cp's OEIS Frontend

Any sequence b(n) satisfying the recurrence b(n) = b(n-1) - b(n-2) can be written as b(n) = b(0)*a(n) + (b(1)-b(0))*a(n-1).

a(n) is the determinant of the n X n matrix M with m(i,j)=1 if |i-j| <= 1 and 0 otherwise. - Mario Catalani (mario.catalani(AT)unito.it), Jan 25 2003

Also row sums of triangle in A108299; a(n)=L(n-1,1), where L is also defined as in A108299; see A061347 for L(n,-1). - Reinhard Zumkeller, Jun 01 2005

Pisano period lengths: 1, 3, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, ... - R. J. Mathar, Aug 10 2012

Periodic sequences of this type can also be calculated as a(n) = c + floor(q/(p^m-1)*p^n) mod p, where c is a constant, q is the number representing the periodic digit pattern and m is the period. c, p and q can be calculated as follows: Let D be the array representing the number pattern to be repeated, m = size of D, max = maximum value of elements in D, min = minimum value of elements in D. Then c := min, p := max - min + 1 and q := p^m*Sum_{i=1..m} (D(i)-min)/p^i. Example: D = (1, 1, 0, -1, -1, 0), c = -1, m = 6, p = 3 and q = 676 for this sequence. - Hieronymus Fischer, Jan 04 2013

B(n) = a(n+5) = S(n-1, 1) appears, together with a(n) = A057079(n+1), in the formula 2*exp(Pi*n*i/3) = A(n) + B(n)*sqrt(3)*i with i = sqrt(-1). For S(n, x) see A049310. See also a Feb 27 2014 comment on A099837. - Wolfdieter Lang, Feb 27 2014

a(n) (for n>=1) is the difference between numbers of even and odd permutations p of 1,2,...,n such that |p(i)-i|<=1 for i=1,2,...,n. - Dmitry Efimov, Jan 08 2016

From Tom Copeland, Jan 31 2016: (Start)

Specialization of the o.g.f. 1 / ((x - w1)(x-w2)) = (1/(w1-w2)) ((w1-w2) + (w1^2 - w2^2) x + (w1^3-w2^3) x^2 + ...) with w1*w2 = (1/w1) + (1/w2) = 1. Then w1 = q = e^(i*Pi/3) and w2 = 1/q = e^(-i*Pi/3), giving the o.g.f. 1 /(1-x+x^2) for this entry with a(n) = (2/sqrt(3)) sin((n+1)Pi/3). See the Copeland link for more relations.

a(n) = (q^(n+1) - q^(-(n+1))) / (q - q^(-1)), so this entry gives the o.g.f. for an instance of the quantum integers denoted by [m]_q in Morrison et al. and Tingley. (End)

WARNING: It is unclear whether this sequence should start at offset 1 (as written) or offset 0 (in analogy to many similar sequences, which seems to be assumed in many of the given formulas).

Inverse binomial transform of A057079. - Paul Barry, May 15 2003

The unsigned version, with g.f. (1 + x + 2*x^2)/(1 - x^3), has a(n) = 4/3 -cos(2*Pi*n/3)/3 - sqrt(3)*sin(2*Pi*n/3)/3 = gcd(Fib(n+4), Fib(n+1)). - Paul Barry, Apr 02 2004

a(n) = L(n-2,-1), where L is defined as in A108299; see also A010892 for L(n,+1). - Reinhard Zumkeller, Jun 01 2005

From the Taylor expansion of log(1 + x + x^2) at x = 1, Sum_{k > 0} a(k)/k = log(3) = A002391. This is case n = 3 of the general expression Sum_{k > 0} (1-n*!(k mod n))/k = log(n). - Jaume Oliver Lafont, Oct 16 2009

If used with offset zero, a non-simple continued fraction representation of 2+sqrt(2). - R. J. Mathar, Mar 08 2012

Periodic sequences of this type can be also calculated by a(n) = c + floor(q/(p^m-1)*p^n) mod p, where c is a constant, q is the number representing the periodic digit pattern and m is the period length. c, p and q can be calculated as follows: Let D be the array representing the number pattern to be repeated, m = size of D, max = maximum value of elements in D, min = minimum value of elements in D. Than c := min, p := max - min + 1 and q := p^m*Sum_{i=1..m} (D(i)-min)/p^i. Example: D = (1, 1, -2), c = -2, p = 4 and q = 60 for this sequence. - Hieronymus Fischer, Jan 04 2013

This is the Dirichlet inverse of A117997. - Petros Hadjicostas, Jul 25 2020

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

Riordan array ((1+x)/(1-x)^2, x/(1-x)^2). Row sums are A002878. Diagonal sums are A003945. Inverse is A113187. An interesting factorization is (1/(1-x), x/(1-x))(1+2*x, x*(1+x)). - Paul Barry, Oct 17 2005

Central coefficients of rows with odd numbers of term are A052227.

From Wolfdieter Lang, Jun 26 2011: (Start)

T(k,s) appears as T_s(k) in the Knuth reference, p. 285.

This triangle is related to triangle A156308(n,m), appearing in this reference as U_m(n) on p. 285, by T(k,s) - T(k-1,s) = A156308(k,s), k>=s>=1 (identity on p. 286). T(k,s) = A156308(k+1,s+1) - A156308(k,s+1), k>=s>=0 (identity on p. 286).

(End)

A111125 is jointly generated with A208513 as an array of coefficients of polynomials v(n,x): initially, u(1,x)= v(1,x)= 1; for n>1, u(n,x)= u(n-1,x) +x*(x+1)*v(n-1) and v(n,x)= u(n-1,x) +x*v(n-1,x) +1. See the Mathematica section. The columns of A111125 are identical to those of A208508. Here, however, the alternating row sums are periodic (with period 1,2,1,-1,-2,-1). - Clark Kimberling, Feb 28 2012

This triangle T(k,s) (with signs and columns scaled with powers of 5) appears in the expansion of Fibonacci numbers F=A000045 with multiples of odd numbers as indices in terms of odd powers of F-numbers. See the Jennings reference, p. 108, Theorem 1. Quoted as Lemma 3 in the Ozeki reference given in A111418. The formula is: F_{(2*k+1)*n} = Sum_{s=0..k} ( T(k,s)*(-1)^((k+s)*n)*5^s*F_{n}^(2*s+1) ), k >= 0, n >= 0. - Wolfdieter Lang, Aug 24 2012

From Wolfdieter Lang, Oct 18 2012: (Start)

This triangle T(k,s) appears in the formula x^(2*k+1) - x^(-(2*k+1)) = Sum_{s=0..k} ( T(k,s)*(x-x^(-1))^(2*s+1) ), k>=0. Prove the inverse formula (due to the Riordan property this will suffice) with the binomial theorem. Motivated to look into this by the quoted paper of Wang and Zhang, eq. (1.4).

Alternating row sums are A057079.

The Z-sequence of this Riordan array is A217477, and the A-sequence is (-1)^n*A115141(n). For the notion of A- and Z-sequences for Riordan triangles see a W. Lang link under A006232. (End)

The signed triangle ((-1)^(k-s))*T(k,s) gives the coefficients of (x^2)^s of the polynomials C(2*k+1,x)/x, with C the monic integer Chebyshev T-polynomials whose coefficients are given in A127672 (C is there called R). See the odd numbered rows there. This signed triangle is the Riordan array ((1-x)/(1+x)^2, x/(1+x)^2). Proof by comparing the o.g.f. of the row polynomials where x is replaced by x^2 with the odd part of the bisection of the o.g.f. for C(n,x)/x. - Wolfdieter Lang, Oct 23 2012

From Wolfdieter Lang, Oct 04 2013: (Start)

The signed triangle S(k,s) := ((-1)^(k-s))*T(k,s) (see the preceding comment) is used to express in a (4*(k+1))-gon the length ratio s(4*(k+1)) = 2*sin(Pi/4*(k+1)) = 2*cos((2*k+1)*Pi/(4*(k+1))) of a side/radius as a polynomial in rho(4*(k+1)) = 2*cos(Pi/4*(k+1)), the length ratio (smallest diagonal)/side:

s(4*(k+1)) = Sum_{s=0..k} ( S(k,s)*rho(4*(k+1))^(2*s+1) ).

This is to be computed modulo C(4*(k+1), rho(4*(k+1)) = 0, the minimal polynomial (see A187360) in order to obtain s(4*(k+1)) as an integer in the algebraic number field Q(rho(4*(k+1))) of degree delta(4*(k+1)) (see A055034). Thanks go to Seppo Mustonen for asking me to look into the problem of the square of the total length in a regular n-gon, where this formula is used in the even n case. See A127677 for the formula in the (4*k+2)-gon. (End)

From Wolfdieter Lang, Aug 14 2014: (Start)

The row polynomials for the signed triangle (see the Oct 23 2012 comment above), call them todd(k,x) = Sum_{s=0..k} ( (-1)^(k-s)*T(k,s)*x^s ) = S(k, x-2) - S(k-1, x-2), k >= 0, with the Chebyshev S-polynomials (see their coefficient triangle (A049310) and S(-1, x) = 0), satisfy the recurrence todd(k, x) = (-1)^(k-1)*((x-4)/2)*todd(k-1, 4-x) + ((x-2)/2)*todd(k-1, x), k >= 1, todd(0, x) = 1. From the Aug 03 2014 comment on A130777.

This leads to a recurrence for the signed triangle, call it S(k,s) as in the Oct 04 2013 comment: S(k,s) = (1/2)*(1 + (-1)^(k-s))*S(k-1,s-1) + (2*(s+1)*(-1)^(k-s) - 1)*S(k-1,s) + (1/2)*(-1)^(k-s)*Sum_{j=0..k-s-2} ( binomial(j+s+2,s)*4^(j+2)* S(k-1, s+1+j) ) for k >= s >= 1, and S(k,s) = 0 if k < s and S(k,0) = (-1)^k*(2*k+1). Note that the recurrence derived from the Riordan A-sequence A115141 is similar but has simpler coefficients: S(k,s) = sum(A115141(j)*S(k-1,s-1+j), j=0..k-s), k >= s >=1.

(End)

From Tom Copeland, Nov 07 2015: (Start)

Rephrasing notes here: Append an initial column of zeros, except for a 1 at the top, to A111125 here. Then the partial sums of the columns of this modified entry are contained in A208513. Append an initial row of zeros to A208513. Then the difference of consecutive pairs of rows of the modified A208513 generates the modified A111125. Cf. A034807 and A127677.

For relations among the characteristic polynomials of Cartan matrices of the Coxeter root groups, Chebyshev polynomials, cyclotomic polynomials, and the polynomials of this entry, see Damianou (p. 20 and 21) and Damianou and Evripidou (p. 7).

As suggested by the equations on p. 7 of Damianou and Evripidou, the signed row polynomials of this entry are given by (p(n,x))^2 = (A(2*n+1, x) + 2)/x = (F(2*n+1, (2-x), 1, 0, 0, ... ) + 2)/x = F(2*n+1, -x, 2*x, -3*x, ..., (-1)^n n*x)/x = -F(2*n+1, x, 2*x, 3*x, ..., n*x)/x, where A(n,x) are the polynomials of A127677 and F(n, ...) are the Faber polynomials of A263196. Cf. A127672 and A127677.

(End)

The row polynomials P(k, x) of the signed triangle S(k, s) = ((-1)^(k-s))*T(k, s) are given from the row polynomials R(2*k+1, x) of triangle A127672 by

P(k, x) = R(2*k+1, sqrt(x))/sqrt(x). - Wolfdieter Lang, May 02 2021

A Chebyshev transform of A007877(n+1). The g.f. is transformed to (1+x)/((1-x)(1+x^2)) under the mapping G(x)->(1/(1+x^2))G(1/(1+x^2)). - Paul Barry, Oct 12 2004

a(n-1) = 2*C(n-2, floor((n-2)/2)) is also the number of bit strings of length n in which the number of 00 substrings is equal to the number of 11 substrings. For example, when n = 4 we have 4 such bit strings: 0011, 0101, 1010, and 1100. - Angel Plaza, Apr 23 2009

Hankel transform is A120617. - Paul Barry, Aug 10 2009

The Hankel transform of a(n) is (-2)^C(n+1,2). The Hankel transform of (-1)^C(n+1,2)*a(n) is (-1)^C(n+1,2)*A164584(n). - Paul Barry, Aug 17 2009

For n > 1, a(n) is also the number of n-step walks starting from the origin and returning to it exactly once. - Geoffrey Critzer, Jan 24 2010

-a(n) is the Z-sequence for the Riordan array A130777. (See the W. Lang link under A006232 for A- and Z-sequences for Riordan matrices). - Wolfdieter Lang, Jul 12 2011

Number of subsets of {1,...,n} in which the even elements appear as often at even positions as at odd positions. - Gus Wiseman, Mar 17 2018

a(n+1) is the number of labeled incomplete ternary trees on n vertices in which each left child has a larger label than its parent. - Brian Drake, Jul 28 2008

Put a(0) = 1. For n > 0, let x(n,k) = 2*cos((2*k-1)*Pi/(2*n+1)), k=1..n. Define the recurrences S(n;0,x(n,k)) = 1, S(n;1,x(n,k)) = x(n,k), S(n;r,x(n,k)) = x(n,k)*S(n;r-1,x(n,k)) - S(n;r-2,x(n,k)), r > 1 an integer, k=1..n. CONJECTURE: For n > 0, a(n) = Product_{k=1..n} (Sum_{m=0..n-1} S(n;2*m,x(n,k))^2). - L. Edson Jeffery, Sep 11 2013

From Wolfdieter Lang, Dec 16 2013: (Start)

Discriminants of the first difference of Chebyshev S-polynomials.

The coefficient table for the first difference polynomials P(n, x) = S(n, x) - S(n-1, x), n >= 0, S(-1, x) = 0, with the Chebyshev S polynomials (see A049310), is given in A130777.

For the discriminant of a polynomial in terms of the square of a determinant of a Vandermonde matrix build from the zeros of the polynomial see, e.g., A127670.

For the proof that D(n) := discriminant(P(n,x)) = (2*n + 1)^(n - 1), n >= 1, use the formula given e.g., in the Rivlin reference, p. 218, Theorem 5.13, eq. (5.3), namely D(n) = (-1)^(n*(n-1)/2)*Product_{j=1..n} P'(n, x(n,j)), with the zeros x(n,j) = -2*cos(2*Pi*j/(2*n+1)) of P(n, x) (see A130777). P'(n, x(n,j)) = (2*n+1)*P(n-1, x(n,j))/(2*sin(Pi*j/(2*n+1))*2*cos(Pi*j/(2*n+1)))^2. P(n-1, x(n,j)) = (-1)^(n+j)*2*cos(Pi*j/(2*n+1)). Product_{j=1..n} 2*sin(Pi*j/(2*n+1)) = 2*n+1 (see the Oct 10 2013 formula in A005408. Product_{j=1..n} 2*cos(Pi*j/(2*n+1)) = 1, because S(2*n, 0) = (-1)^n.

(End)

a(n) is the number of labeled 2-trees with n+2 vertices, rooted at a given edge. - Nikos Apostolakis, Nov 30 2018

a(n) is also the number of 2-trees with n labeled triangles and with a distinguished oriented edge. - Nikos Apostolakis, Dec 14 2018

From Herb Conn, Jan 28 2005: (Start)

"Letting x = 2 Cos 2A, we have the following trigonometric identities:

"Sin 3A = 3*Sin A - 4*Sin^3 A

"Sin 5A = 5*Sin A - 20*Sin^3 A + 16*Sin^5 A

"Sin 7A = 7*Sin A - 56*Sin^3 A + 112*Sin^5 A - 64*Sin^7 A

"Sin 9A = 9*Sin A - 120*Sin^3 A + 432*Sin^5 A - 576*Sin^7 A + 256*Sin^9 A, etc." (End)

Cayley (1876) states "Write sin u = x, then we have sin u = x, [...] sin 3u = 3x - 4x^3, [...] sin 5u = 5x - 20x^3 + 16 x^5, [...]". Since T_n(cos(u)) = cos(nu) for all integer n, sin(u) = cos(u - Pi/2), and sin(u + k*Pi) = (-1)^k sin(u) it follows that T_n(sin(u)) = (-1)^((n-1)/2) sin(nu) for all odd integer n. - Michael Somos, Jun 19 2012

From Wolfdieter Lang, Aug 05 2014: (Start)

The coefficient triangle t(n,k) for the row polynomials Todd(n, x) := T_{2*n+1}(sqrt(x))/sqrt(x) = sum(t(n,k)*x^k, k=0..n) is the Riordan triangle ((1-z)/(1+z)^2, 4*z/(1+z)^2) (rewrite the g.f. for the present triangle a(n,k) given in the formula section). The triangle entries t(n,k) = a(n,k), but the interpretation of the row polynomials is different for both cases.

From the relation Todd(n, x) = S(n, 2*(2*x-1)) - S(n-1, 2*(2*x-1)) with the Chebyshev S-polynomials (see A049310 and the formula section of A130777) follows the recurrence: Todd(n, x) = 2*(-1)^n*(1-x)*Todd(n-1, 1-x) + (2x-1)*Todd(n-1, x), n >= 1, Todd(0, x) = 1.

This corresponds to the triangle recurrence t(n,k) = (2*(k+1)*(-1)^(n-k) - 1)*t(n-1,k) + 2*(1 +(-1)^(n-k))*t(n-1,k-1) + 2*(-1)^(n-k)*sum(binomial(l+1,k)*t(n-1,l), l=k+1..n-1), n >= k >= 1, t(n,k) = 0 if n < k, t(n,0) = (-1)^n*(2*n+1). Compare this with the shorter recurrence involving the rational A-sequence for this Riordan triangle which has g.f. x^2/(2-x-2*sqrt(1-x)). t(n,k) = sum(A(j)*t(n-1,k-1+j), j=0..n-k), n >= k >= 1. The Z-sequence has g.f. -(1 + 2/sqrt(1-x)). For the A- and Z-sequence see a link under A006232. (End)

Conjecture: (i) Let n > 1 and N=2*n+1. Row n of T gives the coefficients of the characteristic polynomial p_N(x)=Sum_{k=0..n} T(n,k)*x^(n-k) of the n X n Danzer matrix D_{N,n-1} = {{0,...,0,1}, {0,...,0,1,1}, ..., {0,1,...,1}, {1,...,1}}. (ii) Let S_0(t)=1, S_1(t)=t and S_r(t)=t*S_(r-1)(t)-S_(r-2)(t), r > 1 (cf. A049310). Then p_N(x)=0 has solutions w_{N,j}=S_(n-1)(phi_{N,j}), where phi_{N,j}=2*(-1)^(j+1)*cos(j*Pi/N), j = 1..n. - L. Edson Jeffery, Dec 18 2011

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