 10:39:32 2008
ja> Date: Wed, 14 May 2008 09:27:16 +0200
ja> From: Joerg Arndt <arndt at jjj.de>
ja> To: seqfan at ext.jussieu.fr
ja> Subject: A005045 thanks and A002817
ja> ... 
ja> Thanks everybody for the cleanup, code, and extension of A005045!
ja> ...
ja> If a similar idea can be applied for A005045 one might
ja> come up with a proof.  I am not enough of a combinatorial
ja> person to do this and don't want to bother Brendan again.

I wrote down 2 of the 12 sub-cases of the proof of equivalence
between the A005045 o.g.f. and sum formula
in http://www.strw.leidenuniv.nl/~mathar/progs/a005045.pdf .
It is all done in a high-school math, grass-root approach.

Richard