cp's OEIS Frontend

Row sums are Fibonacci(n+2). Diagonal sums are A016116. - Paul Barry, Jul 07 2004

Riordan array (1/(1-x), x/(1-x^2)). Matrix inverse is A106180. - Paul Barry, Apr 24 2005

As an infinite lower triangular matrix * [1,2,3,...] = A055244. - Gary W. Adamson, Dec 23 2008

From Emeric Deutsch, Jun 18 2010: (Start)

T(n,k) is the number of alternating parity increasing subsequences of {1,2,...,n} of size k, starting with an odd number (Terquem's problem, see the Riordan reference, p. 17). Example: T(8,5)=6 because we have 12345, 12347, 12367, 12567, 14567, and 34567.

T(n,k) is the number of alternating parity increasing subsequences of {1,2,...,n,n+1} of size k, starting with an even number. Example: T(7,4)=5 because we have 2345, 2347, 2367, 2567, and 4567. (End)

From L. Edson Jeffery, Mar 01 2011: (Start)

This triangle can be constructed as follows. Interlace two copies of the table of binomial coefficients to get the preliminary table

1

1

1 1

1 1

1 2 1

1 2 1

1 3 3 1

1 3 3 1

...,

then shift each entire r-th column up r rows, r=0,1,2,.... Also, a signed version of this sequence (A187660 in tabular form) begins with

1;

1, -1;

1, -1, -1;

1, -2, -1, 1;

1, -2, -3, 1, 1;

...

(compare with A066170, A130777). Let T(N,k) denote the k-th entry in row N of the signed table. Then, for N>1, row N gives the coefficients of the characteristic function p_N(x) = Sum_{k=0..N} T(N,k)*x^(N-k) = 0 of the N X N matrix U_N=[(0 ... 0 1);(0 ... 0 1 1);...;(0 1 ... 1);(1 ... 1)]. Now let Q_r(t) be a polynomial with recurrence relation Q_r(t)=t*Q_(r-1)(t)-Q_(r-2)(t) (r>1), with Q_0(t)=1 and Q_1(t)=t. Then p_N(x)=0 has solutions Q_(N-1)(phi_j), where phi_j=2*(-1)^(j-1)*cos(j*Pi/(2*N+1)), j=1,2,...,N.

For example, row N=3 is {1,-2,-1,1}, giving the coefficients of the characteristic function p_3(x) = x^3-2*x^2-x+1 = 0 for the 3 X 3 matrix U_3=[(0 0 1);(0 1 1);(1 1 1)], with eigenvalues Q_2(phi_j)=[2*(-1)^(j-1)*cos(j*Pi/7)]^2-1, j=1,2,3. (End)

Given the signed polynomials (+--++--,...) of the triangle, the largest root of the n-th row polynomial is the longest (2n+1) regular polygon diagonal length, with edge = 1. Example: the largest root to x^3 - 2x^2 - x + 1 = 0 is 2.24697...; the longest heptagon diagonal, sin(3*Pi/7)/sin(Pi/7). - Gary W. Adamson, Sep 06 2011

Given the signed polynomials from Gary W. Adamson's comment, the largest root of the n-th polynomial also equals the length from the center to a corner (vertex) of a regular 2*(2*n+1)-sided polygon with side (edge) length = 1. - L. Edson Jeffery, Jan 01 2012

Put f(x,0) = 1 and f(x,n) = x + 1/f(x,n-1). Then f(x,n) = u(x,n)/v(x,n), where u(x,n) and v(x,n) are polynomials. The flattened triangles of coefficients of u and v are both essentially A046854, as indicated by the Mathematica program headed "Polynomials". - Clark Kimberling, Oct 12 2014

From Jeremy Dover, Jun 07 2016: (Start)

T(n,k) is the number of binary strings of length n+1 starting with 0 that have exactly k pairs of consecutive 0's and no pairs of consecutive 1's.

T(n,k) is the number of binary strings of length n+2 starting with 1 that have exactly k pairs of consecutive 0's and no pairs of consecutive 1's. (End)

Matrix inverse of A124645.

Let L(n,x) = Sum_{k=0..n} T(n,k)*x^(n-k) and Pi=3.14...:

L(n,x) = Product_{k=1..n} (x - 2*cos((2*k-1)*Pi/(2*n+1)));

Sum_{k=0..n} T(n,k) = L(n,1) = A010892(n+1);

Sum_{k=0..n} abs(T(n,k)) = A000045(n+2);

abs(T(n,k)) = A065941(n,k), T(n,k) = A065941(n,k)*A087960(k);

T(2*n,k) + T(2*n+1,k+1) = 0 for 0 <= k <= 2*n;

T(n,0) = A000012(n) = 1; T(n,1) = -1 for n > 0;

T(n,2) = -(n-1) for n > 1; T(n,3) = A000027(n)=n for n > 2;

T(n,4) = A000217(n-3) for n > 3; T(n,5) = -A000217(n-4) for n > 4;

T(n,6) = -A000292(n-5) for n > 5; T(n,7) = A000292(n-6) for n > 6;

T(n,n-3) = A058187(n-3)*(-1)^floor(n/2) for n > 2;

T(n,n-2) = A008805(n-2)*(-1)^floor((n+1)/2) for n > 1;

T(n,n-1) = A008619(n-1)*(-1)^floor(n/2) for n > 0;

T(n,n) = L(n,0) = (-1)^floor((n+1)/2);

L(n,1) = A010892(n+1); L(n,-1) = A061347(n+2);

L(n,2) = 1; L(n,-2) = A005408(n)*(-1)^n;

L(n,3) = A001519(n); L(n,-3) = A002878(n)*(-1)^n;

L(n,4) = A001835(n+1); L(n,-4) = A001834(n)*(-1)^n;

L(n,5) = A004253(n); L(n,-5) = A030221(n)*(-1)^n;

L(n,6) = A001653(n); L(n,-6) = A002315(n)*(-1)^n;

L(n,7) = A049685(n); L(n,-7) = A033890(n)*(-1)^n;

L(n,8) = A070997(n); L(n,-8) = A057080(n)*(-1)^n;

L(n,9) = A070998(n); L(n,-9) = A057081(n)*(-1)^n;

L(n,10) = A072256(n+1); L(n,-10) = A054320(n)*(-1)^n;

L(n,11) = A078922(n+1); L(n,-11) = A097783(n)*(-1)^n;

L(n,12) = A077417(n); L(n,-12) = A077416(n)*(-1)^n;

L(n,13) = A085260(n);

L(n,14) = A001570(n); L(n,-14) = A028230(n)*(-1)^n;

L(n,n) = A108366(n); L(n,-n) = A108367(n).

Row n of the matrix inverse (A124645) has g.f.: x^floor(n/2)*(1-x)^(n-floor(n/2)). - Paul D. Hanna, Jun 12 2005

From L. Edson Jeffery, Mar 12 2011: (Start)

Conjecture: Let N=2*n+1, with n > 2. Then T(n,k) (0 <= k <= n) gives the k-th coefficient in the characteristic function p_N(x)=0, of degree n in x, for the n X n tridiagonal unit-primitive matrix G_N (see [Jeffery]) of the form

G_N=A_{N,1}=

(0 1 0 ... 0)

(1 0 1 0 ... 0)

(0 1 0 1 0 ... 0)

...

(0 ... 0 1 0 1)

(0 ... 0 1 1),

with solutions phi_j = 2*cos((2*j-1)*Pi/N), j=1,2,...,n. For example, for n=3,

G_7=A_{7,1}=

(0 1 0)

(1 0 1)

(0 1 1).

We have {T(3,k)}=(1,-1,-2,1), while the characteristic function of G_7 is p(x) = x^3-x^2-2*x+1 = 0, with solutions phi_j = 2*cos((2*j-1)*Pi/7), j=1,2,3. (End)

The triangle sums, see A180662 for their definitions, link A108299 with several sequences, see the crossrefs. - Johannes W. Meijer, Aug 08 2011

The roots to the polynomials are chaotic using iterates of the operation (x^2 - 2), with cycle lengths L and initial seeds returning to the same term or (-1)* the seed. Periodic cycle lengths L are shown in A003558 such that for the polynomial represented by row r, the cycle length L is A003558(r-1). The matrices corresponding to the rows as characteristic polynomials are likewise chaotic [cf. Kappraff et al., 2005] with the same cycle lengths but substituting 2*I for the "2" in (x^2 - 2), where I = the Identity matrix. For example, the roots to x^3 - x^2 - 2x + 1 = 0 are 1.801937..., -1.246979..., and 0.445041... With 1.801937... as the initial seed and using (x^2 - 2), we obtain the 3-period trajectory of 8.801937... -> 1.246979... -> -0.445041... (returning to -1.801937...). We note that A003558(2) = 3. The corresponding matrix M is: [0,1,0; 1,0,1; 0,1,1,]. Using seed M with (x^2 - 2*I), we obtain the 3-period with the cycle completed at (-1)*M. - Gary W. Adamson, Feb 07 2012

The original name of this sequence was: Triangle giving coefficients of characteristic function of n X n matrix in which the left upper half and the antidiagonal are filled with 1's and the right lower half is filled with 0's. As was pointed out by L. Edson Jeffery this is only correct if we multiply each triangle row by (-1)^n. For the straightforward version of the coefficients of the characteristic polynomials see A187660. - Johannes W. Meijer, Aug 08 2011

From Kyle Petersen, Jun 02 2024: (Start)

Coefficients of the "P-Eulerian" polynomial of a naturally labeled zig-zag poset, which counts linear extensions according to number of descents. T(n, k) is the number of linear extensions of the n-element zig-zag poset with k descents.

Also T(n, k) is the number of up-down permutations of length n with k "big returns". A big return is a pair (i, i+1) for which i appears more than one place to the right of i+1 in the permutation. This interpretation implies row sums are given by A000111. (End)

From L. Edson Jeffery, Jan 27 2012: (Start)

(Previous name:) Omitting the first two ones, a rectangular array M read by antidiagonals in which entry M_{n-k, k} in row n-k and column k, 0 <= k <= n, gives the coefficient of x^k in the numerator of the conjectured generating function for row n + 3 of the tabular form of A050446.

In the following, let M_{n, k} denote the entry in row n and column k of M, n, k in {0, 1, ...}.

Conjecture: 1. M_{n, k} = M_{k, n}, for all n and k; that is, M is symmetric about the central terms {1, 3, 31, 623,...}. (This has been verified for the first 100 antidiagonals of M.)

Conjecture: 2. For m in {3, 4,...}, row m of array A050446 has generating function of the form H_m(x)/(1 - x)^m, in which the numerator H_m(x) is a polynomial of degree m - 3 in x with coefficients given by the entries of the (m - 3)-th antidiagonal of M containing the sequence of entries {M_{m-3-j,j}}, j=0..m-3 (see the example below). It is known that H_1(x) = H_2(x) = 1.

Conjecture: 3. Define the Chebyshev polynomials of the second kind by U_0(t) = 1, U_1(t) = 2*t and U_r(t) = 2*t*U_(r-1)(t) - U_(r-2)(t) (r > 1). Assuming Conjecture 1, lim_{n -> infinity} M_{n+1, k}/M_{n, k} = U_k(cos(Pi/(2*k+3))) = spectral radius of the (k+1) X (k+1) unit-primitive matrix (see [Jeffery]) A_{2*k+3, k} = [0,...,0,1; 0,...,0,1,1; ...; 0,1,...,1; 1,...,1], with identical limits for the columns of the transpose M^T of M.

Conjecture: 4. Let S(u, v) denote the entry in row u and column v of triangle S = A187660, 0 <= v <= u. Define the polynomials P_u(x) = Sum[S(u, v)*x^v]. Assuming Conjecture 1, then (i) the generating function for row (or column) n of M is of the form

G_n(x)/((P_1(x))^(n+1) * (P_2(x))^n * ... * (P_n(x))^2 * P_(n+1)(x)),

in which (ii) the numerator G_n(x) is a polynomial of degree A005586(n), and (iii) the denominator is a polynomial of degree A000292(n+1).

Remarks: The coefficients in the numerators G_n(x) appear to have no pattern. The polynomial P_j(x), j in {1,...,n+1}, of Conjecture 4 is also obtained from the characteristic polynomial of the unit-primitive matrix A_{2*j+3,j} of Conjecture 3 by taking the exponents of the latter in reverse order; and P_j(x) is otherwise identical to the characteristic polynomial of the unit-primitive matrix A_{2*j+3,1}.

(End)

Conjecture: The Eulerian zig-zag polynomials have only negative and simple real roots and form a Sturm sequence, that is, p(n+1, x) has n real roots separated by the roots of p(n, x). This property was proved by Frobenius for the classical Eulerian polynomials. - Peter Luschny, Jun 04 2024

The entry a(n,k), n >= 0, k = 0,1,...,g, where g=n, of this irregular triangle is the coefficient of (1+q^2)^k*q^(g-k) in the representation of the Gaussian polynomial [2n+1,1]q = Sum{k=0..g) a(n,k)*(1+q^2)^k*q^(g-k).

The sequence arises in the formal derivation of the stability polynomial B(x) = Sum_{i=0..N} d_i T(iM,x) of rank N, and degree L, where T(iM,x) denotes the Chebyshev polynomial of the first kind of degree iM. The coefficients d_i are determined by order conditions on the stability polynomial.

Conjecture: More generally, the Gaussian polynomial [2*n+m+1-(m mod 2),m]q = Sum{k=0..g(m;n)} a(m;n,k)*(1+q^2)^k*q^(g(m;n)-k), for m >= 0, n >= 0, where g(m;n) = m*n if m is odd and (2*n+1)*m/2 if m is even, and the tabf array entries a(m;n,k) are the coefficients of the g.f. for the row n polynomials G(m;n,x) = (d^m/dt^m)G(m;n,t,x)/m!|{t=0}, with G(m;n,t,x) = (1+t)*Product{k=1..n+(m - m (mod 2))/2}(1 + t^2 + 2*t*T(k,x/2) (Chebyshev's T-polynomials). Hence a(m;n,k) = [x^k]G(m;n,x), for k=0..g(m;n). The present entry is the instance m = 2. (Thanks to Wolfdieter Lang for clarifying the text on the general prescription of a(m;n,k).)

Signed version of A046854, A130777.

Conjecture: row n is U(n, x/2) + U(n-1, x/2) where U is the sequence of Chebyshev polynomials of the second kind. - Thomas Baruchel, Jun 03 2018 [For a proof see the following comment.]

From Wolfdieter Lang, Oct 19 2019: (Start)

The row polynomial R(n, x) = Sum_{k=0..n} a(n, k)*x^k = [2*n+1]_q / q^n with the q-number [2*n+1]_q := (1 - q^n)/(1 - q), which for q = 1 becomes 2*n+1, and x = x(q) = q + q^(-1). See the simplified Name and the first comment. In terms of Chebyshev S polynomials (A049310) this q-number is written as [2*n+1]_q = q^n*S(2*n, q^(1/2) + q^(-1/2)), hence R(n, x) = S(2*n, sqrt(2+x)) = S(n, x) + S(n-1, x) (which proves the conjecture of the previous comment).

For the o.g.f. of R(n, x) see the formula section.

My motivation for looking at this sequence came from the Brändli and Beyne paper's recurrence for the polynomial P_m(s) which coincides with R(n, x), with m -> n and s -> x. (End)

A294099(n, k) = Sum_{j=0..k} n^j * T(n, j) for all n, k in Z. - Michael Somos, Jun 19 2023

For definition of Danzer matrix see [Jeffery] (notation differs there!).

Conjecture 1. Let F_n(x)=sum_{j=0..n} A187660(n,j)*x^{(n-1)*j}. Let f_n in Z[x] be any polynomial in x of degree d such that 0<=d<=(n-1)*(n-2). Then the sequence of coefficients of the series expansion of f_n(x)/F_n(x), when taken over Z/kZ, is periodic with period p <= (n-1)*A220555(n,k), for all n,k > 1. (Cf. [Coleman, et al.] for the case for n=2 (generalized Fibonacci).)

Conjecture 2. If G a cyclic multiplicative group generated by an n X n integer matrix over Z/kZ, then |G|<=T(r,k), for some r<=n.

Definition. If T(n,k)>=(k^n-1)/(k-1), for some k>1, then T(n,k) is said to be "optimal."

Conjecture 3. If T(n,k) is optimal, then n is a Queneau number (A054639).

Sequence is read from antidiagonals of array T which begins as

.1...1....1....1......1.......1......1....1.....1.........1

.1...3....8....6.....20......24.....16...12....24........60

.1...7...26...14.....62.....182.....42...28....78.......434

.1...7...18...14.....62.....126....114...28....54.......434

.1..31..121...62....781....3751...2801..124...363.....24211

.1..63...26..126.....24....1638..13072..252....78.......504

.1..15...24...30.....20.....120....400...60....72........60

.1..15.1640...30..32552....4920.240200...60..4920....488280

.1.511.9841.1022.488281.5028751....342.2044.29523.249511591

.1..63...78..126....124....1638.....42..252...234......7812

Rows might be related to Jordan totient functions J_n(k), however, some entries T(n,k) are products of factors of the form (j^n-1)/(j-1).