Thx M Somos,the desired seqce.is prod triangular no....

[Don McDonald]

In message <LFOb8Qj034n at paradise.net.nz> you write:
(Seqfan list seems to be down.)

> x To: se16 at btinternet.com,seqfan,tonyaj
> x Subject: RE: superseeker
> x References: <000001c28c39$320c8120$df4a7ad5 at knbwuaxl>
> 
> Greeting Michael Somos, Seq Fans,
(Tony AJ already deliv,)
> Thank you Neil,  (your original msg below)
> 
> (Mich reports some kind of error in Maple..  the sequence
> should not be hard.)
> 
> I haven't done an exercise like that before.
> /  don.mcdonald.
> 17.11.02  16:59   23:30.
my file > vanderMond.e

>
> the desired problem sequence (below).
>          is a product of a triangular number
>          x a square pyramidal no.
>          x (8n+19) /15.     ??  = (8k+3)/...
> 
>             GP/PARI CALCULATOR Version 1.36
>                  (Archimedes version)
> 
> ? for(n=1, 10, t=n*(n+1)/2;sq=(n+1)*(n+2)*(2*n+3)/6;
> q=(8*n+19);print(t,"  ",sq,"  ",q,"  ",t*sq*q/15))
> 
> T x  SQP x (8n+19) /15 = Required.
> 1    5        27         9
> 3    14        35         98
> 6    30        43         516
> 10    55     51         1870
> 15    91     59         5369
> 21    140     67       13132
> 28    204     75       28560
> 36    285     83       56772
> 45    385     91       105105
> 55    506     99       183678
> 
> I haven't done an exercise like that before.
> So it was a good test for me.

But I would like to generate a cross-product matrix in pari.
and solve regression equation
betahat = (X'X)^-1 * (X'Y)

where for example X = may be a 7x7 vandemonde matrix,
[1 1 1 1,
1 2 4 8
1 3 9 27
1 4 16 64]

i would like to find out how this is done. etc.

ss=[];for(p=0,12,s=0;
for(x=-3,3,s+=x^p);ss=concat(ss,s) )

??
a pari program, ( under the anti-diagonals sequence of
the above matrix,) could give me a start, please.

i have acorn archimedes pari-gp.
RISC-OS reduced instruction set computer operating system.

i haven't run anything in the realms of...
for(n=1,100,print1(vecmax(component(factor(
n),1)),","))

i must try that sequence = largest prime that divides n.

i do not have any ibm compatible computer or win9x
or dos programs.
> 
> cheers
> /  don.mcdonald.          (delete the rest.)
> 17.11.02  16:59
> 
> ? for(n=1, 13, t=n*(n+1)/2;sq=(n+1)*(n+2)*(2*n+3)/6;q=(8*n+19);
> print("trian =",t," sqpyramid = ",sq," q = ",q,
> " prod/15=  ",t*sq*q/15))
> 
> trian =1 sqpyramid = 5 q = 27 prod/15=  9
> trian =3 sqpyramid = 14 q = 35 prod/15=  98
> trian =6 sqpyramid = 30 q = 43 prod/15=  516
> trian =10 sqpyramid = 55 q = 51 prod/15=  1870
> trian =15 sqpyramid = 91 q = 59 prod/15=  5369
> trian =21 sqpyramid = 140 q = 67 prod/15=  13132
> trian =28 sqpyramid = 204 q = 75 prod/15=  28560
> trian =36 sqpyramid = 285 q = 83 prod/15=  56772
> trian =45 sqpyramid = 385 q = 91 prod/15=  105105
> trian =55 sqpyramid = 506 q = 99 prod/15=  183678
> trian =66 sqpyramid = 650 q = 107 prod/15=  306020
> trian =78 sqpyramid = 819 q = 115 prod/15=  489762
> trian =91 sqpyramid = 1015 q = 123 prod/15=  757393
> ? \q.
> 
> 
> ? x=(8*n+19)*n*(n+2)*(2*n+3)*(n+1)^2/180
> %1 = 4/45*n^6 + 7/10*n^5 + 77/36*n^4 + 19/6*n^3 + 409/180*n^2 + 19/30*n
> 
> ? factor(%1)    // Pari?? does not give constant multiplier?
> %2 = 
> |8*n + 19 1 |
> 
> |n 1 |
> 
> |n + 2 1 |
> 
> |2*n + 3 1 |
> 
> |n + 1 2 |   (this should means (n+1)^2.    .)
> 
> 
> ? x/(n*(n+1)/2)
> %3 = 8/45*n^4 + 11/9*n^3 + 55/18*n^2 + 59/18*n + 19/15
> 
> ? x/ (n*(n+1)/2) / ((n+1)*(n+2)*(2*n+3)/6)
> %4 = 8/15*n + 19/15
> 
> 
> In message <000001c28c39$320c8120$df4a7ad5 at knbwuaxl> you write:
> 
> > The sequence looks like a simple 6th order polynomial starting with
> > 4n^6/45+...; if so then the previous three terms are all 0.
> > 
> > > -----Original Message-----
> > > From: N. J. A. Sloane
> > > Sent: 14 November 2002 22:45
> > > To: seqfan at ext.jussieu.fr; Michael Somos
> > > Subject: Re: superseeker
> > >
> > > the sequence that michael somos was waiting for was this:
> > >
> > >
> > > lookup 9 98 516 1870 5369 13132 28560 56772 105105 183678 306020
> > > 489762 757393
> > >
>