A053117 -id:A053117 - cp's OEIS frontend

A049310 Triangle of coefficients of Chebyshev's S(n,x) := U(n,x/2) polynomials (exponents in increasing order).

1, 0, 1, -1, 0, 1, 0, -2, 0, 1, 1, 0, -3, 0, 1, 0, 3, 0, -4, 0, 1, -1, 0, 6, 0, -5, 0, 1, 0, -4, 0, 10, 0, -6, 0, 1, 1, 0, -10, 0, 15, 0, -7, 0, 1, 0, 5, 0, -20, 0, 21, 0, -8, 0, 1, -1, 0, 15, 0, -35, 0, 28, 0, -9, 0, 1, 0, -6, 0, 35, 0, -56, 0, 36, 0, -10, 0, 1, 1, 0, -21, 0, 70, 0, -84, 0

Offset: 0

Wolfdieter Lang

G.f. for row polynomials S(n,x) (signed triangle): 1/(1-x*z+z^2). Unsigned triangle |a(n,m)| has Fibonacci polynomials F(n+1,x) as row polynomials with g.f. 1/(1-x*z-z^2). |a(n,m)| triangle has rows of Pascal's triangle A007318 in the even-numbered diagonals (odd-numbered ones have only 0's).
Row sums (unsigned triangle) A000045(n+1) (Fibonacci). Row sums (signed triangle) S(n,1) sequence = periodic(1,1,0,-1,-1,0) = A010892.
Alternating row sums A049347(n) = S(n,-1) = periodic(1,-1,0). - Wolfdieter Lang, Nov 04 2011
S(n,x) is the characteristic polynomial of the adjacency matrix of the n-path. - Michael Somos, Jun 24 2002
S(n,x) is also the matching polynomial of the n-path. - Eric W. Weisstein, Apr 10 2017
|T(n,k)| = number of compositions of n+1 into k+1 odd parts. Example: |T(7,3)| = 10 because we have (1,1,3,3), (1,3,1,3), (1,3,3,1), (3,1,1,3), (3,1,3,1), (3,3,1,1), (1,1,1,5), (1,1,5,1), (1,5,1,1) and (5,1,1,1). - Emeric Deutsch, Apr 09 2005
S(n,x)= R(n,x) + S(n-2,x), n >= 2, S(-1,x)=0, S(0,x)=1, R(n,x):=2*T(n,x/2) = Sum_{m=0..n} A127672(n,m)*x^m (monic integer Chebyshev T-Polynomials). This is the rewritten so-called trace of the transfer matrix formula for the T-polynomials. - Wolfdieter Lang, Dec 02 2010
In a regular N-gon inscribed in a unit circle, the side length is d(N,1) = 2*sin(Pi/N). The length ratio R(N,k):=d(N,k)/d(N,1) for the (k-1)-th diagonal, with k from {2,3,...,floor(N/2)}, N >= 4, equals S(k-1,x) = sin(k*Pi/N)/sin(Pi/N) with x=rho(N):=R(N,2) = 2*cos(Pi/N). Example: N=7 (heptagon), rho=R(7,2), sigma:=R(N,3) = S(2,rho) = rho^2 - 1. Motivated by the quoted paper by P. Steinbach. - Wolfdieter Lang, Dec 02 2010
From Wolfdieter Lang, Jul 12 2011: (Start)
In q- or basic analysis, q-numbers are [n]_q := S(n-1,q+1/q) = (q^n-(1/q)^n)/(q-1/q), with the row polynomials S(n,x), n >= 0.
The zeros of the row polynomials S(n-1,x) are (from those of Chebyshev U-polynomials):
  x(n-1;k) = +- t(k,rho(n)), k = 1..ceiling((n-1)/2), n >= 2, with t(n,x) the row polynomials of A127672 and rho(n):= 2*cos(Pi/n). The simple vanishing zero for even n appears here as +0 and -0.
Factorization of the row polynomials S(n-1,x), x >= 1, in terms of the minimal polynomials of cos(2 Pi/2), called Psi(n,x), with coefficients given by A181875/A181876:
  S(n-1,x) = (2^(n-1))*Product_{n>=1}(Psi(d,x/2), 2 < d | 2n).
(From the rewritten eq. (3) of the Watkins and Zeitlin reference, given under A181872.) [See the W. Lang ArXiv link, Proposition 9, eq. (62). - Wolfdieter Lang, Apr 14 2018]
  (End)
The discriminants of the S(n,x) polynomials are found in A127670. - Wolfdieter Lang, Aug 03 2011
This is an example for a subclass of Riordan convolution arrays (lower triangular matrices) called Bell arrays. See the L. W. Shapiro et al. reference under A007318. If a Riordan array is named (G(z),F(z)) with F(z)=z*Fhat(z), the o.g.f. for the row polynomials is G(z)/(1-x*z*Fhat(z)), and it becomes a Bell array if G(z)=Fhat(z). For the present Bell type triangle G(z)=1/(1+z^2) (see the o.g.f. comment above). This leads to the o.g.f. for the column no. k, k >= 0, x^k/(1+x^2)^(k+1) (see the formula section), the one for the row sums and for the alternating row sums (see comments above). The Riordan (Bell) A- and Z-sequences (defined in a W. Lang link under A006232, with references) have o.g.f.s 1-x*c(x^2) and -x*c(x^2), with the o.g.f. of the Catalan numbers A000108. Together they lead to a recurrence given in the formula section. - Wolfdieter Lang, Nov 04 2011
The determinant of the N x N matrix S(N,[x[1], ..., x[N]]) with elements S(m-1,x[n]), for n, m = 1, 2, ..., N, and for any x[n], is identical with the determinant of V(N,[x[1], ..., x[N]]) with elements x[n]^(m-1) (a Vandermondian, which equals Product_{1 <= i < j<= N} (x[j] - x[i])). This is a special instance of a theorem valid for any N >= 1 and any monic polynomial system p(m,x), m>=0, with p(0,x) = 1. For this theorem see the Vein-Dale reference, p. 59. Thanks to L. Edson Jeffery for an email asking for a proof of the non-singularity of the matrix S(N,[x[1], ...., x[N]]) if and only if the x[j], j = 1..N, are pairwise distinct. - Wolfdieter Lang, Aug 26 2013
These S polynomials also appear in the context of modular forms. The rescaled Hecke operator T*n = n^((1-k)/2)*T_n acting on modular forms of weight k satisfies T*(p^n) = S(n, T*p), for each prime p and positive integer n. See the Koecher-Krieg reference, p. 223. - _Wolfdieter Lang, Jan 22 2016
For a shifted o.g.f. (mod signs), its compositional inverse, and connections to Motzkin and Fibonacci polynomials, non-crossing partitions and other combinatorial structures, see A097610. - Tom Copeland, Jan 23 2016
From M. Sinan Kul, Jan 30 2016; edited by Wolfdieter Lang, Jan 31 2016 and Feb 01 2016: (Start)
Solutions of the Diophantine equation u^2 + v^2 - k*u*v = 1 for integer k given by (u(k,n), v(k,n)) = (S(n,k), S(n-1,k)) because of the Cassini-Simson identity: S(n,x)^2 - S(n+1,x)*S(n-1, x) = 1, after use of the S-recurrence. Note that S(-n, x) = -S(-n-2, x), n >= 1, and the periodicity of some S(n, k) sequences.
Hence another way to obtain the row polynomials would be to take powers of the matrix [x, -1; 1,0]: S(n, x) = (([x, -1; 1, 0])^n)[1,1], n >= 0.
See also a Feb 01 2016 comment on A115139 for a well-known S(n, x) sum formula.
Then we have with the present T triangle
  A039834(n) = -i^(n+1)*T(n-1, k) where i is the imaginary unit and n >= 0.
  A051286(n) = Sum_{i=0..n} T(n,i)^2 (see the Philippe Deléham, Nov 21 2005 formula),
  A181545(n) = Sum_{i=0..n+1} abs(T(n,i)^3),
  A181546(n) = Sum_{i=0..n+1} T(n,i)^4,
  A181547(n) = Sum_{i=0..n+1} abs(T(n,i)^5).
S(n, 0) = A056594(n), and for k = 1..10 the sequences S(n-1, k) with offset n = 0 are A128834, A001477, A001906, A001353, A004254, A001109, A004187, A001090, A018913, A004189.
(End)
For more on the Diophantine equation presented by Kul, see the Ismail paper. - Tom Copeland, Jan 31 2016
The o.g.f. for the Legendre polynomials L(n,x) is 1 / sqrt(1- 2x*z + z^2), and squaring it gives the o.g.f. of U(n,x), A053117, so Sum_{k=0..n} L(k,x/2) L(n-k,x/2) = S(n,x). This gives S(n,x) = L(n/2,x/2)^2 + 2*Sum_{k=0..n/2-1} L(k,x/2) L(n-k,x/2) for n even and S(n,x) = 2*Sum_{k=0..(n-1)/2} L(k,x/2) L(n-k,x/2) for odd n. For a connection to elliptic curves and modular forms, see A053117. For the normalized Legendre polynomials, see A100258. For other properties and relations to other polynomials, see Allouche et al. - Tom Copeland, Feb 04 2016
LG(x,h1,h2) = -log(1 - h1*x + h2*x^2) = Sum_{n>0} F(n,-h1,h2,0,..,0) x^n/n is a log series generator of the bivariate row polynomials of A127672 with A127672(0,0) = 0 and where F(n,b1,b2,..,bn) are the Faber polynomials of A263916. Exp(LG(x,h1,h2)) = 1 / (1 - h1*x + h2*x^2 ) is the o.g.f. of the bivariate row polynomials of this entry. - Tom Copeland, Feb 15 2016 (Instances of the bivariate o.g.f. for this entry are on pp. 5 and 18 of Sunada. - Tom Copeland, Jan 18 2021)
For distinct odd primes p and q the Legendre symbol can be written as Legendre(q,p) = Product_{k=1..P} S(q-1, 2*cos(2*Pi*k/p)), with P = (p-1)/2. See the Lemmermeyer reference, eq. (8.1) on p. 236. Using the zeros of S(q-1, x) (see above) one has S(q-1, x) = Product_{l=1..Q} (x^2 - (2*cos(Pi*l/q))^2), with Q = (q-1)/2. Thus S(q-1, 2*cos(2*Pi*k/p)) = ((-4)^Q)*Product_{l=1..Q} (sin^2(2*Pi*k/p) - sin^2(Pi*l/q)) = ((-4)^Q)*Product_{m=1..Q} (sin^2(2*Pi*k/p) - sin^2(2*Pi*m/q)). For the proof of the last equality see a W. Lang comment on the triangle A057059 for n = Q and an obvious function f. This leads to Eisenstein's proof of the quadratic reciprocity law Legendre(q,p) = ((-1)^(P*Q)) * Legendre(p,q), See the Lemmermeyer reference, pp. 236-237. - Wolfdieter Lang, Aug 28 2016
For connections to generalized Fibonacci polynomials, compare their generating function on p. 5 of the Amdeberhan et al. link with the o.g.f. given above for the bivariate row polynomials of this entry. - Tom Copeland, Jan 08 2017
The formula for Ramanujan's tau function (see A000594) for prime powers is tau(p^k) = p^(11*k/2)*S(k, p^(-11/2)*tau(p)) for k >= 1, and p = A000040(n), n >= 1. See the Hardy reference, p. 164, eqs. (10.3.4) and (10.3.6) rewritten in terms of S. - Wolfdieter Lang, Jan 27 2017
From Wolfdieter Lang, May 08 2017: (Start)
The number of zeros Z(n) of the S(n, x) polynomials in the open interval (-1,+1) is 2*b(n) for even n >= 0 and 1 + 2*b(n) for odd n >= 1, where b(n) = floor(n/2) - floor((n+1)/3). This b(n) is the number of integers k in the interval (n+1)/3 < k <= floor(n/2). See a comment on the zeros of S(n, x) above, and b(n) = A008615(n-2), n >= 0. The numbers Z(n) have been proposed (with a conjecture related to A008611) by Michel Lagneau, as the number of zeros of Fibonacci polynomials on the imaginary axis (-I,+I), with I=sqrt(-1). They are Z(n) = A008611(n-1), n >= 0, with A008611(-1) = 0. Also Z(n) = A194960(n-4), n >= 0. Proof using the A008611 version. A194960 follows from this.
In general the number of zeros Z(a;n) of S(n, x) for n >= 0 in the open interval (-a,+a) for a from the interval (0,2) (x >= 2 never has zeros, and a=0 is trivial: Z(0;n) = 0) is with b(a;n) = floor(n//2) - floor((n+1)*arccos(a/2)/Pi), as above Z(a;n) = 2*b(a;n) for even n >= 0 and 1 + 2*b(a;n) for odd n >= 1. For the closed interval [-a,+a] Z(0;n) = 1 and for a from (0,1) one uses for Z(a;n) the values b(a;n) = floor(n/2) - ceiling((n+1)*arccos(a/2)/Pi) + 1. (End)
The Riordan row polynomials S(n, x) (Chebyshev S) belong to the Boas-Buck class (see a comment and references in A046521), hence they satisfy the Boas-Buck identity: (E_x - n*1)*S(n, x) = (E_x + 1)*Sum_{p=0..n-1} (1 - (-1)^p)*(-1)^((p+1)/2)*S(n-1-p, x), for n >= 0, where E_x = x*d/dx (Euler operator). For the triangle T(n, k) this entails a recurrence for the sequence of column k, given in the formula section. - Wolfdieter Lang, Aug 11 2017
The e.g.f. E(x,t) := Sum_{n>=0} (t^n/n!)*S(n,x) for the row polynomials is obtained via inverse Laplace transformation from the above given o.g.f. as E(x,t) = ((1/xm)*exp(t/xm) - (1/xp)*exp(t/xp) )/(xp - xm) with xp = (x + sqrt(x^2-4))/2 and xm = (x - sqrt(x^2-4))/2. - Wolfdieter Lang, Nov 08 2017
From Wolfdieter Lang, Apr 12 2018: (Start)
Factorization of row polynomials S(n, x), for n >= 1, in terms of C polynomials (not Chebyshev C) with coefficients given in A187360. This is obtained from the factorization into Psi polynomials (see the Jul 12 2011 comment above) but written in terms of minimal polynomials of 2*cos(2*Pi/n) with coefficients in A232624:
S(2*k, x) = Product_{2 <= d | (2*k+1)} C(d, x)*(-1)^deg(d)*C(d, -x), with deg(d) = A055034(d) the degree of C(d, x).
S(2*k+1, x) = Product_{2 <= d | 2*(k+1)} C(d, x) * Product_{3 <= 2*d + 1 | (k+1)} (-1)^(deg(2*d+1))*C(2*d+1, -x).
Note that (-1)^(deg(2*d+1))*C(2*d+1, -x)*C(2*d+1, x) pairs always appear.
The number of C factors of S(2*k, x), for k >= 0, is 2*(tau(2*k+1) - 1) = 2*(A099774(k+1) - 1) = 2*A095374(k), and for S(2*k+1, x), for k >= 0, it is tau(2*(k+1)) + tau_{odd}(k+1) - 2 = A302707(k), with tau(2*k+1) = A099774(k+1), tau(n) = A000005 and tau(2*(k+1)) = A099777(k+1).
For the reverse problem, the factorization of C polynomials into S polynomials, see A255237. (End)
The S polynomials with general initial conditions S(a,b;n,x) = x*S(a,b;n-1,x) - S(a,b;n-2,x), for n >= 1, with S(a,b;-1,x) = a and S(a,b;0,x) = b are S(a,b;n,x) = b*S(n, x) - a*S(n-1, x), for n >= -1. Recall that S(-2, x) = -1 and S(-1, x) = 0. The o.g.f. is G(a,b;z,x) = (b - a*z)/(1 - x*z + z^2). - Wolfdieter Lang, Oct 18 2019
Also the convolution triangle of A101455. - Peter Luschny, Oct 06 2022
From Wolfdieter Lang, Apr 26 2023: (Start)
Multi-section of S-polynomials: S(m*n+k, x) = S(m+k, x)*S(n-1, R(m, x)) - S(k, x)*S(n-2, R(m, x)), with R(n, x) = S(n, x) - S(n-2, x) (see A127672), S(-2, x) = -1, and S(-1, x) = 0, for n >= 0, m >= 1, and k = 0, 1, ..., m-1.
O.g.f. of {S(m*n+k, y)}_{n>=0}: G(m,k,y,x) = (S(k, y) - (S(k, y)*R(m, y) - S(m+k, y))*x)/(1 - R(m,y)*x + x^2).
See eqs. (40) and (49), with r = x or y and s =-1, of the G. Detlefs and W. Lang link at A034807. (End)
S(n, x) for complex n and complex x: S(n, x) = ((-i/2)/sqrt(1 - (x/2)^2))*(q(x/2)*exp(+n*log(q(x/2))) - (1/q(x/2))*exp(-n*log(q(x/2)))), with q(x) = x + sqrt(1 - x^2)*i. Here log(z) = |z| + Arg(z)*i, with Arg(z) from [-Pi,+Pi) (principal branch). This satisfies the recurrence relation for S because it is derived from the Binet - de Moivre formula for S. Examples: S(n/m, 0) = cos((n/m)*Pi/4), for n >= 0 and m >= 1. S(n*i, 0) = (1/2)*(1 + exp(n*Pi))*exp(-(n/2)*Pi), for n >= 0. S(1+i, 2+i) = 0.6397424847... + 1.0355669490...*i. Thanks to Roberto Alfano for asking a question leading to this formula. - Wolfdieter Lang, Jun 05 2023
Lim_{n->oo} S(n, x)/S(n-1, x) = r(x) = (x - sqrt(x^2 -4))/2, for |x| >= 2. For x = +-2, this limit is +-1. - Wolfdieter Lang, Nov 15 2023

			The triangle T(n, k) begins:
  n\k  0  1   2   3   4   5   6    7   8   9  10  11
  0:   1
  1:   0  1
  2:  -1  0   1
  3:   0 -2   0   1
  4:   1  0  -3   0   1
  5:   0  3   0  -4   0   1
  6:  -1  0   6   0  -5   0   1
  7:   0 -4   0  10   0  -6   0    1
  8:   1  0 -10   0  15   0  -7    0   1
  9:   0  5   0 -20   0  21   0   -8   0   1
  10: -1  0  15   0 -35   0  28    0  -9   0   1
  11:  0 -6   0  35   0 -56   0   36   0 -10   0   1
  ... Reformatted and extended by _Wolfdieter Lang_, Oct 24 2012
For more rows see the link.
E.g., fourth row {0,-2,0,1} corresponds to polynomial S(3,x)= -2*x + x^3.
From _Wolfdieter Lang_, Jul 12 2011: (Start)
Zeros of S(3,x) with rho(4)= 2*cos(Pi/4) = sqrt(2):
  +- t(1,sqrt(2)) = +- sqrt(2) and
  +- t(2,sqrt(2)) = +- 0.
Factorization of S(3,x) in terms of Psi polynomials:
S(3,x) = (2^3)*Psi(4,x/2)*Psi(8,x/2) = x*(x^2-2).
(End)
From _Wolfdieter Lang_, Nov 04 2011: (Start)
A- and Z- sequence recurrence:
T(4,0) = - (C(0)*T(3,1) + C(1)*T(3,3)) = -(-2 + 1) = +1,
T(5,3) = -3 - 1*1 = -4.
(End)
Boas-Buck recurrence for column k = 2, n = 6: S(6, 2) = (3/4)*(0 - 2* S(4 ,2) + 0 + 2*S(2, 2)) = (3/4)*(-2*(-3) + 2) = 6. - _Wolfdieter Lang_, Aug 11 2017
From _Wolfdieter Lang_, Apr 12 2018: (Start)
Factorization into C polynomials (see the Apr 12 2018 comment):
S(4, x) = 1 - 3*x^2 + x^4 = (-1 + x + x^2)*(-1 - x + x^2) = (-C(5, -x)) * C(5, x); the number of factors is 2 = 2*A095374(2).
S(5, x) = 3*x - 4*x^3 + x^5 = x*(-1 + x)*(1 + x)*(-3 + x^2) = C(2, x)*C(3, x)*(-C(3, -x))*C(6, x); the number of factors is 4 = A302707(2). (End)

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Index entries for sequences related to Chebyshev polynomials.
Index entries for "core" sequences

Cf. A000005, A000217, A000292, A000332, A000389, A001227, A007318, A008611, A008615, A101455, A010892, A011973, A053112 (without zeros), A053117, A053119 (reflection), A053121 (inverse triangle), A055034, A097610, A099774, A099777, A100258, A112552 (first column clipped), A127672, A168561 (absolute values), A187360. A194960, A232624, A255237.

Triangles of coefficients of Chebyshev's S(n,x+k) for k = 5, 4, 3, 2, 1, 0, -1, -2, -3, -4, -5: A207824, A207823, A125662, A078812, A101950, A049310, A104562, A053122, A207815, A159764, A123967.

A049310:= func< n,k | ((n+k) mod 2) eq 0 select (-1)^(Floor((n+k)/2)+k)*Binomial(Floor((n+k)/2), k) else 0 >;
[A049310(n,k): k in [0..n], n in [0..15]]; // G. C. Greubel, Jul 25 2022

A049310 := proc(n,k): binomial((n+k)/2,(n-k)/2)*cos(Pi*(n-k)/2)*(1+(-1)^(n-k))/2 end: seq(seq(A049310(n,k), k=0..n),n=0..11); # Johannes W. Meijer, Aug 08 2011
# Uses function PMatrix from A357368. Adds a row above and a column to the left.
PMatrix(10, n -> ifelse(irem(n, 2) = 0, 0, (-1)^iquo(n-1, 2))); # Peter Luschny, Oct 06 2022

t[n_, k_] /; EvenQ[n+k] = ((-1)^((n+k)/2+k))*Binomial[(n+k)/2, k]; t[n_, k_] /; OddQ[n+k] = 0; Flatten[Table[t[n, k], {n, 0, 12}, {k, 0, n}]][[;; 86]] (* Jean-François Alcover, Jul 05 2011 *)
Table[Coefficient[(-I)^n Fibonacci[n + 1, - I x], x, k], {n, 0, 10}, {k, 0, n}] //Flatten (* Clark Kimberling, Aug 02 2011; corrected by Eric W. Weisstein, Apr 06 2017 *)
CoefficientList[ChebyshevU[Range[0, 10], -x/2], x] // Flatten (* Eric W. Weisstein, Apr 06 2017 *)
CoefficientList[Table[(-I)^n Fibonacci[n + 1, -I x], {n, 0, 10}], x] // Flatten (* Eric W. Weisstein, Apr 06 2017 *)

{T(n, k) = if( k<0 || k>n || (n + k)%2, 0, (-1)^((n + k)/2 + k) * binomial((n + k)/2, k))} /* Michael Somos, Jun 24 2002 */

@CachedFunction
def A049310(n,k):
    if n< 0: return 0
    if n==0: return 1 if k == 0 else 0
    return A049310(n-1,k-1) - A049310(n-2,k)
for n in (0..9): [A049310(n,k) for k in (0..n)] # Peter Luschny, Nov 20 2012

T(n,k) := 0 if n < k or n+k odd, otherwise ((-1)^((n+k)/2+k))*binomial((n+k)/2, k); T(n, k) = -T(n-2, k)+T(n-1, k-1), T(n, -1) := 0 =: T(-1, k), T(0, 0)=1, T(n, k)= 0 if n < k or n+k odd; g.f. k-th column: (1 / (1 + x^2)^(k + 1)) * x^k. - Michael Somos, Jun 24 2002
T(n,k) = binomial((n+k)/2, (n-k)/2)*cos(Pi*(n-k)/2)*(1+(-1)^(n-k))/2. - Paul Barry, Aug 28 2005
Sum_{k=0..n} T(n,k)^2 = A051286(n). - Philippe Deléham, Nov 21 2005
Recurrence for the (unsigned) Fibonacci polynomials: F(1)=1, F(2)=x; for n > 2, F(n) = x*F(n-1) + F(n-2).
From Wolfdieter Lang, Nov 04 2011: (Start)
The Riordan A- and Z-sequences, given in a comment above, lead together to the recurrence:
T(n,k) = 0 if n < k, if k=0 then T(0,0)=1 and
  T(n,0)= -Sum_{i=0..floor((n-1)/2)} C(i)*T(n-1,2*i+1), otherwise T(n,k) = T(n-1,k-1) - Sum_{i=1..floor((n-k)/2)} C(i)*T(n-1,k-1+2*i), with the Catalan numbers C(n)=A000108(n).
(End)
The row polynomials satisfy also S(n,x) = 2*(T(n+2, x/2) - T(n, x/2))/(x^2-4) with the Chebyshev T-polynomials. Proof: Use the trace formula 2*T(n, x/2) = S(n, x) - S(n-2, x) (see the Dec 02 2010 comment above) and the S-recurrence several times. This is a formula which expresses the S- in terms of the T-polynomials. - Wolfdieter Lang, Aug 07 2014
From Tom Copeland, Dec 06 2015: (Start)
The non-vanishing, unsigned subdiagonals Diag_(2n) contain the elements D(n,k) = Sum_{j=0..k} D(n-1,j) = (k+1) (k+2) ... (k+n) / n! = binomial(n+k,n), so the o.g.f. for the subdiagonal is (1-x)^(-(n+1)). E.g., Diag_4 contains D(2,3) = D(1,0) + D(1,1) + D(1,2) + D(1,3) = 1 + 2 + 3 + 4 = 10 = binomial(5,2). Diag_4 is shifted A000217; Diag_6, shifted A000292: Diag_8, shifted A000332; and Diag_10, A000389.
The non-vanishing antidiagonals are signed rows of the Pascal triangle A007318.
For a reversed, unsigned version with the zeros removed, see A011973. (End)
The Boas-Buck recurrence (see a comment above) for the sequence of column k is: S(n, k) = ((k+1)/(n-k))*Sum_{p=0..n-1-k} (1 - (-1)^p)*(-1)^((p+1)/2) * S(n-1-p, k), for n > k >= 0 and input S(k, k) = 1. - Wolfdieter Lang, Aug 11 2017
The m-th row consecutive nonzero entries in order are (-1)^c*(c+b)!/c!b! with c = m/2, m/2-1, ..., 0 and b = m-2c if m is even and with c = (m-1)/2, (m-1)/2-1, ..., 0 with b = m-2c if m is odd. For the 8th row starting at a(36) the 5 consecutive nonzero entries in order are 1,-10,15,-7,1 given by c = 4,3,2,1,0 and b = 0,2,4,6,8. - Richard Turk, Aug 20 2017
O.g.f.: exp( Sum_{n >= 0} 2*T(n,x/2)*t^n/n ) = 1 + x*t + (-1 + x^2)*t^2 + (-2*x + x^3)*t^3 + (1 - 3*x^2 + x^4)*t^4 + ..., where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. - Peter Bala, Aug 15 2022

A111125 Triangle read by rows: T(k,s) = ((2k+1)/(2s+1))binomial(k+s,2s), 0 <= s <= k.

Original entry on oeis.org

1, 3, 1, 5, 5, 1, 7, 14, 7, 1, 9, 30, 27, 9, 1, 11, 55, 77, 44, 11, 1, 13, 91, 182, 156, 65, 13, 1, 15, 140, 378, 450, 275, 90, 15, 1, 17, 204, 714, 1122, 935, 442, 119, 17, 1, 19, 285, 1254, 2508, 2717, 1729, 665, 152, 19, 1, 21, 385, 2079, 5148, 7007, 5733, 2940, 952, 189, 21, 1

Offset: 0

N. J. A. Sloane, Oct 16 2005

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

			Triangle T(k,s) begins:
k\s  0    1     2     3     4     5     6    7    8   9 10
0:   1
1:   3    1
2:   5    5     1
3:   7   14     7     1
4:   9   30    27     9     1
5:  11   55    77    44    11     1
6:  13   91   182   156    65    13     1
7:  15  140   378   450   275    90    15    1
8:  17  204   714  1122   935   442   119   17    1
9:  19  285  1254  2508  2717  1729   665  152   19   1
10: 21  385  2079  5148  7007  5733  2940  952  189  21  1
... Extended and reformatted by _Wolfdieter Lang_, Oct 18 2012
Application for Fibonacci numbers F_{(2*k+1)*n}, row k=3:
F_{7*n} = 7*(-1)^(3*n)*F_n + 14*(-1)^(4*n)*5*F_n^3 + 7*(-1)^(5*n)*5^2*F_n^5 + 1*(-1)^(6*n)*5^3*F_n^7, n>=0. - _Wolfdieter Lang_, Aug 24 2012
Example for the  Z- and A-sequence recurrences  of this Riordan triangle: Z = A217477 = [3,-4,12,-40,...]; T(4,0) = 3*7 -4*14 +12*7 -40*1 = 9. A =  [1, 2, -1, 2, -5, 14, ..]; T(5,2) = 1*30 + 2*27 - 1*9 + 2*1= 77. _Wolfdieter Lang_, Oct 18 2012
Example for the (4*(k+1))-gon length ratio s(4*(k+1))(side/radius) as polynomial in the ratio rho(4*(k+1)) ((smallest diagonal)/side): k=0, s(4) = 1*rho(4) = sqrt(2); k=1, s(8) = -3*rho(8) + rho(8)^3 = sqrt(2-sqrt(2)); k=2, s(12) = 5*rho(12) - 5*rho(12)^3 + rho(12)^5, and C(12,x) = x^4 - 4*x^2 + 1, hence rho(12)^5 = 4*rho(12)^3 - rho(12), and s(12) = 4*rho(12) - rho(12)^3 = sqrt(2 - sqrt(3)). - _Wolfdieter Lang_, Oct 04 2013
Example for the recurrence for the signed triangle S(k,s)= ((-1)^(k-s))*T(k,s) (see the Aug 14 2014 comment above):
S(4,1) = 0 + (-2*2 - 1)*S(3,1) - (1/2)*(3*4^2*S(3,2) + 4*4^3*S(3,3)) = - 5*14 - 3*8*(-7) - 128*1 = -30. The recurrence from the Riordan A-sequence A115141 is S(4,1) = -7 -2*14 -(-7) -2*1 = -30. - _Wolfdieter Lang_, Aug 14 2014

G. C. Greubel, Rows n = 0..50 of the triangle, flattened
R. Andre-Jeannin, A generalization of Morgan-Voyce polynomials, The Fibonacci Quarterly 32.3 (1994): 228-31.
Alexander Burstein and Louis W. Shapiro, Pseudo-involutions in the Riordan group, arXiv:2112.11595 [math.CO], 2021.
Alexander Burstein and Louis W. Shapiro, Pseudo-involutions in the Riordan group and Chebyshev polynomials, arXiv:2502.13673 [math.CO], 2025.
K. Dilcher and K. B. Stolarsky, A Pascal-type triangle characterizing twin primes, Amer. Math. Monthly, 112 (2005), 673-681.
P. Damianou , On the characteristic polynomials of Cartan matrices and Chebyshev polynomials, arXiv preprint arXiv:1110.6620 [math.RT], 2014.
P. Damianou and C. Evripidou, Characteristic and Coxeter polynomials for affine Lie algebras, arXiv preprint arXiv:1409.3956 [math.RT], 2014.
D. Jennings, Some Polynomial Identities for the Fibonacci and Lucas Numbers, Fib. Quart., 31(2) (1993), 134-137.
D. E. Knuth, Johann Faulhaber and sums of powers, Math. Comp. 61 (1993), no. 203, 277-294.
Yidong Sun, Numerical triangles and several classical sequences, Fib. Quart., Nov. 2005, pp. 359-370.
T. Wang and W. Zhang, Some identities involving Fibonacci, Lucas polynomials and their applications, Bull. Math. Soc. Sci. Math. Roumanie, Tome 55(103), No.1, (2012) 95-103.
Eric Weisstein's World of Mathematics, Morgan-Voyce polynomials

Mirror image of A082985, which see for further references, etc.
Also closely related to triangles in A098599 and A100218.
Cf. A052227, A085478, A208513, A208508, A211957
Cf. A034807, A127672, A127677, A263196.

[((2*n+1)/(n+k+1))*Binomial(n+k+1, 2*k+1): k in [0..n], n in [0..12]];  // G. C. Greubel, Feb 01 2022

(* First program *)
u[1, x_]:=1; v[1, x_]:=1; z=16;
u[n_, x_]:= u[n-1, x] + x*v[n-1, x];
v[n_, x_]:= u[n-1, x] + (x+1)*v[n-1, x] + 1;
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]  (* A208513 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]  (* A111125 *) (* Clark Kimberling, Feb 28 2012 *)
(* Second program *)
T[n_, k_]:= ((2*n+1)/(2*k+1))*Binomial[n+k, 2*k];
Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 01 2022 *)

@CachedFunction
def T(n,k):
    if n< 0: return 0
    if n==0: return 1 if k == 0 else 0
    h = 3*T(n-1,k) if n==1 else 2*T(n-1,k)
    return T(n-1,k-1) - T(n-2,k) - h
A111125 = lambda n,k: (-1)^(n-k)*T(n,k)
for n in (0..9): [A111125(n,k) for k in (0..n)] # Peter Luschny, Nov 20 2012

T(k,s) = ((2*k+1)/(2*s+1))*binomial(k+s,2*s), 0 <= s <= k.
From Peter Bala, Apr 30 2012: (Start)
T(n,k) = binomial(n+k,2*k) + 2*binomial(n+k,2*k+1).
The row generating polynomials P(n,x) are a generalization of the Morgan-Voyce polynomials b(n,x) and B(n,x). They satisfy the recurrence equation P(n,x) = (x+2)*P(n-1,x) - P(n-2,x) for n >= 2, with initial conditions P(0,x) = 1, P(1,x) = x+r+1 and with r = 2. The cases r = 0 and r = 1 give the Morgan-Voyce polynomials A085478 and A078812 respectively. Andre-Jeannin has considered the case of general r.
P(n,x) = U(n+1,1+x/2) + U(n,1+x/2), where U(n,x) denotes the Chebyshev polynomial of the second kind - see A053117. P(n,x) = (2/x)*(T(2*n+2,u)-T(2*n,u)), where u = sqrt((x+4)/4) and T(n,x) denotes the Chebyshev polynomial of the first kind - see A053120. P(n,x) = Product_{k = 1..n} ( x + 4*(sin(k*Pi/(2*n+1)))^2 ). P(n,x) = 1/x*(b(n+1,x) - b(n-1,x)) and P(n,x) = 1/x*{(b(2*n+2,x)+1)/b(n+1,x) - (b(2*n,x)+1)/b(n,x)}, where b(n,x) := Sum_{k = 0..n} binomial(n+k,2*k)*x^k are the Morgan-Voyce polynomials of A085478. Cf. A211957.
(End)
From Wolfdieter Lang, Oct 18 2012 (Start)
O.g.f. column No. s: ((1+x)/(1-x)^2)*(x/(1-x)^2)^s, s >= 0. (from the Riordan data given in a comment above).
O.g.f. of the row polynomials R(k,x):= Sum_{s=0..k} ( T(k,s)*x^s ), k>=0: (1+z)/(1-(2+x)*z+z^2) (from the Riordan property).
(End)
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k), T(0,0) = 1, T(1,0) = 3, T(1,1) = 1, T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Nov 12 2013

More terms from Paul Barry, Oct 17 2005

A100258 Triangle of coefficients of normalized Legendre polynomials, with increasing exponents.

Original entry on oeis.org

1, 0, 1, -1, 0, 3, 0, -3, 0, 5, 3, 0, -30, 0, 35, 0, 15, 0, -70, 0, 63, -5, 0, 105, 0, -315, 0, 231, 0, -35, 0, 315, 0, -693, 0, 429, 35, 0, -1260, 0, 6930, 0, -12012, 0, 6435, 0, 315, 0, -4620, 0, 18018, 0, -25740, 0, 12155, -63, 0, 3465, 0, -30030, 0, 90090, 0, -109395, 0, 46189

Offset: 0

Ralf Stephan, Nov 13 2004

For a relation to Jacobi quartic elliptic curves, see the MathOverflow link. For a self-convolution of the polynomials relating them to the Chebyshev and Fibonacci polynomials, see A049310 and A053117. For congruences and connections to other polynomials (Jacobi, Gegenbauer, and Chebyshev) see the Allouche et al. link. For relations to elliptic cohomology and modular forms, see references in Copeland link.- Tom Copeland, Feb 04 2016

			Triangle begins:
   1;
   0,   1;
  -1,   0,     3;
   0,  -3,     0,   5;
   3,   0,   -30,   0,   35;
   0,  15,     0, -70,    0,   63;
  -5,   0,   105,   0, -315,    0,    231;
   0, -35,     0, 315,    0, -693,      0, 429;
  35,   0, -1260,   0, 6930,    0, -12012,   0, 6435;
  ...

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 798.

T. D. Noe, Rows n=0..100 of triangle, flattened
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
J. Allouche and G. Skordev, Schur congruences, Carlitz sequences of polynomials and automaticity, Discrete Mathematics, Vol. 214, Issue 1-3, 21 March 2000, p. 21-49.
Tom Copeland, The Elliptic Lie Triad: Riccati and KdV Equations, Infinigens, and Elliptic Genera
H. N. Laden, An historical, and critical development of the theory of Legendre polynomials before 1900, Master of Arts Thesis, University of Maryland 1938.
Shi-Mei Ma, On gamma-vectors and the derivatives of the tangent and secant functions, arXiv:1304.6654 [math.CO], 2013.
MathOverflow, Geometric picture of invariant differential of an elliptic curve, Dec 4 2011.

Without zeros: A008316. Row sums are A060818.
Columns (with interleaved zeros and signs) include A001790, A001803, A100259. Diagonals include A001790, A001800, A001801, A001802.
Cf. A049310, A005187, A049600, A053117.

row[n_] := CoefficientList[ LegendreP[n, x], x]*2^IntegerExponent[n!, 2]; Table[row[n], {n, 0, 10}] // Flatten (* Jean-François Alcover, Jan 15 2015 *)

a(k,n)=polcoeff(pollegendre(k,x),n)*2^valuation(k!,2)

from mpmath import *
mp.dps=20
def a007814(n):
    return 1 + bin(n - 1)[2:].count('1') - bin(n)[2:].count('1')
for n in range(11):
    y=2**sum(a007814(i) for i in range(2, n+1))
    l=chop(taylor(lambda x: legendre(n, x), 0, n))
    print([int(i*y) for i in l]) # Indranil Ghosh, Jul 02 2017

The n-th normalized Legendre polynomial is generated by 2^(-n-a(n)) (d/dx)^n (x^2-1)^n / n! with a(n) = A005187(n/2) for n even and a(n) = A005187((n-1)/2) for n odd. The non-normalized polynomials have the o.g.f. 1 / sqrt(1 - 2xz + z^2). - Tom Copeland, Feb 07 2016

The consecutive nonzero entries in the m-th row are, in order, (c+b)!/(c!(m-b)!(2b-m)!*A048896(m-1)) with sign (-1)^b where c = m/2-1, m/2, m/2+1, ..., (m-1) and b = c+1 if m is even and sign (-1)^c with c = (m-1)/2, (m-1)/2+1, (m-1)/2+2, ..., (m-1) with b = c+1 if m is odd. For the 9th row the 5 consecutive nonzero entries are 315, -4620, 18018, -25740, 12155 given by c = 4,5,6,7,8 and b = 5,6,7,8,9. - Richard Turk, Aug 22 2017

A133156 Irregular triangle read by rows: coefficients of U(n,x), Chebyshev polynomials of the second kind with exponents in decreasing order.

Original entry on oeis.org

1, 2, 4, -1, 8, -4, 16, -12, 1, 32, -32, 6, 64, -80, 24, -1, 128, -192, 80, -8, 256, -448, 240, -40, 1, 512, -1024, 672, -160, 10, 1024, -2304, 1792, -560, 60, -1, 2048, -5120, 4608, -1792, 280, -12, 4096, -11264, 11520, -5376, 1120, -84, 1

Offset: 0

Gary W. Adamson, Dec 16 2007

The Chebyshev polynomials of the second kind are defined by the recurrence relation: U(0,x) = 1; U(1,x) = 2x; U(n+1,x) = 2x*U(n,x) - U(n-1,x).
From Gary W. Adamson, Nov 28 2008: (Start)
Triangle read by rows, unsigned = A000012 * A028297.
Row sums of absolute values give the Pell series, A000129.
(End)
The row sums are {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ...}.
Triangle, with zeros omitted, given by (2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Dec 27 2011
Coefficients in the expansion of sin((n+1)*x)/sin(x) in descending powers of cos(x). The length of the n-th row is A008619(n). - Jianing Song, Nov 02 2018

			The first few Chebyshev polynomials of the second kind are
    1;
    2x;
    4x^2 -    1;
    8x^3 -    4x;
   16x^4 -   12x^2 +   1;
   32x^5 -   32x^3 +   6x;
   64x^6 -   80x^4 +  24x^2 -   1;
  128x^7 -  192x^5 +  80x^3 -   8x;
  256x^8 -  448x^6 + 240x^4 -  40x^2 +  1;
  512x^9 - 1024x^7 + 672x^5 - 160x^3 + 10x;
  ...
From _Roger L. Bagula_ and _Gary W. Adamson_: (Start)
     1;
     2;
     4,    -1;
     8,    -4;
    16,   -12,    1;
    32,   -32,    6;
    64,   -80,   24,   -1;
   128,  -192,   80,   -8;
   256,  -448,  240,  -40,  1;
   512, -1024,  672, -160, 10;
  1024, -2304, 1792, -560, 60, -1; (End)
From  _Philippe Deléham_, Dec 27 2011: (Start)
Triangle (2, 0, 0, 0, 0, ...) DELTA (0, -1/2, 1/2, 0, 0, 0, 0, 0, ...) begins:
   1;
   2,   0;
   4,  -1,  0;
   8,  -4,  0,  0;
  16, -12,  1,  0,  0;
  32, -32,  6,  0,  0,  0;
  64, -80, 24, -1,  0,  0,  0; (End)

Tracale Austin, Hans Bantilan, Isao Jonas and Paul Kory, The Pfaffian Transformation, Journal of Integer Sequences, Vol. 12 (2009), page 25
P. Damianou, On the characteristic polynomials of Cartan matrices and Chebyshev polynomials, arXiv preprint arXiv:1110.6620 [math.RT], 2014. - From _Tom Copeland_, Oct 11 2014
Pantelis A. Damianou, A Beautiful Sine Formula, Amer. Math. Monthly 121 (2014), no. 2, 120-135. MR3149030
Caglar Koca and Ozgur B. Akan, Modelling 1D Partially Absorbing Boundaries for Brownian Molecular Communication Channels, arXiv:2402.15888 [q-bio.MN], 2024. See p. 9.
Wikipedia, Chebyshev polynomials

Cf. A038207, A053117.
Cf. A018297, A000129. - Gary W. Adamson, Nov 28 2008
Cf. A000079, A001787, A001788, A001789, A003472, A054849, A002409, A054851. - Philippe Deléham, Sep 12 2009
Cf. A091894, A097610, A099089, A207538.

t[n_, m_] = (-1)^m*Binomial[n - m, m]*2^(n - 2*m);
Table[Table[t[n, m], {m, 0, Floor[n/2]}], {n, 0, 10}];
Flatten[%] (* Roger L. Bagula, Dec 19 2008 *)

A generating function for U(n) is 1/(1 - 2tx + t^2). Given A038207, shift down columns to allow for (1, 1, 2, 2, 3, 3, ...) terms in each row, then insert alternate signs.
T(n,m) = (-1)^m*binomial(n - m, m)*2^(n - 2*m). - Roger L. Bagula and Gary W. Adamson, Dec 19 2008
From Tom Copeland, Feb 11 2016: (Start)
Shifted o.g.f.: G(x,t) = x/(1 - 2x + tx^2).
A053117 is a reflected, aerated version of this entry; A207538, an unsigned version; and A099089, a reflected, shifted version.
The compositional inverse of G(x,t) is Ginv(x,t) = ((1 + 2x) - sqrt((1 + 2x)^2 - 4tx^2))/(2tx) = x - 2x^2 + (4 + t)x^3 - (8 + 6t)x^4 + ..., a shifted o.g.f. for A091894 (mod signs with A091894(0,0) = 0.). Cf. A097610 with h_1 = -2 and h_2 = t. (End)

More terms from Philippe Deléham, Sep 12 2009

A207538 Triangle of coefficients of polynomials v(n,x) jointly generated with A207537; see Formula section.

Original entry on oeis.org

1, 2, 4, 1, 8, 4, 16, 12, 1, 32, 32, 6, 64, 80, 24, 1, 128, 192, 80, 8, 256, 448, 240, 40, 1, 512, 1024, 672, 160, 10, 1024, 2304, 1792, 560, 60, 1, 2048, 5120, 4608, 1792, 280, 12, 4096, 11264, 11520, 5376, 1120, 84, 1, 8192, 24576, 28160, 15360

Offset: 1

Clark Kimberling, Feb 18 2012

As triangle T(n,k) with 0<=k<=n and with zeros omitted, it is the triangle given by (2, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1/2, -1/2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 04 2012
The numbers in rows of the triangle are along "first layer" skew diagonals pointing top-left in center-justified triangle given in A013609 ((1+2*x)^n) and  along (first layer) skew diagonals pointing top-right in center-justified triangle given in A038207 ((2+x)^n), see links. - Zagros Lalo, Jul 31 2018
If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 2.414213562373095... (A014176: Decimal expansion of the silver mean, 1+sqrt(2)), when n approaches infinity. - Zagros Lalo, Jul 31 2018

			First seven rows:
1
2
4...1
8...4
16..12..1
32..32..6
64..80..24..1
(2, 0, 0, 0, 0, ...) DELTA (0, 1/2, -1/2, 0, 0, 0, ...) begins:
    1
    2,   0
    4,   1,  0
    8,   4,  0, 0
   16,  12,  1, 0, 0
   32,  32,  6, 0, 0, 0
   64,  80, 24, 1, 0, 0, 0
  128, 192, 80, 8, 0, 0, 0, 0

Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 80-83, 357-358.

Jean-Luc Baril and José Luis Ramírez, Descent distribution on Catalan words avoiding ordered pairs of Relations, arXiv:2302.12741 [math.CO], 2023.
S. Halici, On some Pell polynomials , Acta Universitatis Apulensis, No. 29/2012, pp. 105-112.
Zagros Lalo, First layer skew diagonals in center-justified triangle of coefficients in expansion of (1 + 2x)^n
Zagros Lalo, First layer skew diagonals in center-justified triangle of coefficients in expansion of (2 + x)^n

Cf. A028297, A207537, A133156, A038207, A099089.
Cf. A053117, A097610, A091894.
Cf. A013609, A038207.
Cf. A128099.

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := u[n - 1, x] + (x + 1)*v[n - 1, x]
v[n_, x_] := u[n - 1, x] + v[n - 1, x]
Table[Factor[u[n, x]], {n, 1, z}]
Table[Factor[v[n, x]], {n, 1, z}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]  (* A207537, |A028297| *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]  (* A207538, |A133156| *)
t[0, 0] = 1; t[n_, k_] := t[n, k] = If[n < 0 || k < 0, 0, 2 t[n - 1, k] + t[n - 2, k - 1]]; Table[t[n, k], {n, 0, 15}, {k, 0, Floor[n/2]}] // Flatten (* Zagros Lalo, Jul 31 2018 *)
t[n_, k_] := t[n, k] = 2^(n - 2 k) * (n -  k)!/((n - 2 k)! k!) ; Table[t[n, k], {n, 0, 15}, {k, 0, Floor[n/2]} ]  // Flatten (* Zagros Lalo, Jul 31 2018 *)

u(n,x) = u(n-1,x)+(x+1)*v(n-1,x), v(n,x) = u(n-1,x)+v(n-1,x), where u(1,x) = 1, v(1,x) = 1. Also, A207538 = |A133156|.
From Philippe Deléham, Mar 04 2012: (Start)
With 0<=k<=n:
Mirror image of triangle in A099089.
Skew version of A038207.
Riordan array (1/(1-2*x), x^2/(1-2*x)).
G.f.: 1/(1-2*x-y*x^2).
Sum_{k, 0<=k<=n} T(n,k)*x^k = A190958(n+1), A127357(n), A090591(n), A089181(n+1), A088139(n+1), A045873(n+1), A088138(n+1), A088137(n+1), A099087(n), A000027(n+1), A000079(n), A000129(n+1), A002605(n+1), A015518(n+1), A063727(n), A002532(n+1), A083099(n+1), A015519(n+1), A003683(n+1), A002534(n+1), A083102(n), A015520(n+1), A091914(n) for x = -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 respectively.
T(n,k) = 2*T(n-1,k) + T(-2,k-1) with T(0,0) = 1, T(1,0) = 2, T(1,1) = 0 and T(n, k) = 0 if k<0 or if k>n. (End)
T(n,k) = A013609(n-k, n-2*k+1). - Johannes W. Meijer, Sep 05 2013
From Tom Copeland, Feb 11 2016: (Start)
A053117 is a reflected, aerated and signed version of this entry. This entry belongs to a family discussed in A097610 with parameters h1 = -2 and h2 = -y.
Shifted o.g.f.: G(x,t) = x / (1 - 2 x - t x^2).
The compositional inverse of G(x,t) is Ginv(x,t) = -[(1 + 2x) - sqrt[(1+2x)^2 + 4t x^2]] / (2tx) = x - 2 x^2 + (4-t) x^3 - (8-6t) x^4 + ..., a shifted o.g.f. for A091894 (mod signs with A091894(0,0) = 0).
(End)

A137276 Triangle T(n,k), read by rows: T(n,k)= 0 if n-k odd. T(n,k)= 2(-1)^((n-k)/2)(2k-n)/(n+k)*binomial((n+k)/2,(n-k)/2) if n-k even.

Original entry on oeis.org

1, 0, 1, 2, 0, 1, 0, 1, 0, 1, -2, 0, 0, 0, 1, 0, -3, 0, -1, 0, 1, 2, 0, -3, 0, -2, 0, 1, 0, 5, 0, -2, 0, -3, 0, 1, -2, 0, 8, 0, 0, 0, -4, 0, 1, 0, -7, 0, 10, 0, 3, 0, -5, 0, 1, 2, 0, -15, 0, 10, 0, 7, 0, -6, 0, 1, 0, 9, 0, -25, 0, 7, 0, 12, 0, -7, 0, 1, -2, 0, 24, 0, -35, 0, 0, 0, 18, 0, -8, 0, 1, 0, -11, 0, 49, 0, -42, 0, -12, 0

Offset: 0

Roger L. Bagula and Gary W. Adamson, Mar 13 2008

Polynomial coefficients of P(n,x) in increasing powers, read by rows, where P(0,x)=1, P(1,x)=x, P(2,x)=2+x^2, P(3,x)=x+x^3, P(4,x)=-2+x^4, and  P(n,x) = x*P(n-1,x) - P(n-2,x) for n>=5.
The row-reversed version of A135929.
Row sums are repeating 1, 1, 3, 2, -1, -3, -2, 1, 3, 2, -1..., see A138034 and A119910.

			{1}, = 1
{0, 1}, = x
{2, 0, 1}, = 2+x^2
{0, 1, 0, 1}, = x+x^3
{-2, 0, 0, 0, 1}, = -2+x^4
{0, -3, 0, -1, 0, 1}, = -3x-x^3+x^5
{2, 0, -3, 0, -2, 0, 1},
{0, 5, 0, -2, 0, -3, 0, 1},
{-2, 0, 8, 0, 0, 0, -4, 0, 1},
{0, -7, 0, 10, 0, 3, 0, -5, 0, 1},
{2, 0, -15, 0, 10, 0, 7, 0, -6, 0, 1},
{0, 9, 0, -25, 0, 7, 0, 12, 0, -7, 0, 1}

P. Steinbach, Golden fields: a case for the heptagon, Math. Mag. 70 (1997), no. 1, 22-31, MR 1439165

Cf. A123956, A137289.

A137276 := proc(n,k) local nmk,npk; if n = 0 then 1; elif (n-k) mod 2 <> 0 then 0; else nmk := (n-k)/2 ; npk := (n+k)/2 ; (-1)^nmk*(2*k-n)/npk*binomial(npk,nmk) ; fi; end:
seq( seq(A137276(n,k),k=0..n),n=0..13) ;

T(n,k)= 0 if n-k odd. T(n,k)= 2*(-1)^((n-k)/2)*(2k-n)/(n+k)*binomial((n+k)/2,(n-k)/2) if n-k even.
P(n,x) = x*P(n-1,x)-P(n-2,x), n>=5.
P(n,2*x) = -2*T(n,x)+4*x*U(n-1,x), where T(n,x) is A053120 and U(n,x) is A053117.

Fourth row inserted by the Associate Editors of the OEIS, Aug 27 2009

A187596 Array T(m,n) read by antidiagonals: number of domino tilings of the m X n grid (m>=0, n>=0).

Original entry on oeis.org

1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 2, 0, 1, 1, 1, 3, 3, 1, 1, 1, 0, 5, 0, 5, 0, 1, 1, 1, 8, 11, 11, 8, 1, 1, 1, 0, 13, 0, 36, 0, 13, 0, 1, 1, 1, 21, 41, 95, 95, 41, 21, 1, 1, 1, 0, 34, 0, 281, 0, 281, 0, 34, 0, 1, 1, 1, 55, 153, 781, 1183, 1183, 781, 153, 55, 1, 1, 1, 0, 89, 0, 2245, 0, 6728, 0, 2245, 0, 89, 0, 1, 1, 1, 144, 571, 6336

Offset: 0

N. J. A. Sloane, Mar 11 2011

A099390 supplemented by an initial row and column of 1's.
See A099390 (the main entry for this array) for further information.
If we work with the row index starting at 1 then every row of the array is a divisibility sequence, i.e., the terms satisfy the property that if n divides m then a(n) divide a(m) provided a(n) != 0. Row k satisfies a linear recurrence of order 2^floor(k/2) (Stanley, Ex. 36 p. 273). - Peter Bala, Apr 30 2014

			Array begins:
  1,  1,  1,  1,   1,    1,     1,     1,       1,      1,        1, ...
  1,  0,  1,  0,   1,    0,     1,     0,       1,      0,        1, ...
  1,  1,  2,  3,   5,    8,    13,    21,      34,     55,       89, ...
  1,  0,  3,  0,  11,    0,    41,     0,     153,      0,      571, ...
  1,  1,  5, 11,  36,   95,   281,   781,    2245,   6336,    18061, ...
  1,  0,  8,  0,  95,    0,  1183,     0,   14824,      0,   185921, ...
  1,  1, 13, 41, 281, 1183,  6728, 31529,  167089, 817991,  4213133, ...
  1,  0, 21,  0, 781,    0, 31529,     0, 1292697,      0, 53175517, ...

R. P. Stanley, Enumerative Combinatorics, Vol. 1, Cambridge University Press, 1997.

Alois P. Heinz, Antidiagonals n = 0..80, flattened
James Propp, Enumeration of Matchings: Problems and Progress, arXiv:math/9904150 [math.CO], 1999.
Eric Weisstein's World of Mathematics, Chebyshev Polynomial of the second kind.
Eric Weisstein's World of Mathematics, Fibonacci Polynomial.

Cf. A099390.
See A187616 for a triangular version, and A187617, A187618 for the sub-array T(2m,2n).
See also A049310, A053117.

with(LinearAlgebra):
T:= proc(m,n) option remember; local i, j, t, M;
      if m<=1 or n<=1 then 1 -irem(n*m, 2)
    elif irem(n*m, 2)=1 then 0
    elif mAlois P. Heinz, Apr 11 2011

t[m_, n_] := Product[2*(2+Cos[2*j*Pi/(m+1)]+Cos[2*k*Pi/(n+1)]), {k, 1, n/2}, {j, 1, m/2}]; t[?OddQ, ?OddQ] = 0; Table[t[m-n, n] // FullSimplify, {m, 0, 13}, {n, 0, m}] // Flatten (* Jean-François Alcover, Jan 07 2014, after A099390 *)

From Peter Bala, Apr 30 2014: (Start)
T(n,k)^2 = absolute value of Product_{b=1..k} Product_{a=1..n} ( 2*cos(a*Pi/(n+1)) + 2*i*cos(b*Pi/(k+1)) ), where i = sqrt(-1). See Propp, Section 5.
Equivalently, working with both the row index n and column index k starting at 1 we have T(n,k)^2 = absolute value of Resultant (F(n,x), U(k-1,x/2)), where U(n,x) is a Chebyshev polynomial of the second kind and F(n,x) is a Fibonacci polynomial defined recursively by F(0,x) = 0, F(1,x) = 1 and F(n,x) = x*F(n-1,x) + F(n-2,x) for n >= 2. The divisibility properties of the array entries mentioned in the Comments are a consequence of this result. (End)

A008312 Triangle of coefficients of Chebyshev polynomials U_n(x).

Original entry on oeis.org

1, 2, -1, 4, -4, 8, 1, -12, 16, 6, -32, 32, -1, 24, -80, 64, -8, 80, -192, 128, 1, -40, 240, -448, 256, 10, -160, 672, -1024, 512, -1, 60, -560, 1792, -2304, 1024, -12, 280, -1792, 4608, -5120, 2048

Offset: 0

N. J. A. Sloane

Version with zeros in A053117. - Philippe Deléham, Nov 27 2013

			From _Philippe Deléham_, Nov 27 2013: (Start)
Triangle begins:
    1;
    2;
   -1,    4;
   -4,    8;
    1,  -12,    16;
    6,  -32,    32;
   -1,   24,   -80,    64;
   -8,   80,  -192,   128;
    1,  -40,   240,  -448,   256;
   10, -160,   672, -1024,   512;
   -1,   60,  -560,  1792, -2304, 1024;
  -12,  280, -1792,  4608, -5120, 2048;
  ...
With zeros, triangle begins:
   1;
   0,   2;
  -1,   0,   4;
   0,  -4,   0,    8;
   1,   0, -12,    0,   16;
   0,   6,   0,  -32,    0,    32;
  -1,   0,  24,    0,  -80,     0,   64;
   0,  -8,   0,   80,    0,  -192,    0,   128;
   1,   0, -40,    0,  240,     0, -448,     0,   256;
   0,  10,   0, -160,    0,   672,    0, -1024,     0,   512;
  -1,   0,  60,    0, -560,     0, 1792,     0, -2304,     0, 1024;
   0, -12,   0,  280,    0, -1792,    0,  4608,     0, -5120,    0, 2048;
  ...
(End)

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 796.

T. D. Noe, Rows n = 0..100 of triangle, flattened
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
D. Foata and G.-N. Han, Nombres de Fibonacci et polynomes orthogonaux.
Valentin Ovsienko, Towards quantized complex numbers: q-deformed Gaussian integers and the Picard group, arXiv:2103.10800 [math.QA], 2021.
M. Janjic and B. Petkovic, A Counting Function, arXiv preprint arXiv:1301.4550, 2013. - From _N. J. A. Sloane_, Feb 13 2013
Index entries for sequences related to Chebyshev polynomials.

Reflected version with zeros: A053118. Cf. A008310, A053112, A053117.

a[n_, k_] := Coefficient[ ChebyshevU[n, x], x, k]; row[n_] := Table[a[n, k], {k, Mod[n, 2], n, 2}]; Table[row[n], {n, 0, 11}] // Flatten (* Jean-François Alcover, Oct 03 2012 *)

A239473 Triangle read by rows: signed version of A059260: coefficients for expansion of partial sums of sequences a(n,x) in terms of their binomial transforms (1+a(.,x))^n ; Laguerre polynomial expansion of the truncated exponential.

Original entry on oeis.org

1, 0, 1, 1, -1, 1, 0, 2, -2, 1, 1, -2, 4, -3, 1, 0, 3, -6, 7, -4, 1, 1, -3, 9, -13, 11, -5, 1, 0, 4, -12, 22, -24, 16, -6, 1, 1, -4, 16, -34, 46, -40, 22, -7, 1, 0, 5, -20, 50, -80, 86, -62, 29, -8, 1, 1, -5, 25, -70, 130, -166, 148, -91, 37, -9, 1, 0, 6, -30, 95, -200, 296, -314, 239, -128, 46, -10, 1

Offset: 0

Tom Copeland, Mar 19 2014

With T the lower triangular array above and the Laguerre polynomials L(k,x) = Sum_{j=0..k} (-1)^j binomial(k, j) x^j/j!, the following identities hold:
(A) Sum_{k=0..n} (-1)^k L(k,x) = Sum_{k=0..n} T(n,k) x^k/k!;
(B) Sum_{k=0..n} x^k/k! = Sum_{k=0..n} T(n,k) L(k,-x);
(C) Sum_{k=0..n} x^k = Sum_{k=0..n} T(n,k) (1+x)^k = (1-x^(n+1))/(1-x).
More generally, for polynomial sequences,
(D) Sum_{k=0..n} P(k,x) = Sum_{k=0..n} T(n,k) (1+P(.,x))^k,
where, e.g., for an Appell sequence, such as the Bernoulli polynomials, umbrally, (1+ Ber(.,x))^k = Ber(k,x+1).
Identity B follows from A through umbral substitution of j!L(j,-x) for x^j in A. Identity C, related to the cyclotomic polynomials for prime index, follows from B through the Laplace transform.
Integrating C gives Sum_{k=0..n} T(n,k) (2^(k+1)-1)/(k+1) = H(n+1), the harmonic numbers.
Identity A >= 0 for x >= 0 (see MathOverflow link for evaluation in terms of Hermite polynomials).
From identity C, W(m,n) = (-1)^n Sum_{k=0..n} T(n,k) (2-m)^k = number of walks of length n+1 between any two distinct vertices of the complete graph K_m for m > 2.
Equals A112468 with the first column of ones removed. - Georg Fischer, Jul 26 2023

			Triangle begins:
   1
   0    1
   1   -1    1
   0    2   -2    1
   1   -2    4   -3    1
   0    3   -6    7   -4    1
   1   -3    9  -13   11   -5    1
   0    4  -12   22  -24   16   -6    1
   1   -4   16  -34   46  -40   22   -7    1
   0    5  -20   50  -80   86  -62   29   -8    1
   1   -5   25  -70  130 -166  148  -91   37   -9    1

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
J. Adams, On the groups J(x)-II, Topology, Vol. 3, p. 137-171, Pergamon Press, (1965).
MathOverflow, Cyclotomic Polynomials in Combinatorics
Mathoverflow, Inequality for Laguerre polynomials

For column 2: A001057, A004526, A008619, A140106.
Column 3: A002620, A087811.
Column 4: A002623, A173196.
Column 5: A001752.
Column 6: A001753.
Cf. Bottomley's cross-references in A059260.
Embedded in alternating antidiagonals of T are the reversals of arrays A071921 (A225010) and A210220.
Cf. A049310, A112468, A341091.

[[(&+[(-1)^(j+k)*Binomial(j,k): j in [0..n]]): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Feb 06 2018

A239473 := proc(n,k)
    add(binomial(j,k)*(-1)^(j+k),j=k..n) ;
end proc; # R. J. Mathar, Jul 21 2016

Table[Sum[(-1)^(j+k)*Binomial[j,k], {j,0,n}], {n,0,10}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 06 2018 *)

for(n=0,10, for(k=0,n, print1(sum(j=0,n, (-1)^(j+k)*binomial(j, k)), ", "))) \\ G. C. Greubel, Feb 06 2018

Trow = lambda n: sum((x-1)^j for j in (0..n)).list()
for n in (0..10): print(Trow(n)) # Peter Luschny, Jul 09 2019

T(n, k) = Sum_{j=0..n} (-1)^(j+k) * binomial(j, k).
E.g.f: (exp(t) - (x-1)*exp((x-1)*t))/(2-x).
O.g.f. (n-th row): (1-(x-1)^(n+1))/(2-x).
Associated operator identities:
With D=d/dx, :xD:^n=x^n*D^n, and :Dx:^n=D^n*x^n, then bin(xD,n)= binomial(xD,n)=:xD:^n/n! and L(n,-:xD:)=:Dx:^n/n!=bin(xD+n,n)=(-1)^n bin(-xD-1,n),
A-o) Sum_{k=0..n} (-1)^k L(k,-:xD:) = Sum_{k=0..n} :-Dx:^k/k!
     = Sum_{k=0..n} T(n,k) :-xD:^k/k! = Sum_{k=0..n} (-1)^k T(n,k)bin(xD,k)
B-o) Sum_{k=0..n} :xD:^k/k! = Sum_{k=0..n}, T(n,k) L(k,-:xD:)
     = Sum_{k=0..n} T(n,k) :Dx:^k/k! = Sum_{k=0..n}, bin(xD,k).
Associated binomial identities:
A-b) Sum_{k=0..n} (-1)^k bin(s+k,k) = Sum_{k=0..n} (-1)^k T(n,k) bin(s,k)
     = Sum_{k=0..n} bin(-s-1,k) = Sum{k=0..n} T(n,k) bin(-s-1+k,k)
B-b) Sum_{k=0..n} bin(s,k) = Sum_{k=0..n} T(n,k) bin(s+k,k)
     = Sum_{k=0..n} (-1)^k bin(-s-1+k,k)
     = Sum_{k=0..n} (-1)^k T(n,k) bin(-s-1,k).
In particular, from B-b with s=n, Sum_{k=0..n} T(n,k) bin(n+k,k) = 2^n. From B-b with s=0, row sums are all 1.
From identity C with x=-2, the unsigned row sums are the Jacobsthal sequence, i.e., Sum_{k=0..n} T(n,k) (1+(-2))^k = (-1)^n A001045(n+1); for x=2, the Mersenne numbers A000225; for x=-3, A014983 or signed A015518; for x=3, A003462; for x=-4, A014985 or signed A015521; for x=4, A002450; for x=-5, A014986 or signed A015531; and for x=5, A003463; for x=-6, A014987 or signed A015540; and for x=6, A003464.
With -s-1 = m = 0,1,2,..., B-b gives finite differences (recursions):
Sum_{k=0..n} (-1)^k T(n,k) bin(m,k) = Sum_{k=0..n} (-1)^k bin(m+k,k) = T(n+m,m), i.e., finite differences of the columns of T generate shifted columns of T. The columns of T are signed, shifted versions of sequences listed in the cross-references. Since the finite difference is an involution, T(n,k) = Sum_{j=0..k} (-1)^j T(n+j,j) bin(k,j)}. Gauss-Newton interpolation can be applied to give a generalized T(n,s) for s noninteger.
From identity C, S(n,m) = Sum_{k=0..n} T(n,k) bin(k,m) = 1 for m < n+1 and 0 otherwise, i.e., S = T*P, where S = A000012, as a lower triangular matrix and P = Pascal = A007318, so T = S*P^(-1), where P^(-1) = A130595, the signed Pascal array (see A132440), the inverse of P, and T^(-1) = P*S^(-1) = P*A167374 = A156644.
U(n,cos(x)) = e^(-n*i*x)*Sum_{k=0..n} T(n,k)*(1+e^(2*i*x))^k = sin((n+1)x)/sin(x), where U is the Chebyschev polynomial of the second kind A053117 and i^2 = -1. - Tom Copeland, Oct 18 2014
From Tom Copeland, Dec 26 2015: (Start)
With a(n,x) = e^(nx), the partial sums are 1+e^x+...+e^(nx) = Sum_{k=0..n} T(n,k) (1+e^x)^k = [ x / (e^x-1) ] [ e^((n+1)x) -1 ] / x = [ (x / (e^x-1)) e^((n+1)x) -  (x / (e^x-1)) ] / x =  Sum_{k>=0} [ (Ber(k+1,n+1) - Ber(k+1,0)) / (k+1) ] * x^k/k!, where Ber(n,x) are the Bernoulli polynomials (cf. Adams p. 140). Evaluating (d/dx)^m at x=0 of these expressions gives relations among the partial sums of the m-th powers of the integers, their binomial transforms, and the Bernoulli polynomials.
With a(n,x) = (-1)^n e^(nx), the partial sums are 1-e^x+...+(-1)^n e^(nx) = Sum_{k=0..n} T(n,k) (1-e^x)^k = [ (-1)^n e^((n+1)x) + 1 ] / (e^x+1) = [ (-1)^n (2 / (e^x+1)) e^((n+1)x) + (2 / (e^x+1)) ] / 2 = (1/2) Sum_{k>=0} [ (-1)^n Eul(k,n+1) + Eul(k,0) ] * x^k/k!, where Eul(n,x) are the Euler polynomials. Evaluating (d/dx)^m at x=0 of these expressions gives relations among the partial sums of signed m-th powers of the integers; their binomial transforms, related to the Stirling numbers of the second kind and face numbers of the permutahedra; and the Euler polynomials. (End)
As in A059260, a generator in terms of bivariate polynomials with the coefficients of this entry is given by (1/(1-y))*1/(1 + (y/(1-y))*x - (1/(1-y))*x^2) = 1 + y + (x^2 - x*y + y^2) + (2*x^2*y - 2*x*y^2 + y^3) + (x^4 - 2*x^3*y + 4*x^2*y^2 - 3*x*y^3 + y^4) + ... . This is of the form -h2 * 1 / (1 + h1*x + h2*x^2), related to the bivariate generator of A049310 with h1 = y/(1-y) and h2 = -1/(1-y) = -(1+h1). - Tom Copeland, Feb 16 2016
From Tom Copeland, Sep 05 2016: (Start)
Letting P(k,x) = x in D gives Sum_{k=0..n} T(n,k)*Sum_{j=0..k} binomial(k,j) = Sum_{k=0..n} T(n,k) 2^k = n + 1.
The quantum integers [n+1]q = (q^(n+1) - q^(-n-1)) / (q - q^(-1)) = q^(-n)*(1 - q^(2*(n+1))) / (1 - q^2) = q^(-n)*Sum{k=0..n} q^(2k) = q^(-n)*Sum_{k=0..n} T(n,k)*(1 + q^2)^k. (End)
T(n, k) = [x^k] Sum_{j=0..n} (x-1)^j. - Peter Luschny, Jul 09 2019
a(n) = -n + Sum_{k=0..n} A341091(k). - Thomas Scheuerle, Jun 17 2022

Inverse array added by Tom Copeland, Mar 26 2014

Formula re Euler polynomials corrected by Tom Copeland, Mar 08 2024

cp's OEIS Frontend

A137862 Triangular sequence of coefficients of the expansion of a degenerate partition of Chebyshev U(x,n);A053117 and Hermite H(x,n);A060821 functions: 1) f(x,t)=1/(1-2*x*t+t^2); 2) g(x,t)=Exp[2*x*t-t^2]; to give: p(x,t)=Exp[2*x*t-t^2]/(1-2*x*t+t^2).

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A049310 Triangle of coefficients of Chebyshev's S(n,x) := U(n,x/2) polynomials (exponents in increasing order).

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A111125 Triangle read by rows: T(k,s) = ((2*k+1)/(2*s+1))*binomial(k+s,2*s), 0 <= s <= k.

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A100258 Triangle of coefficients of normalized Legendre polynomials, with increasing exponents.

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A133156 Irregular triangle read by rows: coefficients of U(n,x), Chebyshev polynomials of the second kind with exponents in decreasing order.

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A207538 Triangle of coefficients of polynomials v(n,x) jointly generated with A207537; see Formula section.

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A137276 Triangle T(n,k), read by rows: T(n,k)= 0 if n-k odd. T(n,k)= 2*(-1)^((n-k)/2)*(2k-n)/(n+k)*binomial((n+k)/2,(n-k)/2) if n-k even.

A137862 Triangular sequence of coefficients of the expansion of a degenerate partition of Chebyshev U(x,n);A053117 and Hermite H(x,n);A060821 functions: 1) f(x,t)=1/(1-2xt+t^2); 2) g(x,t)=Exp[2xt-t^2]; to give: p(x,t)=Exp[2xt-t^2]/(1-2xt+t^2).

A111125 Triangle read by rows: T(k,s) = ((2k+1)/(2s+1))binomial(k+s,2s), 0 <= s <= k.

A137276 Triangle T(n,k), read by rows: T(n,k)= 0 if n-k odd. T(n,k)= 2(-1)^((n-k)/2)(2k-n)/(n+k)*binomial((n+k)/2,(n-k)/2) if n-k even.