A053117 -id:A053117 - cp's OEIS frontend

A053118 Triangle of coefficients of Chebyshev's U(n,x) polynomials (exponents in decreasing order).

1, 2, 0, 4, 0, -1, 8, 0, -4, 0, 16, 0, -12, 0, 1, 32, 0, -32, 0, 6, 0, 64, 0, -80, 0, 24, 0, -1, 128, 0, -192, 0, 80, 0, -8, 0, 256, 0, -448, 0, 240, 0, -40, 0, 1, 512, 0, -1024, 0, 672, 0, -160, 0, 10, 0, 1024, 0, -2304, 0, 1792, 0, -560, 0, 60, 0, -1, 2048, 0, -5120, 0, 4608, 0, -1792, 0, 280, 0, -12, 0, 4096, 0, -11264, 0, 11520, 0, -5376

Offset: 0

Wolfdieter Lang

a(n,m)= A053117(n,n-m) = 2^(n-m)*A049310(n,n-m).
G.f. for row polynomials U(n,x) (signed triangle): 1/(1-2*x*z+z^2). Unsigned triangle |a(n,m)| has Fibonacci polynomials F(n+1,2*x) as row polynomials with G.f. 1/(1-2*x*z-z^2).
Row sums (unsigned triangle) A000129(n+1) (Pell). Row sums (signed triangle) A000027(n+1) (natural numbers).

			1;
2,0;
4,0,-1;
8,0,-4,0;
16,0,-12,0,1;
... E.g. fourth row (n=3) {8,0,-4,0} corresponds to polynomial U(3,x)= 8*x^3-4*x.

Theodore J. Rivlin, Chebyshev polynomials: from approximation theory to algebra and number theory, 2. ed., Wiley, New York, 1990.

Cf. A053117, A049310, A000129.

Triangle reflected without zeros: A008312 (the main entry).

Flatten[ Table[ Reverse[ CoefficientList[ ChebyshevU[n, x], x]], {n, 0, 12}]] (* Jean-François Alcover, Jan 20 2012 *)

a(n, m) := 0 if n= 0 is even else 0.

A121448 Triangle read by rows: T(n,k) is the number of binary trees with n edges and having k vertices of outdegree 1 (n>=0, k>=0). A binary tree is a rooted tree in which each vertex has at most two children and each child of a vertex is designated as its left or right child.

Original entry on oeis.org

1, 0, 2, 1, 0, 4, 0, 6, 0, 8, 2, 0, 24, 0, 16, 0, 20, 0, 80, 0, 32, 5, 0, 120, 0, 240, 0, 64, 0, 70, 0, 560, 0, 672, 0, 128, 14, 0, 560, 0, 2240, 0, 1792, 0, 256, 0, 252, 0, 3360, 0, 8064, 0, 4608, 0, 512, 42, 0, 2520, 0, 16800, 0, 26880, 0, 11520, 0, 1024, 0, 924, 0, 18480, 0

Offset: 0

Emeric Deutsch, Jul 31 2006

T(2n,0) = binomial(2n,n)/(n+1) (the Catalan numbers; A000108); T(2n+1,0)=0. T(n,n)=2^n (A000079). Sum(k*T(n,k),k=0..n)=2*binomial(2n,n-1)=2*A001791(n). After deleting the zeros, reflection of A091894.
From Tom Copeland, Feb 07 2016: (Start)
A shifted o.g.f. is OG(x,t) = [1 - 2tx - sqrt[(1-2tx)^2-4x^2]] / (2x) = x + 2t x^2 + (1+4t^2) x^3 + ... with compositional inverse OGinv(x,t) =  x / (1 + 2tx + x^2), the shifted o.g.f. for A053117 (mod signs).
For x > 0 and choosing the positive square root, OG(x^2,t) = H(x,t) = x^2 + 2t x^4 + (1+4t^2) x^6 + ... has the compositional inverse Hinv(x,t) = sqrt[x / (1 + 2tx + x^2)] , which satisfies Hinv(H(x, t), t) = x, and which is the generating function for the Legendre polynomials (mod signs, cf. A008316) times sqrt(x).
In general, GB(x,t,b) =  [x / (1 - 2tx + x^2)]^b is a generator for the Gegenbauer polynomials times x^b for positive roots with compositional inverse about the origin GBinv(x,t,b) = OG(x^(1/b),-t) for x>0. Cf. A097610.
(End)
From Tom Copeland, Feb 09 2016: (Start)
z1 = OG(x,t) is the zero that vanishes for x=0 for the quadratic polynomial Q(z;z1(x,t),z2(x,t)) =(z-z1)(z-z2) = z^2 - (z1+z2) z + (z1*z2) = z^2 - e1 z + e2 = z^2 - [(1-2tx)/x] z + 1, where e1 and e2 are the elementary symmetric polynomials for two indeterminates.
The other zero is given by z2(x,t) = [1 - 2tx + sqrt[(1-2tx)^2-4x^2]] / (2x) = (1 - 2tx)/x - z1(x,t).
The two are zeros of the elliptic curve in Legendre normal form y^2 = z (z-z1)(z-z2). (Added Feb 13 2016. See Landweber et al., p 14. Cf. A097610.)
(End)

			T(2,2)=4 because, denoting by L (R) an edge going from a vertex to a left (right) child, we have the paths: LL, LR, RL and RR.
Triangle starts:
  1;
  0,2;
  1,0,4;
  0,6,0,8;
  2,0,24,0,16;

Colin Defant, Postorder Preimages, arXiv preprint arXiv:1604.01723 [math.CO], 2016.
FindStat - Combinatorial Statistic Finder, The number of vertices with out-degree 1 in a binary tree.
P. Landweber, D. Ravenel, and R. Stong, Periodic cohomology theories defined by elliptic curves

Cf. A000108, A000079, A001791, A091894.
Cf. A038207, A053117, A097610, A126120, A145271, A180874, A218272.
Cf. A008316.

T:=proc(n,k) if n-k mod 2 = 0 then 2^k*binomial(n+1,k)*binomial(n+1-k,(n-k)/2)/(n+1) else 0 fi end: for n from 0 to 12 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form

nn=10;Drop[CoefficientList[Series[(1-2x y - ((-4x^2+(1-2x y)^2))^(1/2))/(2 x),{x,0,nn}],{x,y}],1]//Grid  (* Geoffrey Critzer, Feb 20 2013 *)

T(n,k) = 2^k*binomial(n+1,k)binomial(n+1-k,(n-k)/2)/(n+1) if n-k is even; otherwise, T(n,k) = 0. G.f. G=G(t,z) satisfies G=1+2tzG+z^2*G^2.
T(n,k) = 2^k*A097610(n,k). - Philippe Deléham, Aug 17 2006
From Tom Copeland, Feb 09 2016: (Start)
The following is from the formalism in A097610 with h1 = 2t, h2 = 1, and MT(n,h1,h2) = MT(n,2t,1) and with OG(x,t) defined above.
E.g.f.: M(x,t) = e^(2tx) AC(x) = exp[x MT(.,2t,1)] = exp[x P(.,t)], where AC(x) = I_1(2x)/x = Sum_{n>=0} x^(2n)/(n!(n+1)!) = exp(c.x) is the e.g.f. of A126120.
P(n,t) = MT(n,2t,1) = (c. + 2t)^n = Sum_{k=0..n} binomial(n,k) c(n-k) (2t)^k with c(k) = A126120(k). P(n,t+s) = (c. + 2t + 2s)^n = (P(.,t) + 2s)^n.
P(n,t) = t^n FC(n,c./t) = t^n (2 + c./t)^n, where FC(n,t) = (2 + t)^n are the face polynomials (vectors) of the hypercubes of A038207, i.e., the row polynomials of this entry can be obtained as the umbral composition of the reverse face polynomials with the aerated Catalan numbers of A000108.
The lowering and raising operators for the row polynomials P(n,t) of this entry are L = (1/2) d/dt = (1/2) D and R = 2t + dlog{AC(L)}/dL = 2t + Sum_{n>=0} b(n) L^(2n+1)/(2n+1)! = 2t + L - L^3/3! + 5 L^5/5! - ...  with b(n) = (-1)^n A180874(n+1).
Let CP(n,t) = P(n+1,t) with CP(0,t) = 0. Then the infinitesimal generator for CP(n,t) is g(x) d/dx with g(x) = 1 /[dOGinv(x,t)/dx] = x^2 / [(OGinv(x,t))^2 (1 - x^2)] = (1 + 2t x + x^2)^2 / (1 - x^2) so that [g(x)d/dx]^n/n! x evaluated at x = 0 gives the row polynomial CP(n,t), i.e., exp[x g(u)d/du] u |_(u=0) = OG(x,t) = 1 /[1 - x P(.,t)]. Cf. A145271.
g(x) = 1 + 4t x + (3+4t) x^2 + 8t x^3 + 4(1+t^2) x^4 + 8t x^5 + 4(1+t^2) x^6 + 8t x^7 + ... has the repeating coefficients of the vector V = (1, 4t, 3+4t, 8t, 4(1+t^2), 8t, 4(1+t^2), 8t, ...). Form the lower triangular matrix U with all ones on the diagonal and below. Multiply the n-th diagonal of U by V(n), giving the matrix VU with VU(n,k) = V(n-k). Then (1,0,0,0,..) [VU * DM]^n/n! (0,1,0,0,..)^T = CP(n,t) = P(n-1,t) for n>0 with DM being the matrix A218272 representing differentiation of a power series.
(End)

A123956 Triangle of coefficients of polynomials P(k) = 2XP(k - 1) - P(k - 2), P(0) = -1, P(1) = 1 + X, with twisted signs.

Original entry on oeis.org

-1, 1, 1, -1, -2, -2, 1, -3, 4, 4, -1, 4, 8, -8, -8, 1, 5, -12, -20, 16, 16, -1, -6, -18, 32, 48, -32, -32, 1, -7, 24, 56, -80, -112, 64, 64, -1, 8, 32, -80, -160, 192, 256, -128, -128, 1, 9, -40, -120, 240, 432, -448, -576, 256, 256, -1, -10, -50, 160, 400, -672, -1120, 1024, 1280, -512, -512

Offset: 0

Roger L. Bagula and Gary W. Adamson, Oct 27 2006

Up to signs also the coefficients of polynomials y(n+1) = y(n-1) - 2*h*y(n), arising when the ODE y' = -y is numerically solved with the leapfrog (a.k.a. two-step Nyström) method, with y(0) = 1, y(1) = 1 - h. In this case, the coefficients are negative exactly for the odd powers of h. - M. F. Hasler, Nov 30 2022

			Triangle begins:
  {-1},
  { 1,   1},
  {-1,  -2,  -2},
  { 1,  -3,   4,    4},
  {-1,   4,   8,   -8,   -8},
  { 1,   5, -12,  -20,   16,   16},
  {-1,  -6, -18,   32,   48,  -32,   -32},
  { 1,  -7,  24,   56,  -80, -112,    64,   64},
  {-1,   8,  32,  -80, -160,  192,   256, -128, -128},
  { 1,   9, -40, -120,  240,  432,  -448, -576,  256,  256},
  {-1, -10, -50,  160,  400, -672, -1120, 1024, 1280, -512, -512},
  ...

CRC Standard Mathematical Tables and Formulae, 16th ed. 1996, p. 484.
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 799.

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].
Lutterbach, Approximating ODE y' = f(t,y) by using leapfrog method, Mathematics Stack Exchange, Nov 21 2019
Alastair MacDougall, 83.31 A Pascal-like triangle for coefficients of Chebyshev polynomials">, The Mathematical Gazette, Vol. 83, Issue 497 (Jul 1999), pp. 276-280.

Cf. A008312, A028297, A053117, A123235.

p[ -1, x] = 0; p[0, x] = 1; p[1, x] = x + 1;
p[k_, x_] := p[k, x] = 2*x*p[k - 1, x] - p[k - 2, x];
w = Table[CoefficientList[p[n, x], x], {n, 0, 10}];
An[d_] := Table[If[n == d && m  1/y)] /. 1/y -> 1, y], {d, 1, 11}] // Flatten

P=List([-1,1-'x]); {A123956(n,k)=for(i=#P, n+1, listput(P, P[i-1]-2*'x*P[i])); polcoef(P[n+1],k)*(-1)^((n-k-1)\2+!k*n\2)} \\ M. F. Hasler, Nov 30 2022

From M. F. Hasler, Nov 30 2022: (Start)
a(n,0) = (-1)^(n+1), a(n,1) = (-1)^floor(n/2)*n,
a(n,2) = (-1)^floor((n+1)/2)*A007590(n) = (-1)^floor((n+1)/2)*floor(n^2 / 2),
a(n,n) = a(n,n-1) = (-2)^(n-1) (n > 0),
a(n,3) / a(n,2) = { n/3 if n odd, -4*(n+2)/n if n even },
a(n,4) / a(n,3) = n/4 if n is even. (End)
From Peter Bala, Feb 06 2025: (Start)
Let T(n, x) and U(n, x) denote the n-th Chebyshev polynomial of the first and second kind. It appears that the row g.f.'s are as follows: for n >= 0,
row 4*n+1: T(4*n+1, x) + U(4*n, x); row 4*n+2: - 2 - T(4*n+2, x) - U(4*n+1, x);
row 4*n+3: 2 + T(4*n+3, x) + U(4*n+2, x); row 4*n+4: - T(4*n+4, x) - U(4*n+3, x). (End)

Offset changed to 0 by M. F. Hasler, Nov 30 2022

A100551 Coefficient list of ChebyshevU(n, 1-x).

Original entry on oeis.org

1, 2, -2, 3, -8, 4, 4, -20, 24, -8, 5, -40, 84, -64, 16, 6, -70, 224, -288, 160, -32, 7, -112, 504, -960, 880, -384, 64, 8, -168, 1008, -2640, 3520, -2496, 896, -128, 9, -240, 1848, -6336, 11440, -11648, 6720, -2048, 256, 10, -330, 3168, -13728, 32032, -43680, 35840, -17408, 4608, -512

Offset: 0

Wouter Meeussen, Nov 27 2004

			Triangle begins as:
  1;
  2,   -2;
  3,   -8,    4;
  4,  -20,   24,    -8;
  5,  -40,   84,   -64,    16;
  6,  -70,  224,  -288,   160,    -32;
  7, -112,  504,  -960,   880,   -384,   64;
  8, -168, 1008, -2640,  3520,  -2496,  896,  -128;
  9, -240, 1848, -6336, 11440, -11648, 6720, -2048, 256;

Cf. A001353, A001787, A053117, A056594, A122803.

[Binomial(n+k+1, n-k)*(-2)^k: k in [0..n], n in [0..12]]; // G. C. Greubel, Mar 27 2023

Table[CoefficientList[ChebyshevU[n, 1-x], x], {n, 0, 12}]

row(n) = Vecrev(polchebyshev(n, 2, 1-x)); \\ Michel Marcus, Apr 27 2020

def A100551(n,k): return binomial(n+k+1, n-k)*(-2)^k
flatten([[A100551(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Mar 27 2023

G.f.: ChebyshevU(n, 1-x).
From G. C. Greubel, Mar 27 2023: (Start)
T(n, k) = binomial(n+k+1, n-k)*(-2)^k.
T(n, n) = A122803(n).
T(n, n-1) = 2*(-1)^(n-1)*A001787(n), n >= 1.
Sum_{k=0..n} T(n, k) = A056594(n).
Sum_{k=0..n} (-1)^k*T(n, k) = A001353(n+1). (End)

Keyword tabl from Michel Marcus, Apr 27 2020

A115322 Triangle of coefficients of Pell polynomials.

Original entry on oeis.org

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

Offset: 1

Eric W. Weisstein, Jan 20 2006

Aside from signs, same as A053117.

Row n gives the coefficients in the expansion of the Alexander-Conway polynomial for the barrier-free Celtic link CK_4^(2n). - Franck Maminirina Ramaharo, Aug 06 2025

			1, 2*x, 1 + 4*x^2, 4*x + 8*x^3, 1 + 12*x^2 + 16*x^4, ...
Triangle begins:
  [n\k] 0   1   2    3    4    5     6    7    8     9    10
  ----------------------------------------------------------
  [1 ]  1;
  [2 ]  0,  2;
  [3 ]  1,  0,  4;
  [4 ]  0,  4,  0,   8;
  [5 ]  1,  0, 12,   0,  16;
  [6 ]  0,  6,  0,  32,   0,  32;
  [7 ]  1,  0, 24,   0,  80,   0,   64;
  [8 ]  0,  8,  0,  80,   0, 192,    0,  128;
  [9 ]  1,  0, 40,   0, 240,   0,  448,    0,  256;
  [10]  0, 10,  0, 160,   0, 672,    0, 1024,    0, 512;
  [11]  1,  0, 60,   0, 560,   0, 1792,    0, 2304,   0, 1024;
... - _Franck Maminirina Ramaharo_, Aug 06 2025

Jonathan L. Gross and Thomas W. Tucker, A Celtic Framework for Knots and Links, Discrete & Computational Geometry 46 (2011), 86-99.
Eric Weisstein's World of Mathematics, Pell Polynomial

Cf. A053117, A049310, A000129 (row sums).

A115322 := (n, k) -> ifelse(irem(n - k, 2) = 0, 0, binomial((n + k - 1)/2, k)*2^k);
seq(print(seq(A115322(n, k), k = 0..n-1)), n = 1..10);  # Peter Luschny, Aug 06 2025

Flatten[Table[CoefficientList[Fibonacci[n, 2 x], x], {n, 0, 20}]] (* Emanuele Munarini, Dec 01 2017 *)

T(n, k) := if mod(n - k, 2) = 0 then 0 else  binomial((n + k - 1)/2, k)*2^k$
create_list(T(n, k), n, 1, 10, k, 0, n - 1); /* Franck Maminirina Ramaharo, Aug 06 2025 */

G.f. for n-th row is Fibonacci(n, 2*x).
From Franck Maminirina Ramaharo, Aug 06 2025: (Start)
Row n = coefficients in the expansion of ((x + srt(x^2 + 1))^n - (x - sqrt(x^2 + 1))^n)/(2*sqrt(x^2 + 1)).
T(n,k) = 0 if n == k (mod 2) else T(n,k) = (2^k)*binomial((n + k - 1)/2, k).
G.f.: y/(1 - 2 x*y - y^2).
(End)

A121872 Triangle T(n, k) = (kChebyshevU(n, (k+2)/2) + 2ChebyshevT(n+1, (k+2)/2))/2.

Original entry on oeis.org

5, 13, 41, 34, 153, 436, 89, 571, 2089, 5741, 233, 2131, 10009, 33461, 90481, 610, 7953, 47956, 195025, 620166, 1663585, 1597, 29681, 229771, 1136689, 4250681, 13097377, 34988311, 4181, 110771, 1100899, 6625109, 29134601, 103115431, 310957991, 828931049

Offset: 1

Roger L. Bagula and Gary W. Adamson, Sep 09 2006

			Triangle begins as:
    5;
   13,   41;
   34,  153,   436;
   89,  571,  2089,  5741;
  233, 2131, 10009, 33461, 90481;

G. C. Greubel, Rows n = 1..100 of triangle, flattened

Cf. A094954, A162997.

Cf. A053117, A053120.

T:= func< n,k | ( k*Sinh((n+1)*Argcosh((k+2)/2))/Sinh(Argcosh((k+2)/2)) + 2*Cosh((n+1)*Argcosh((k+2)/2)) )/2 >;
[Round(T(n,k)): k in [1..n], n in [1..10]]; // G. C. Greubel, Oct 08 2019

seq(seq(simplify(( k*ChebyshevU(n, (k+2)/2) + 2*ChebyshevT(n+1, (k+2)/2) )/2), k=1..n), n=1..10); # G. C. Greubel, Oct 09 2019

f[k_]:= Sqrt[k*(k+4)]; T[n_, m_]:= T[n, m]= FullSimplify[((m+f[m])*(m+2 - f[m])^(n+2) - (m-f[m])*(m+2 + f[m])^(n+2))/(2^(n+3)*f[m])]; Table[T[n, m], {n,10}, {m,n}]//Flatten (* modified by G. C. Greubel, Oct 08 2019 *)
T[n_, k_]:= T[n, k]= (k*ChebyshevU[n, (k+2)/2] + 2*ChebyshevT[n+1, (k+ 2)/2])/2; Table[T[n, k], {n,10}, {k,n}]/Flatten (* G. C. Greubel, Oct 08 2019 *)

T(n,k)= ( k*sin((n+1)*acos((k+2)/2))/sin(acos((k+2)/2)) + 2*cos((n+1)*acos((k+2)/2)) )/2;
for(n=1,10, for(k=1,n, print1(round(T(n,k)), ", "))) \\ G. C. Greubel, Oct 08 2019

[[( k*chebyshev_U(n,(k+2)/2) + 2*chebyshev_T(n+1, (k+2)/2) )/2 for k in (1..n)] for n in (1..10)] # G. C. Greubel, Oct 08 2019

T(n, m) = ((m+f(m))*(m+2 - f(m))^(n+2) - (m-f(m))*(m+2 + f(m))^(n+2))/( 2^(n+3)*f(m)), where f(m) = sqrt(m*(m+4)).
From G. C. Greubel, Oct 08 2019: (Start)
T(n, k) = (k*ChebyshevU(n, (k+2)/2) + 2*ChebyshevT(n+1, (k+2)/2))/2;
T(n, k) = (k*Fibonacci(n+2, m+2, -1) + Lucas(n+2, m+2, -1))/2, where Fibonacci(n, x, y) and Lucas(n, x, y) are the bi-variate Fibonacci an Lucas polynomials, respectively. (End)

Major edit and new name, G. C. Greubel, Oct 08 2019

A136388 Triangle read by rows of coefficients of Chebyshev-like polynomials P_{n,2}(x) with 0 omitted (exponents in increasing order).

Original entry on oeis.org

1, -2, 2, 1, -5, 4, 4, -12, 8, -1, 13, -28, 16, -6, 38, -64, 32, 1, -25, 104, -144, 64, 8, -88, 272, -320, 128, -1, 41, -280, 688, -704, 256, -10, 170, -832, 1696, -1536, 512, 1, -61, 620, -2352, 4096, -3328, 1024, 12, -292, 2072, -6400, 9728, -7168, 2048

Offset: 2

Milan Janjic, Mar 30 2008, entry revised Apr 05 2008

If U_n(x), T_n(x) are Chebyshev's polynomials then U_n(x)=P_{n,0}(x), T_n(x)=P_{n,1}(x).

Let n>=2 and k be of the same parity. Consider a set X consisting of (n+k)/2-2 main blocks of the size 2 and an additional block of the size 2, then (-1)^((n-k)/2)a(n,k) is the number of n-2-subsets of X intersecting each main block.

			Rows are (1),(-2,2),(1,-5,4),(4,-12,8),(-1,13,-28,16),...
since P_{2,2}=x^2, P_{3,2}=-2x+2x^3, P_{4,2}=1-5x^2+4x^4,...

Michael De Vlieger, Table of n, a(n) for n = 2..10199 (rows 2 <= n <= 200, flattened).
Milan Janjic, Two enumerative functions.
M. Janjic, On a class of polynomials with integer coefficients, JIS 11 (2008) 08.5.2.
Milan Janjić, On Restricted Ternary Words and Insets, arXiv:1905.04465 [math.CO], 2019.

Cf. A008310, A053117.

if modp(n-k, 2)=0 then a[n,k]:=(-1)^((n-k)/2)*sum((-1)^i*binomial((n+k)/2-2, i)*binomial(n+k-2-2*i, n-2), i=0..(n+k)/2-2); end if;

Rest@ Flatten@ Table[If[SameQ @@ Mod[{n, k}, 2], (-1)^((n - k)/2)*Sum[(-1)^i*Binomial[(n + k)/2 - 2, i]*Binomial[n + k - 2 - 2 i, n - 2], {i, 0, (n + k)/2 - 2}], 0], {n, 2, 13}, {k, Boole@ OddQ@ n, n, 2}] (* Michael De Vlieger, Jul 02 2019 *)

If n>=2 and k are of the same parity then a(n,k)= (-1)^((n-k)/2)*sum((-1)^i*binomial((n+k)/2-2, i)*binomial(n+k-2-2*i, n-2), i=0..(n+k)/2-2) and a(n,k)=0 if n and k are of different parity.

A136389 Triangle read by rows of coefficients of Chebyshev-like polynomials P_{n,3}(x) with 0 omitted (exponents in increasing order).

Original entry on oeis.org

1, -3, 2, 3, -7, 4, -1, 9, -16, 8, -5, 25, -36, 16, 1, -19, 66, -80, 32, 7, -63, 168, -176, 64, -1, 33, -192, 416, -384, 128, -9, 129, -552, 1008, -832, 256, 1, -51, 450, -1520, 2400, -1792, 512, 11, -231, 1452, -4048, 5632, -3840, 1024, -1, 73, -912, 4424, -10496, 13056, -8192, 2048

Offset: 3

Milan Janjic, Mar 30 2008, revised Apr 05 2008

If U_n(x), T_n(x) are Chebyshev's polynomials then U_n(x)=P_{n,0}(x), T_n(x)=P_{n,1}(x).

Let n>=3 and k be of the same parity. Consider a set X consisting of (n+k)/2-3 blocks of the size 2 and an additional block of the size 3, then (-1)^((n-k)/2)a(n,k) is the number of n-3-subsets of X intersecting each block of the size 2.

			Rows are (1),(-3,2),(3,-7,4),(-1,9,-16,8),(-5,25,-36,16),...
since P_{3,3}=x^3, P_{4,3}=-3x^2+2x^4, P_{5,3}=3x-7x^3+4x^5,...

Michael De Vlieger, Table of n, a(n) for n = 3..10197 (rows 3 <= n <= 200, flattened).
Milan Janjic, Two enumerative functions.
Milan Janjic, On a class of polynomials with integer coefficients, JIS 11 (2008) 08.5.2
Milan Janjić, On Restricted Ternary Words and Insets, arXiv:1905.04465 [math.CO], 2019.

Cf. A008310, A053117.

if modp(n-k, 2)=0 then a[n, k]:=(-1)^((n-k)/2)*sum((-1)^i*binomial((n+k)/2-3, i)*binomial(n+k-3-2*i, n-3), i=0..(n+k)/2-3); end if;

DeleteCases[#, 0] &@ Flatten@ Table[(-1)^((n - k)/2)*Sum[(-1)^i*Binomial[(n + k)/2 - 3, i] Binomial[n + k - 3 - 2 i, n - 3], {i, 0, (n + k)/2 - 3}], {n, 3, 14}, {k, 0 + Boole[OddQ@ n], n, 2}] (* Michael De Vlieger, Jul 05 2019 *)

If n>=3 and k are of the same parity then a(n,k)= (-1)^((n-k)/2)*sum((-1)^i*binomial((n+k)/2-3, i)*binomial(n+k-3-2*i, n-3), i=0..(n+k)/2-3) and a(n,k)=0 if n and k are of different parity.

A136390 Triangle read by rows of coefficients of Chebyshev-like polynomials P_{n,4}(x) with 0 omitted (exponents in increasing order).

Original entry on oeis.org

1, -4, 2, 6, -9, 4, -4, 16, -20, 8, 1, -14, 41, -44, 16, 6, -44, 102, -96, 32, -1, 26, -129, 248, -208, 64, -8, 96, -360, 592, -448, 128, 1, -42, 321, -968, 1392, -960, 256, 10, -180, 1002, -2528, 3232, -2048, 512, -1, 62, -681, 2972, -6448, 7424, -4352, 1024

Offset: 4

Milan Janjic, Mar 30 2008, revised Apr 05 2008

If U_n(x), T_n(x) are Chebyshev's polynomials then U_n(x)=P_{n,0}(x), T_n(x)=P_{n,1}(x).

Let n>=4 and k be of the same parity. Consider a set X consisting of (n+k)/2-4 blocks of the size 2 and an additional block of the size 4, then (-1)^((n-k)/2)a(n,k) is the number of n-4-subsets of X intersecting each block of the size 2.

			Rows are (1),(-4,2),(6,-9,4),(-4,16,-20,8),... since P_{4,4}=x^4, P_{5,4}=-4x^3+2x^5, P_{6,4}=6x^2-9x^4+4x^6,...

Michael De Vlieger, Table of n, a(n) for n = 4..10194 (rows 4 <= n <= 200, flattened).
Milan Janjic, Two enumerative functions.
M. Janjic, On a class of polynomials with integer coefficients, JIS 11 (2008) 08.5.2
Milan Janjić, On Restricted Ternary Words and Insets, arXiv:1905.04465 [math.CO], 2019.

Cf. A008310, A053117.

if modp(n-k, 2)=0 then a[n, k]:=(-1)^((n-k)/2)*sum((-1)^i*binomial((n+k)/2-4, i)*binomial(n+k-4-2*i, n-4), i=0..(n+k)/2-4); end if;

DeleteCases[#, 0] &@ Flatten@ Table[(-1)^((n - k)/2) * Sum[(-1)^i * Binomial[(n + k)/2 - 4, i] Binomial[n + k - 4 - 2 i, n - 4], {i, 0, (n + k)/2 - 4}], {n, 4, 14}, {k, 0 + Boole[OddQ@ n], n, 2}] (* Michael De Vlieger, Jul 05 2019 *)

If n>=4 and k are of the same parity then a(n,k)= (-1)^((n-k)/2)*sum((-1)^i*binomial((n+k)/2-4, i)*binomial(n+k-4-2*i, n-4), i=0..(n+k)/2-4) and a(n,k)=0 if n and k are of different parity.

A136397 Triangle read by rows of coefficients of Chebyshev-like polynomials P_{n,5}(x) with 0 omitted (exponents in increasing order).

Original entry on oeis.org

1, -5, 2, 10, -11, 4, -10, 25, -24, 8, 5, -30, 61, -52, 16, -1, 20, -85, 146, -112, 32, -7, 70, -231, 344, -240, 64, 1, -34, 225, -608, 800, -512, 128, 9, -138, 681, -1560, 1840, -1088, 256, -1, 52, -501, 1970, -3920, 4192, -2304, 512

Offset: 5

Milan Janjic, Mar 30 2008, revised Apr 05 2008

If U_n(x), T_n(x) are Chebyshev's polynomials then U_n(x)=P_{n,0}(x), T_n(x)=P_{n,1}(x).

Let n>=5 and k be of the same parity. Consider a set X consisting of (n+k)/2-5 blocks of the size 2 and an additional block of the size 5, then (-1)^((n-k)/2)a(n,k) is the number of n-5-subsets of X intersecting each block of the size 2.

			Rows are (1),(-5,2),(10,-11,4),... since P_{5,5}=x^5, P_{6,5}=-5x^4+2x^6, P_{7,5}=10x^3-11x^5+4x^7,...

Michael De Vlieger, Table of n, a(n) for n = 5..10190 (rows 5 <= n <= 200, flattened).
Milan Janjic, Two enumerative functions.
M. Janjic, On a class of polynomials with integer coefficients, JIS 11 (2008) 08.5.2.
Milan Janjić, On Restricted Ternary Words and Insets, arXiv:1905.04465 [math.CO], 2019.

Cf. A008310, A053117.

if modp(n-k, 2)=0 then a[n, k]:=(-1)^((n-k)/2)*sum((-1)^i*binomial((n+k)/2-5, i)*binomial(n+k-5-2*i, n-5), i=0..(n+k)/2-5); end if;

DeleteCases[#, 0] &@ Flatten@ Table[(-1)^((n - k)/2) * Sum[(-1)^i * Binomial[(n + k)/2 - 5, i] Binomial[n + k - 5 - 2 i, n - 5], {i, 0, (n + k)/2 - 5}], {n, 5, 14}, {k, 0 + Boole[OddQ@ n], n, 2}] (* Michael De Vlieger, Jul 05 2019 *)

If n>=5 and k are of the same parity then a(n,k)= (-1)^((n-k)/2)*sum((-1)^i*binomial((n+k)/2-5, i)*binomial(n+k-5-2*i, n-5), i=0..(n+k)/2-5) and a(n,k)=0 if n and k are of different parity.

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A053118 Triangle of coefficients of Chebyshev's U(n,x) polynomials (exponents in decreasing order).

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A121448 Triangle read by rows: T(n,k) is the number of binary trees with n edges and having k vertices of outdegree 1 (n>=0, k>=0). A binary tree is a rooted tree in which each vertex has at most two children and each child of a vertex is designated as its left or right child.

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A123956 Triangle of coefficients of polynomials P(k) = 2*X*P(k - 1) - P(k - 2), P(0) = -1, P(1) = 1 + X, with twisted signs.

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A100551 Coefficient list of ChebyshevU(n, 1-x).

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A115322 Triangle of coefficients of Pell polynomials.

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A121872 Triangle T(n, k) = (k*ChebyshevU(n, (k+2)/2) + 2*ChebyshevT(n+1, (k+2)/2))/2.

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A123956 Triangle of coefficients of polynomials P(k) = 2XP(k - 1) - P(k - 2), P(0) = -1, P(1) = 1 + X, with twisted signs.

A121872 Triangle T(n, k) = (kChebyshevU(n, (k+2)/2) + 2ChebyshevT(n+1, (k+2)/2))/2.