Tuesday, February 23, 2010

shifted down matrices

Below: question posed, reply received by Richard Mathar; the operation being, given an infinite lower triangular matrix with leftmost col offset 0, then for col>0, shifted each down k times, k>1.
Given the original sequence S(n), say = (1, 2, 3,..); then shifted down twice we get
1,
2,
3, 1,
4, 2,
...and so = M. Then take Lim_{n->inf} M^n, shifting a vector to the left, considered as a sequence. Let that seq = A(x) = (1 + ax + bx^2 + ...); then
S(x) = A(x) / A(x^2).
Then shifted down thrice, we obtain sequence (a different sequence A(x)):
S(x) = A(x) / A(x^3)
...and so on.
(C.F. http://www.tinyurl.com/yzw9hrd)
Suggested proof below, sent by Richard Mathar. Relating to A173279 in OEIS
at http://www.tinyurl.com/4zq4q
ga> From qntmpkt@hotmail.com Mon Feb 22 21:26:06 2010ga> Return-Path: ga> From: info unit ga> To: , Maximilian Haslerga> , Gary Adamson ga> Subject: proof needed on "shift down"ga> Date: Mon, 22 Feb 2010 12:26:03 -0800ga> ga> Richard:=20ga> ga> No need to reply to this=2C but if a proof pops into your head spontaneousl=ga> y=3B it might be a helpful comment addition:ga> ga> Given a triangle with offset 0=2C S(n) in every column but shifted down twi=ga> ce for col. >0=3Bga> ga> Let the triangle =3D T=2C thenga> ga> Lim_{n->inf} T^n =3D a polcoeff sequence A(x) (left-shifted vector conside=ga> red as a seq): such thatga> ga> =20ga> ga> S(n) =3D A(x) / A(x^2).ga> ga> =20ga> ga> Then (conjecture): shifted down thrice =3Dga> ga> S(n) =3D A(x) / A(x^3)=3B You have to re-do what I posted for the 2nd power (r=2) inhttp://list.seqfan.eu/pipermail/seqfan/2010-February/003725.html :Write S(n)*A(x^r) = A(x), compare equal powers on both sidesto give without any further complication a recurrence of theform a(n) = sum_{t=0..n, n-t = .... mod r} S(t) *a_{(n-t)/r}and this is basically (because it applies then again to thea_{(n-t)/r} on the right hand sides) the kind of matrix multiplicationset up by these matrices with the rows pulled in chunks of r.I suspect with a little bit care on getting the index limits right

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