[seqfan] Update on triangle related to A050446.

Ed Jeffery lejeffery7 at gmail.com
Mon Dec 19 07:09:30 CET 2011


Seq fans,

First I'd like to wish everyone happy and safe holidays. Second, please
donate to OEIS even if just a small amount.

Third, I am indebted to Chris Gribble for his efforts in writing a Python
program to produce (and send to me) the first 100 rows of my triangle which
I am going to submit soon. In fact, I am working on the text for that
sequence now, and there is at least one other related sequence that I plan
on submitting at the same time.

Neil Sloane authored a sequence A050446 (https://oeis.org/A050446) which in
tabular form begins as

A=
1,1,1,1,1,1,1,...
1,2,3,4,5,6,7,...
1.3.6.10,15,21,28,...
1,5,14,30,55,91,140,...
1,8,31,85,190,371,568,...
etc.

(A050447 produces the same table), and for which I recently described a
related triangle beginning as

T=
1
1,1
1,3,1
1,7,7,1
1,14,31,14,1
etc.

Let T(m,k) denote entry k in row m of my triangle T, where m in {0,1,...}
and 0<=k<=m. Let the rows of table A be indexed by n=1,2,... It is known
that the generating functions for rows n=1 and n=2 of A are F_1(x)=1/(1-x)
and F_2(x)=1/(1-x)^2, respectively. Previously I stated the following

CONJECTURE 1. Let n>=3. Then row n of table A has generating function of
the form

F_n(x)=(T(n-3,0)+T(n-3,1)*x+...+T(n-3,n-3)*x^(n-3))/(1-x)^n,   (end of
conjecture)

This conjecture has now been verified for the first 102 rows of A.
Moreover, thanks to Chris Gribble, I was able to verify for the first 100
rows that T is indeed symmetrical, as I suspected.

Let us now put the triangle in tabular form instead, that is, let

T#=
1,1,1,1,1,...
1,3,7,14,26,...
1,7,31,109,334,...
1,14,109,623,2951,...
1,26,334,2951,20641,...
etc.,
the reason being that problems with offsets can be avoided in the
following. Now row m reads the same as column k when m=k, that is, we take
T# as just an infinite symmetrical matrix: consequently row m has the same
generating function as column k, when m=k. We now focus only on the columns
of table T#.

My search for a pattern in the generating functions for the columns of T#
was motivated by one thing that I noticed, this one thing I was able to
confirm with a reasonable amount of certainty after receiving the triangle
from Chris Gribble, namely, that T#(m+1,k)/T#(m,k) approaches a definite
limit but, for all k in the range I verified, these limits are ones I
happened to recognize from my research in tiling theory. Hence

CONJECTURE 2. Define "quasi-Chebyshev" polynomials Q_r(t) such that

Q_0(t)=1, Q_1(t)=t, and Q_r(t)=t*Q_{r-1}(t)-Q_{r-2}(t)  (r>1).

Let k in {0,1,...} and w_k=2*cos(Pi/(2*k+3)). Then

T#(m+1,k)/T#(m,k) --> Q_k(w_k) as k --> infinity. (end of conjecture)

As it turns out, the (numerical value of the) polynomial Q_k(w_k) equals
the spectral radius of the (k+1) X (k+1) unit-primitive matrix
A_(2*k+3,k+1) (see [
https://oeis.org/wiki/User:L._Edson_Jeffery/UnitPrimitiveMatrix]) with
characteristic polynomial of degree k+1 of the form

(1)   c_(k+1,0)*x^(k+1)+c_(k+1,1)*x^k+...+c_(k+1,k)*x+c_(k+1,k+1),

where c_(k+1,0),...,c_(k+1,k+1) are coefficients given by row k+1 of
triangle A187660 (https://oeis.org/A187660) beginning as

1
1,-1
1,-1,-1
1,-2,-1,1
1,-2,-3,1,1
etc.

Now all of that is related to the following

CONJECTURE 3. Let k in {0,1,...}. Let

p_k(x)=c_(k+1,0)+c_(k+1,1)*x+...+c_(k+1,k+1)*x^(k+1)

be a polynomial as described in the definitions surrounding (1) but with
the exponents taken in reverse order. Column (or row) k of T# has
generating function of the form

G_k(x) = H_k(x) / (p_k(x) * (p_(k-1)(x))^2 * ... * (p_1(x))^k *
(p_0(x))^(k+1)),

where H_k(x) in the numerator is a monic polynomial of degree less than or
equal to the degree of the polynomial in the denominator. (end of
conjecture)

Assuming the conjecture is true, it is obvious that the denominator of
G_k(x) has degree = A000292(k+1) (https://oeis.org/A000292). I stated
Conjecture 3 previously and then retracted it because I thought it must be
wrong. However, it turned out that I had read data from the wrong column of
T#, so naturally my equations didn't work. Although this is still
conjectural, I found that the pattern in the denominators is in fact
consistent for columns k=0,1,2,3,4,5,6, which leads me to believe that the
conjecture must be true after all. The first five of those generating
functions are

G_0(x)=1/(1-x)         in OEIS
G_1(x)=1/((1-x)^2*(1-x-x^2))        in OEIS
G_2(x)=H_2(x)/((1-x)^3*(1-x-x^2)^2*(1-2*x-x^2+x^3))        new sequence
G_3(x)=H_3(x)/((1-x)^4*(1-x-x^2)^3*(1-2*x-x^2+x^3)^2*(1-2*x-3*x^2+x^3+x^4))
new sequence
G_4(x)=H_4(x)/((1-x)^5*(1-x-x^2)^4*(1-2*x-x^2+x^3)^3*(1-2*x-3*x^2+x^3+x^4)^2*(1-3*x-3*x^2+4*x^3+x^4-x^5))
new sequence

Note that the denominators of G_5(x) and G_6(x) are of degrees 56 and 84,
respectively. The coefficients for H_2(x), H_3(x) and H_4(x) are

{1,0,-1,-1,-1,1},

{1,1,-6,-15,21,35,-13,-51,3,21,5,1,-5,-1,1}

and

{1,4,-31,-67,348,418,-1893,-1084,4326,4295,-7680,-9172,9104,11627,-5483,-10773,1108,7255,315,-3085,-228,669,102,-23,-45,-16,11,2,-1},

respectively. H_5(x) and H_6(x) are of degrees 48 and 75, respectively, and
at any rate too long to write out here.

One thing I noticed about the H_k(x) is that the coefficients appear to be
very chaotic with no pattern that I have been able to discover, nor do
those numerators seem to be related in any way. I described the
denominators of G_k(x) as "monster convolutions" in a recent communication
to Neil Sloane.

Finally, I want to also submit the above sequence of convolutions in
tabular form but with H_k(x)=1 for all k. It is also a nice sequence in
which I thought of calling the entries "hyper-binacci" or "hyper-nacci"
numbers. Any thoughts about all of this?

Best regards,

Ed Jeffery



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