Re: help in constructing a binomial consequence

• To: mathgroup at smc.vnet.net
• Subject: [mg98277] Re: help in constructing a binomial consequence
• From: "Sjoerd C. de Vries" <sjoerd.c.devries at gmail.com>
• Date: Sun, 5 Apr 2009 06:33:46 -0400 (EDT)
• References: <gr4jm9\$ac5\$1@smc.vnet.net>

```Hi Galina,

I'm not sure why your topic title uses the word binomial. It suggests
doesn't show any such coefficients. Furthermore, your series is a bit
irregular. The first few start with x having the highest power, then
decreasing, but F[7] - F[9] do not adhere to that pattern.

Assuming you're mistaken here and F[7] should be x^3, F[8] x^2 y, F[9]
x y^2 and F[10] y^3, F can be given by

F[n_] :=
Module[{t, m},
t = \[LeftFloor]1/2 (-1 + Sqrt[1 + 8 (-1 + n)])\[RightFloor];
m = n - 1 - 1/2 t (1 + t);
x ^(t - m) y^m
]

I leave it as an exercise to you to determine why the variables t and
m have these specific values.

A brute-force approach, but easier to understand, would be

F[i_] := Flatten[Table[x ^(n - r) y^r, {n, 0, i}, {r, 0, n}]][[i]].

It does more calculations than necessary, but I assume i will never
get very high.

Cheers -- Sjoerd

On Apr 3, 11:08 am, Galina <Galina.Pil... at gmail.com> wrote:
> Hello all,
> I need  to find an eigenvalues of the matrix M [N*N] where elements
> are of the type F[i]*F[j]. I need help to construct the elements F[i]
> which must be in the following order: F[1]=1, F[2]=x, F[3]=y, F[4]
> =x^2, F[5]=x*y, F[6]=y^2, F[7]=y*x^2, F[8]=x*y^2, F[9]=x^3 an=
d etc....