qntmpkt: 2010

Wednesday, September 1, 2010

Wednesday, August 25, 2010

invert transform, an introduction

Refer to Aug. 24-th 2010 blog; "INVERT transform"

Tuesday, August 24, 2010

The basic links at OEIS, http://www.tinyurl.com/4zq4q showing the
arxiv article listed - by Bernstein and Sloane.
The operations are presented in this blog, by way of example: find the INVERT
transform of (1, 2, 3,..).
...
1. Given an offset of 1 for S(x) = (x + 2x^2 + 3x^3 + 4x^4 +...); we plug about 6 of such terms into Wolfram alpha with S(x) in the following:
INVERT transform = (1 / (1 - S(x)) - 1.
So we plug into the field: 1/(1 - x - 2x^2 - 3x^3 - 4x^4 - 5x^5), enter.
Then subtract 1 getting (x + 3x^2 + 8x^3 + 21x^4 + 55x^5 + ...) where the coefficients = the even indexed Fibonacci numbers (1, 3, 8, 21,...). Therefore, the INVERT transform of (1, 2, 3,...) = (1, 3, 8, 21, 55,...).
....
The reverse operation = the INVERTi transform and can be defined as
(-1 / (1 + S(x)) + 1; and in an operational mode, we enter into WolframAlpha:
-1 / (1 + x + 3x^2 + 8x^3 + 21x^4 + 55x^5), then enter.
Subtract 1, getting (x + 2x^2 + 3x^3 + 4x^4 + ...). Therefore, the INVERTi
transform of (1, 3, 8, 21,...) = (1, 2, 3,...).
...
A property of INVERT transforms (say, B(x) is the INVERT transform of A(x);
then we preface B(x) with a 1 = (1 + B(x)) . Then A(x) * (1 + B(x)) = B(x).
That is, the INVERT transform shifts to the left when prefaced with a 1 and "convolved" with A(x). Example using A(x) = (1, 2, 3,...) and B(x) = (1, 3, 8,..).
...
Again, using WolframAlpha, we plug in:
(x + 2x^2 + 3x^3 + 4x^4 + 5x^5) * (1 + x + 3x^2 + 8x^3 + 21x^4 + 55x^5 + ...).
getting (x + 3x^2 + 8x^3 + 21x^4).
Thus, (1, 1, 3, 8, 21, 55,....) has shifted to the left to (1, 3, 8, 21,...) when
convolved with (1, 2, 3,..).

Monday, July 5, 2010

Proof of Adamson's conjecture on palindromic continued fractions, re: A179238

Proof of Adamson's conjecture: Gary W. Adamson, re: A179238 in OEIS stating that :

a. given t = irrational Tan A, then if t = the convergent to an infinitely periodic palindromic continued fraction, then Tan 2A is rational; and the converse.

b. Tan 2A is rational when Tan A is rational iff, Tan A = t = the convergent to an infinitely periodic palindromic continued fraction.

Example: barover[a], with a = 1 getting [1,1,1,1,...] = .618....; with the inverse 1.618.... = t (either case) = Tan A.; where 1.618... is irrational.. If t = .618...then Tan 2A = 2.0. If t = 1.618..., then Tan 2A = -2.
...
We will present the sections of the proof then put the separate components together.
Part A: Given t = Tan A, then the double angle 2A triangle hs lets of 2t and t^2-1, with hyptenuse t^2+1.
...
The key step here is the relationship Q = t - 1/t = rational; example 1.618... - .618...= 1.
In order for t to be irrational but Tan 2A to be rational, t - 1/t = Q must be rational (to be shown); but for now continuing with this algrbra,; we have
Q = 1 - 1/t, then Qt = t^2 - 1, and
2t * (QT) = 2t * (t^2 - 1); with 2t / (t^2-1) = 2t / Qt = 2/Q; (and Q is rational).
...
Alternatively, using Tan 2A = 2TanA / (1 - Tan^2 A,) , let t = Tan A, say phi^(-3).
Then Tan 2A = 2*(phi^-3) / (1 - phi^-3) =
2 / (phi^3 - phi^(-3); fulfilling our required identity such that phi^3 - phi^(-3) = 4, since
phi^3 = 4.236...; where .236...= [4,4,4,4,4,4,....] ; palindromic
...Thus, a specific case. of Tan 2A = 2 Tan A / (1 - Tan^2 A), where Tan A = phi^3 or it's inverse.
...
Next, we investigate Part B:
States that t - 1/t must be rational, and looking at the "tail" of an infinitely periodic continued fraction (refer to Wolfram's Mathworld); the "tail" using the formula, and applying to barover [1,2,3] = [1,2,3,1,2,3,1,2,3,...] this is not palindromic and is thus a counter example. The "tail" becomes (2x + 7 / (3x + 10) setting this = to x, getting 3x^2 + 8x - 7, the roots to this being the convergent to barover [1,2,3].
...
We can straightaway obtain the formula 3x^2 + 8x - 7 by writing the partial quotients of [1,2,3] underneath as follows:
1,.......2,........3
1/1 2/3 7/10.
We put 2/3 and 7/10 into a 2x2 format as:
2, 7
3, 10 and perform the following operation such that lower left term (3) = a, upper right = c;
and (lower right - upper left) = b, or f(x), (a,b,c) = ax^2 + bx - c = 0 or
3x^2 + 8x - 7 = 0, and we note that a is not equal to c or 3 is not equal to 7.
...
Next, it follows that in constants of the form used in our conjecture c - 1/c = rational; the corresponding quadratic equation must be such that a must = c.; this being the case, dividing through by (a, c) we obtain an equation of the form x^2 - bx - 1 = 0, where b is rational.
This equation is the characteristic polynomial for an infinitely periodic, palindromic continued fraction, equavalent to the statement in the above 2x2 format that the lower left term must equal the upper right term. In this case 3 is not equal to 7, and the roots are not of the form t - 1/t = rational.
....
Part C. In order for the lower left term to equal the upper right term, we again refer to the "tail" of an infinitely periodic continued fraction which in the case of barover[1,2,3] =
2x + 7 / (3x+10), then setting this to x, we get 3x^2 + 8x - 7 = 0; again 3 is not equal to 7.
We need the tail of the form where upper right and lower left terms are equal, in other words,
in the convergents to [1, 2, 3] we have the partial quotients underneath:
1,.......2,.......7 and
1,.......3,......10. since [1,2,3] = 7/10 and the partial quotient to the left = 2/3.
...
Part D. We refer to Mathworld "Continued fraction", theorem 30 and 31 which state that:
pn / p(n-1) = [an, a(n-1), ....ao] = (30) and
qn / (q(n-1) = [an....a1]................=(31)
...In other words, (30) applies to the reversal of [a,b,c] saying we start with an, then procede to a0. For example, given [1,2,3,4] , theorem 30 uses [4, 3, 2, 1]. Where [1,2,3,4] = 30/43 as follows:
1,.......2,.......7,........30
1,.......3,......10,......43
...then reverse using [4, 3, 2, 1] = obtaining
1,.......3,.......7,.......10
4,......13,.....30,.....43.; so in essence, theorem 30 switches the position of the upper left and lower right terms, with the 7 and 43 terms invariant.
...
Now if the reversal of [a,b,c] remains invariant and thus the upper right and lower left terms are unchanged, this could only occur if the reversal is palindromic, such as [1, 2, 2, 1] where our convergents are
1,....2,......5,......7
1,....3,......7,.....10.
Thus only in a palindromic continued fraction is the lower left term = to the upper right.
Part E.
It follows from the formula of the "tail" say given (using different a,b,c,...):
we set x =( ax + b) / (cx + d); but again changing the usage of our (a,b,c,...) the quadratic formula for the tail if the (upper right = lower left) condition holds must be such that in
ax^2 + bx - c, a must = c. where a,b,c are all rational.
...But only when in this quadratic, a = c does the result hold (dividing through by a = c), we obtain
x^2 - bx - 1 = 0 (with an allowance for changing signs); and if b is rational, then we have our necessary equation with roots t and 1/t such that for example 1.618... - .618 = 1, rational, or
t - 1/t is rational, but t is irrational.
...
Combining all of the steps, Q.E.D. especially noting the key step "upper right = lower left term";
we have shown that in given t irrational, t must be the convergent of an infinitely periodic, palindromic continued fraction in order that angle 2A is rational. Of course, if Tan 2A is rational, t can be rational; but our conjecture only applies to the case in which Tan 2A is rational and t (Tan A) is irrational. The converse of the conjecture is also true.

Thursday, March 25, 2010

Ternary Gray code

Enter "Gray code conversion rules" into Google to bring up the general rules; following rules in A147995 in OEIS: http://www.tinyurl.com/4zq4q
then, following the reflection principle for Gray code k, we write out (say k=3) 0,1,2; 2,0,1; 1,2,0; ...in the right column; i.e. we write each term in the series once, then for the next subset of 3, we repeat the last term recorded, or "reflect" it.
...0
...1
...2;
now reflect, starting the series again (2, then 0, 1,...):
...2,
...0,
...1;
now for the next subset we reflect the "1" getting
...1
...2 (continuing with 0,1,2, etc.)
Now for the next column going to the left, we mark down each term 3 times; and the next column after that (column 3), 9 times, so that each term in the series 0,1,2 (and generally, 0,1,2,3,...for any k) is marked down k^0, k^1, k^2,....times along with the reflection principle. Thus, for Ternary Gray code we obtain:
000
001
002
012
010
011
021
022
020
120
121
122
102
101
100
110
111
112
...

Gray code conversion rules

at http://www.tinyurl.com/4zq4q
then access sequence A147995
The reflection principle is as follows, by way of example, ternary Gray code :
First column, going right to left, within each set of 3 terms starting with 0, write 0,1,2, then the next set repeats the 2 (reflection) then continues with the series 0,1,2. Each set reflects, then continues in the series 0->1->2, so we obtain
0,1,2; 2,0,1; 1,2,0; 0,1,2;...; (but these are column 1 terms.). Column 2 terms each term is marked down 3 times: 0,0,0; then 1,1,1; 2,2,2; then reflect: 2,2,2;...
For column 3, each term is written down 9 times along with the reflection principle as before. Generally for Gray code K, each term is recorded K^0, K^1, K^2...times by columns. Putting the rules together for k=3, Ternary, we
obtain (for decimal 0, 1, 2, 3,....):
000
001
002
012
010
011
021
022
020
120
121
122
102
101
100
110
111
112
...
and, we note that the column change from n-th to (n+1)-th term =
A051064 in OEIS: (1, 1, 2, 1, 1, 2, 1, 1, 3, 1, 1, 2, 1, ...) representing
the row (labeled 1,2,3,...) in the following multiplication table:
(heading terms not multiples of 3, * left column = powers of 3):
1,...2,...4,....5,....7,...8,...10...
3,..6..,12,..20, 21,.24,..30,..
9,..;
Then record the row that n=1,2,3,...occurs in, getting A051064 as before:
(1, 1, 2, 1, 1, 2, 1, 1, 3,...) = the ruler sequence for k=3.

Wednesday, March 17, 2010

Genomic matrices

Generalized Genomic Matrices, Silver Means, and Pythagorean Triples"> > is now published on the web;> http://www.scipress.org/journals/forma/frame/24.html>

partitions and the formal power series

But to review, you have given p(x) = polcoeff partition series = A(x)/A(x^2), = B(x)/ B(x^3)....; such that A(x), B(x),...etc for any k = Euler transform of the generalized ruler sequences ("p-adic valuations of kn"). Below we have an important algebraic operation and identity.It states that given A(x), B(x), etc,....for any k = 2, 3,...; A(x) = p(x) * p(x^2) * p(x^4) * p(x^8) * p(x^16) *....; andB(x) = p(x) * p(x^3) * p(x^9) * p(x^27) * p(x^81) * ...etc. Gary
Date: Wed, 17 Mar 2010 17:16:59 -0700From: qntmpkt@yahoo.comSubject: Fw: [seqfan] Re: need algebra proofTo: qntmpkta@gmail.com; qntmpkta@hotmail.com
Conjecture:> Given polcoeff S(x), by way of example given A000041, the partition sequence = (1, 1, 2, 3, 5, 7, 11, 15, 22,...); then> S(x) = (1 + x + 2x^2 + 3x^3 + ...).>> Then S(x) = some A(x) such that S(x) = A(x) / A(x^2).
If you replace the somehow obscure " S(x) = some A(x) " (?)by "there exists A(x)", then this is indeed true,given the hypothesis that S(0) = 1.(Else such an A cannnot exist, because A(0) / A(0) = 1 (or ill definedif A(0)=0).)
> Our conjecture states that A(x) = S(x) * S(x^2) * S(x^4) * S(x^8) * S(x^16) * ...).
Obviously, A(x) is not uniquely defined, so you should say "onesolution is given by ...".The choice of this solution is equivalent to the assumption (or choice) A(0)=1.You could multiply this A by any nonzero constant and you would againget the same A(x) / A(x^2).
The infinite product converges in the topology of formal power series(which means that the coefficients up to a given orderare only affected by a finite number of terms, cf.http://en.wikipedia.org/wiki/Formal_power_series )so it is perfectly well defined.To exclude any ambiguity, you can specifythat the expression ... / A(x^2)means ... * ( A(x^2) )^(-1) ,where ( 1 - T(x) )^(-1) = 1 + T(x) + T(x)^2 + ...(where T(x) is any power series with T(0)=0).
So you can check that the initial formula is true(which means : coefficients of any given power are the sameon the l.h.s. and on the r.h.s.)by writing out the relevant factors of the infinite productsand cancelling those which correspond.
> Similarly, for S(x) = B(x) / B(x^3), our conjecture states that:>> B(x) = S(x) * S(x^3) * S(x^9) * S(x^27) * S(x^81) * ...>> Generally, for any k, in S(x) = C(x) / C(x^k).>
Yes, this is the same, the only necessary condition is again that S(0)= 1, and assuming the choice B(0) = C(0) = 1.
Maximilian
PS: I think Richard Mathar already pointed out roughly the same thing

Partitions and the formal power series

https://mail.google.com/mail/?shva=1#inbox/1276eab0f83c07c3

Friday, March 12, 2010

Hogan's quantum noise

at http://astro.fnal.gov/people/Hogan

Monday, March 1, 2010

Gary W. Adamson articles

https://mail.google.com/mail/?shva=1#inbox

Sunday, February 28, 2010

Articles by Gary W. Adamson, mathematician

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

Friday, February 26, 2010

Quantum physics and the Cartan matrix

From:
info unit (qntmpkt@hotmail.com)
Sent:
Wed 8/26/09 2:47 PM
To: Gary Adamson (qntmpkt@gmail.com); qntmpkt@hotmail.com
1 attachment
JCP II(09...pdf (831.3 KB)

Thursday, February 25, 2010

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
To:
info unit (qntmpkt@hotmail.com)
Cc:
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)
1 attachment
adamson.pdf (71.2 KB)

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
To:
info unit (qntmpkt@hotmail.com)
Cc:
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)
1 attachment
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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

Wednesday, February 24, 2010

Fractal structure of partitions of n

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

Tuesday, February 23, 2010

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

Friday, January 22, 2010

Petoukhov matrices

article at:
http://www.scipress.org/journals/forma/frame/24.html It is in No. 2.

Tuesday, January 12, 2010

Petoukhov matrices - article

check out new article:
http://www.scipress.org/journals/forma/frame/24.html It is in No. 2.

Saturday, January 9, 2010

toothpick sequences

New article:
http://www.scipress.org/journals/forma/frame/24.html
/Then click on to No.2

Date: Thu, 7 Jan 2010 20:23:26 -0800
From: qntmpkt@yahoo.com
Subject: Toothpick / Tetrahedral "closed walks" - proof - Gary

proof of the generalization should be within each reach, where we state that:

...

Theorem: (toothpick sequence )/ ("tends to" sequence of triangle rows) = corresponding "closed walks of length n along edges of (tetrahedron, etc) based at vertex.

First, by way of example, let's cover the left side of the equation.

Given our triangle A160552:

1, 3

1, 3, 5, 7

1, 3, 5, 7, 5, 11 17, 15;

...

where rows tend to A151548: (1, 3, 5, 7, 5, 11 17, 15, 5, 11, 17, 15, 5, 11, 17, 19,...).

Already in your comments section, we have triangle (A160552) as a string * (1, 2, 2, 2, ....) = toothpick, sequence A139250. But

(A160552) * (1, 2, 0, 0, 0, ...) = A151548., the "tends to" sequence.

So we put

A160552 * (1, 2, 2, 2,...) / A160552*(1, 2, 0, 0, 0,...); = (toothpick / "tends to"); then the A160552 cancels., leaving

(1, 2, 2, 2,...) / (1, 2, 0, 0, 0,...).

...

Now we cover the right side of the equation by listing some examples. Case N=1 turns out to be (1, 0, 1, 0, 1, 0, 1,...) trivial.

So next we address A151575: (1, 0, 2, -2, 6, -10, 22, -42, 86,...). But first we access the unsigned version:

(1, 0, 2, 2, 6, 10, 22, 42, 86,....); noting that the g.f. is

(1-x) / (1 - x - 2x^2).

It can be shown that the signed version =

(1, 2, 2, 2,....) / (1, 2, 0, 0, 0,...).; so the previous formula should be tweaked to reflect the alternating signs.

But (1, 2, 2, 2,...) / (1, 2, 0, 0,....) is what we had before on the left side of the formula. Proof is complete for this case N=2.

...

Next, case N=3: Right side of the formula is A054878: (1, 0, 3, 6, 21, 60 183, 546, 1641,...) but with alternating signs.

A054878 g.f. is (1 - 2x)/ (1-2x-3x^2) where from the previous g.f. we just add "1" to coeff of x's. it can be shown that the alternating sign version of A054878 = (1, 3, 3, 3,....) / (1, 3, 0, 0, 0); where again this = (toothpick seq N=3) / ("tends to" sequence); where the generating triangle for N=3 =

1, 4

1, 4, 7, 13;

1, 4, 7, 13, 7, 19, 34, 40;

... again left side of equation = right side.

....

Next, case N=4, right side of equation = A109499: (1, 0, 4, 12, 52, 204, 820, 3276,...) where we just add "1" to coeff of x (but the entry has a slightly different formula. Anyway, the same formatted formula would be (1 - 3x) / (1 - 3x - 4x^2). But we want the version with alternate signs, so that formula can be tweaked to equate to:

(1, 4, 4, 4,....) / (1, 4, 0, 0, 0).; proof should follow.

Next, case N=5 = A109500: (1, 0, 5, 20, 105, 520, 2605, 13020,...); and the analogous formula should be:

(1 - 4x) / (1 - 4x - 5x^2); with the alternating sign version =

(1, 5, 5, 5,...) / (1, 5, 0, 0, 0,...). proof should follow.

...

...but I'm not into proofs so you can have this project, if desired.

...Gary

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