Articles by Gary W. Adamson, mathematician

at http://www.tinyurl.com/yk6ewox

Quantum physics and the Cartan matrix

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To: Gary Adamson (qntmpkt@gmail.com); qntmpkt@hotmail.com
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JCP II(09...pdf (831.3 KB)

Binomial transform

First, the finite difference method, as shown in several websites; but I prefer not to use sequences starting with 0.By way of example, taking finite differences of (1, 4, 9, 16, 25,...) we get:1,.....4,.....9,.....16,.....25,.....36,.........3......5.......7.......9.......11...................2.......2......2.......2...................(when bottom row has all k,k,k,k,...the operation ends.). This implies that the Binomial transform of [1, 3, 2, 0, 0, 0,...] = (1, 4, 9, 16, 25, 36,...); and that theinverse Binomial transform of [1, 4, 9, 16, 25, 36,...] = (1, 3, 2, 0, 0, 0,...)....Pascal's triangle method. (C.f. A007318 in OEIS, http://www.tinyurl.com/4zq4qUsing a pocket calculator or online calculator we plug in rows of Pascal's triangle:11, 11, 2, 11, 3, 3, 11, 4, 6, 4, 1...preferably as many rows as possible. I use 17 rows. Then in the VECTOR section, plug in [1, 3, 2, 0, 0, 0,...], 17 terms. Let the triangle = P and the vector = V. Then simply multiply, P * V, getting (1, 4, 9, 16, 25, 36, 49,...); i.e. the vector in brackets, and (1, 4, 9, 16,...) in parentheses meaning: vector considered as a sequence; whereas Pascal's triangle A007318 is used as an infinite lower triangular matrix.Now for the inverse b.t.: Given the vector [1, 4, 9, 16,...] (17 terms); we have our Pascal's triangle already stored, so first invert it, getting:1-1, 11, -2, 1-1, 3, -3, 1...So, this matrix = M, with [1, 4, 9, 16,...] = V. Then take M * V, getting [1, 3, 2, 0, 0, 0,...] meaning that the inverse b.t. of [1, 4, 9, 16, 25,...] = [1, 3, 2, 0, 0, 0,...]....QUIZ: what's the b.t. of [1, 2, 3,...]. We plug in [1, 2, 3, 4, 5, 6,...] (17 terms) into the vector section and multiply it by our stored Pascal's triangle getting(1, 3, 8, 20, 48, 112, 256,...). Let's go in reverse, using the finite difference method:1,...3,...8,...20,...48,...112,.....2,....5,...12,...28,.....................3,....7,....16,............................4,....9,....................................5,............................But our bottom rows don't terminate in a row with k,k,k,k,...; but no matter.We can see by inspection that given enough rows, the inverse binomial transform of [1, 3, 8, 20, 48,...] is shaping up to be (1, 2, 3, 4, 5,...). But naturally, the more rows we have to deal with, the more certain is our conclusion..
Posted by Yifu Xero at 8:38 PM

Binomial transform

First, the finite difference method, as shown in several websites; but I prefer not to use sequences starting with 0.
By way of example, taking finite differences of (1, 4, 9, 16, 25,...) we get:
1,.....4,.....9,.....16,.....25,.....36,.....
....3......5.......7.......9.......11...........
........2.......2......2.......2................
...
(when bottom row has all k,k,k,k,...the operation ends.). This implies that the Binomial transform of [1, 3, 2, 0, 0, 0,...] = (1, 4, 9, 16, 25, 36,...); and that the
inverse Binomial transform of [1, 4, 9, 16, 25, 36,...] = (1, 3, 2, 0, 0, 0,...).
...
Pascal's triangle method. (C.f. A007318 in OEIS, http://www.tinyurl.com/4zq4q
Using a pocket calculator or online calculator we plug in rows of Pascal's triangle:
1
1, 1
1, 2, 1
1, 3, 3, 1
1, 4, 6, 4, 1
...preferably as many rows as possible. I use 17 rows. Then in the VECTOR section, plug in [1, 3, 2, 0, 0, 0,...], 17 terms. Let the triangle = P and the vector = V. Then simply multiply, P * V, getting (1, 4, 9, 16, 25, 36, 49,...); i.e. the vector in brackets, and (1, 4, 9, 16,...) in parentheses meaning: vector considered as a sequence; whereas Pascal's triangle A007318 is used as an infinite lower triangular matrix.
Now for the inverse b.t.: Given the vector [1, 4, 9, 16,...] (17 terms); we have our Pascal's triangle already stored, so first invert it, getting:
1
-1, 1
1, -2, 1
-1, 3, -3, 1
...
So, this matrix = M, with [1, 4, 9, 16,...] = V. Then take M * V, getting [1, 3, 2, 0, 0, 0,...] meaning that the inverse b.t. of [1, 4, 9, 16, 25,...] = [1, 3, 2, 0, 0, 0,...].
...
QUIZ: what's the b.t. of [1, 2, 3,...]. We plug in [1, 2, 3, 4, 5, 6,...] (17 terms) into the vector section and multiply it by our stored Pascal's triangle getting
(1, 3, 8, 20, 48, 112, 256,...). Let's go in reverse, using the finite difference method:
1,...3,...8,...20,...48,...112,...
..2,....5,...12,...28,...............
......3,....7,....16,..................
..........4,....9,......................
..............5,.........................
...But our bottom rows don't terminate in a row with k,k,k,k,...; but no matter.We can see by inspection that given enough rows, the inverse binomial transform of [1, 3, 8, 20, 48,...] is shaping up to be (1, 2, 3, 4, 5,...). But naturally, the more rows we have to deal with, the more certain is our conclusion..

Prodinger proof of Adamson's conjecture on partitions

Second conjecture in A173238, http://www.tinyurl.com/4zq4q
at OEIS:
solution‏
From:
Prodinger, Helmut (hproding@sun.ac.za)
Sent:
Thu 2/18/10 4:33 AM
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jackson@math.sjsu.edu (jackson@math.sjsu.edu); ruskey@cs.uvic.ca (ruskey@cs.uvic.ca); qntmpkt@gmail.com (qntmpkt@gmail.com); qntmpkt@hotmail.com (qntmpkt@hotmail.com)
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Adamson's conjectures on partitions

in Sequence A173238, http://www.tinyurl.com/4zq4q
Proof by Helmut Prodinger for conjecture #2
solution‏
From:
Prodinger, Helmut (hproding@sun.ac.za)
Sent:
Thu 2/18/10 4:33 AM
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jackson@math.sjsu.edu (jackson@math.sjsu.edu); ruskey@cs.uvic.ca (ruskey@cs.uvic.ca); qntmpkt@gmail.com (qntmpkt@gmail.com); qntmpkt@hotmail.com (qntmpkt@hotmail.com)
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Partitions of n and the ruler function

http://qntmpkt.blogspot.com/2010/02/partitions-and-ruler-function-tower-of.html

Partitions and the ruler function (Tower of Hanoi series)

Conjecture #2 in A173238 at OEIS, http://www.tinyurl.com/4zq4q
with proof submitted by Dr. Helmut Prodinger, 02/28/10:
http://www.tinyurl.com/y9z2a58

Fractal structure of partitions of n

At OEIS, http://www.tinyurl.com/4zq4q
then enter sequence A173301

A173279 and A173280

A173279 and A173280,
http://www.tinyurl.com/y9z2a58

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