A053123 -id:A053123 - cp's OEIS frontend

A011973 Irregular triangle read by rows: T(n,k) = binomial(n-k, k), n >= 0, 0 <= k <= floor(n/2); or, coefficients of (one version of) Fibonacci polynomials.

1, 1, 1, 1, 1, 2, 1, 3, 1, 1, 4, 3, 1, 5, 6, 1, 1, 6, 10, 4, 1, 7, 15, 10, 1, 1, 8, 21, 20, 5, 1, 9, 28, 35, 15, 1, 1, 10, 36, 56, 35, 6, 1, 11, 45, 84, 70, 21, 1, 1, 12, 55, 120, 126, 56, 7, 1, 13, 66, 165, 210, 126, 28, 1, 1, 14, 78, 220, 330, 252, 84, 8, 1, 15, 91, 286, 495, 462

Offset: 0

N. J. A. Sloane

T(n,k) is the number of subsets of {1,2,...,n-1} of size k and containing no consecutive integers. Example: T(6,2)=6 because the subsets of size 2 of {1,2,3,4,5} with no consecutive integers are {1,3},{1,4},{1,5},{2,4},{2,5} and {3,5}. Equivalently, T(n,k) is the number of k-matchings of the path graph P_n. - Emeric Deutsch, Dec 10 2003
T(n,k) = number of compositions of n+2 into k+1 parts, all >= 2. Example: T(6,2)=6 because we have (2,2,4),(2,4,2),(4,2,2),(2,3,3),(3,2,3) and (3,3,2). - Emeric Deutsch, Apr 09 2005
Given any recurrence sequence S(k) = x*a(k-1) + a(k-2), starting (1, x, x^2+1, ...); the (k+1)-th term of the series = f(x) in the k-th degree polynomial: (1, (x), (x^2 + 1), (x^3 + 2x), (x^4 + 3x^2 + 1), (x^5 + 4x^3 + 3x), (x^6 + 5x^4 + 6x^2 + 1), ...). Example: let x = 2, then S(k) = 1, 2, 5, 12, 29, 70, 169, ... such that A000129(7) = 169 = f(x), x^6 + 5x^4 + 6x^2 + 1 = (64 + 80 + 24 + 1). - Gary W. Adamson, Apr 16 2008
Row k gives the nonzero coefficients of U(k,x/2) where U is the Chebyshev polynomial of the second kind. For example, row 6 is 1,5,6,1 and U(6,x/2) = x^6 - 5x^4 + 6x^2 - 1. - David Callan, Jul 22 2008
T(n,k) is the number of nodes at level k in the Fibonacci tree f(k-1). The Fibonacci trees f(k) of order k are defined as follows: 1. f(-1) and f(0) each consist of a single node. 2. For k >= 1, to the root of f(k-1), taken as the root of f(k), we attach with a rightmost edge the tree f(k-2). See the Iyer and Reddy references. These trees are not the same as the Fibonacci trees in A180566. Example: T(3,0)=1 and T(3,1)=2 because in f(2) = /\ we have 1 node at level 0 and 2 nodes at level 1. - Emeric Deutsch, Jun 21 2011
Triangle, with zeros omitted, given by (1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Dec 12 2011
Riordan array (1/(1-x),x^2/(1-x). - Philippe Deléham, Dec 12 2011
This sequence is the elements on the rising diagonals of the Pascal triangle, where the sum of the elements in each rising diagonal represents a Fibonacci number. - Mohammad K. Azarian, Mar 08 2012
If we set F(0;x) = 0, F(1;x) = 1, F(n+1;x) = x*F(n;x) + F(n-1;x), then we obtain the sequence of Vieta-Fibonacci polynomials discussed by Gary W. Adamson above. We note that F(n;x) = (-i)^n * U(n;i*x/2), where U denotes the respective Chebyshev polynomial of the second kind (see David Callan's remark above). Let us fix a,b,f(0),f(1) in C, b is not the zero, and set f(n) = a*f(n-1) + b*f(n-2). Then we deduce the relation: f(n) = b^((n-1)/2) * F(n;a/sqrt(b))*f(1) + b^(n/2) * F(n-1;a/sqrt(b))*f(0), where for a given value of the complex root sqrt(b) we set b^(n/2) = (sqrt(b))^n. Moreover, if b=1 then we get f(n+k) + (-1)^k * f(n-k) = L(k;a)*f(n), for every k=0,1,...,n, and where L(0;a)=2, L(1;a)=a, L(n+1;a)=a*L(n;a) + L(n-1;a) are the Vieta-Lucas polynomials. Let us observe that L(n+2;a) = F(n+2;a) + F(n;a), L(m+n;a) = L(m;a)*F(n;a) + L(m-1;a)*F(n-1;a), which implies also L(n+1;a) = a*F(n;a) + 2*F(n-1;a). Further we have L(n;a) = 2*(-i)^n * T(n;i*x/2), where T(n;x) denotes the n-th Chebyshev polynomial of the first kind. For the proofs, other relations and facts - see Witula-Slota's papers. - Roman Witula, Oct 12 2012
The diagonal sums of this triangle are A000930. - John Molokach, Jul 04 2013
Aside from signs and index shift, the coefficients of the characteristic polynomial of the Coxeter adjacency matrix for the Coxeter group A_n related to the Chebyshev polynomial of the second kind (cf. Damianou link p. 19). - Tom Copeland, Oct 11 2014
For a mirrored, shifted version showing the relation of these coefficients to the Pascal triangle, Fibonacci, and other number triangles, see A030528. See also A053122 for a relation to Cartan matrices. - Tom Copeland, Nov 04 2014
For a relation to a formulation for a universal Lie Weyl algebra for su(1,1), see page 16 of Durov et al. - Tom Copeland, Nov 29 2014
A reversed, signed and aerated version is given by A049310, related to Chebyshev polynomials. - Tom Copeland, Dec 06 2015
For n >= 3, the n-th row gives the coefficients of the independence polynomial of the (n-2)-path graph P_{n-2}. - Eric W. Weisstein, Apr 07 2017
For n >= 2, the n-th row gives the coefficients of the matching-generating polynomial of the (n-1)-path graph P_{n-1}. - Eric W. Weisstein, Apr 10 2017
Antidiagonals of the Pascal matrix A007318 read bottom to top. These are also the antidiagonals read from top to bottom of the numerical coefficients of the Maurer-Cartan form matrix of the Leibniz group L^(n)(1,1) presented on p. 9 of the Olver paper), which is generated as exp[c. * M] with (c.)^n = c_n and M the Lie infinitesimal generator A218272. Reverse is A102426. - Tom Copeland, Jul 02 2018
T(n,k) is the number of Markov equivalence classes with skeleton the path on n+1 nodes having exactly k immoralities. See Theorem 2.1 in the article by A. Radhakrishnan et al. below. - Liam Solus, Aug 23 2018
T(n, k) = number of compositions of n+1 into n+1-2*k odd parts. For example, T(6,2) = 6 because 7 = 5+1+1 = 3+3+1 = 3+1+3 = 1+1+5 = 1+3+3 = 1+1+5. - Michael Somos, Sep 19 2019
From Gary W. Adamson, Apr 25 2022: (Start)
Alternate rows can be parsed into those with odd integer coefficients to the right of the leftmost 1, and those with even integer coefficients to the right of the leftmost 1. The first set is shown in A054142 and are characteristic polynomials of submatrices of an infinite tridiagonal matrix (A332602) with all -1's in the super and subdiagonals and (1,2,2,2,...) as the main diagonal. For example, the characteristic equation of the 3 X 3 submatrix (1,-1,0; -1,2,-1; 0,-1,2) is x^3 - 5x^2 + 6x - 1. The roots are the Beraha constants B(7,1) = 3.24697...; B(7,2) = 1.55495...; and B(7,3) = 0.198062.... For n X n matrices of this form, the largest eigenvalue is B(2n+1, 1). The 3 X 3 matrix has an eigenvalue of 3.24697... = B(7,1).
Polynomials with even integer coefficients to the right of the leftmost 1 are in A053123 with roots being the even-indexed Beraha constants. The generating Cartan matrices are those with (2,2,2,...) as the main diagonal and -1's as the sub- and superdiagonals. The largest eigenvalue of n X n matrices of this form are B(2n+2,1). For example, the largest eigenvalue of (2,-1,0; -1,2,-1; 0,-1,2) is 3.414... = B(8,1) = a root to x^3 - 6x^2 + 10x - 4. (End)
T(n,k) is the number of edge covers of P_(n+2) with (n-k) edges. For example, T(6,2)=6 because among edges 1, 2, ..., 7 of P_8, we can eliminate any two non-consecutive edges among 2-6. These numbers can be found using the recurrence relation for the edge cover polynomial of P_n, which is E(P_n,x) = xE(P_(n-1),x)+xE(P_(n-2),x) and E(P_1,x)=0, E(P_2,x)=x (ref. Akbari and Oboudi). - Feryal Alayont, Jun 03 2022
T(n,k) is the number of ways to tile an n-board (an n X 1 array of 1 X 1 cells) using k dominoes and n-2*k squares. - Michael A. Allen, Dec 28 2022
T(n,k) is the number of positive integer sequences (s(1),s(2),...,s(n-2k)) such that s(i) < s(i+1), s(1) is odd, s(n-2k) <= n, and s(i) and s(i+1) have opposite parity (ref. Donnelly, Dunkum, and McCoy). Example: T(6,0)=1 corresponds to 123456; T(6,1)=5 corresponds to 1234, 1236, 1256, 1456, 3456; T(6,2)=6 corresponds to 12, 14, 16, 34, 36; and T(6,3)=1 corresponds to the empty sequence () with length 0. - Molly W. Dunkum, Jun 27 2023

			The first few Fibonacci polynomials (defined here by F(0,x) = 0, F(1,x) = 1; F(n+1, x) = F(n, x) + x*F(n-1, x)) are:
0: 0
1: 1
2: 1
3: 1 + x
4: 1 + 2*x
5: 1 + 3*x + x^2
6: (1 + x)*(1 + 3*x)
7: 1 + 5*x + 6*x^2 + x^3
8: (1 + 2*x)*(1 + 4*x + 2*x^2)
9: (1 + x)*(1 + 6*x + 9*x^2 + x^3)
10: (1 + 3*x + x^2 )*(1 + 5*x + 5*x^2)
11: 1 + 9*x + 28*x^2 + 35*x^3 + 15*x^4 + x^5
From _Roger L. Bagula_, Feb 20 2009: (Start)
  1
  1
  1   1
  1   2
  1   3   1
  1   4   3
  1   5   6   1
  1   6  10   4
  1   7  15  10   1
  1   8  21  20   5
  1   9  28  35  15   1
  1  10  36  56  35   6
  1  11  45  84  70  21   1
  1  12  55 120 126  56   7 (End)
For n=9 and k=4, T(9,4) = C(5,4) = 5 since there are exactly five size-4 subsets of {1,2,...,8} that contain no consecutive integers, namely, {1,3,5,7}, {1,3,5,8}, {1,3,6,8}, {1,4,6,8}, and {2,4,6,8}. - _Dennis P. Walsh_, Mar 31 2011
When the rows of the triangle are displayed as centered text, the falling diagonal sums are A005314. The first few terms are row1 = 1 = 1; row2 = 1+1 = 2; row3 = 2+1 = 3; row4 = 1+3+1 = 5; row5 = 1+3+4+1 = 9; row6 = 4+6+5+1 = 16; row7 = 1+10+10+6+1 = 28; row8 = 1+5+20+15+7+1 = 49; row9 = 6+15+35+21+8+1 = 86; row10 = 1+21+35+56+28+9+1 = 151. - _John Molokach_, Jul 08 2013
In the example, you can see that the n-th row of Pascal's triangle is given by T(n, 0), T(n+1, 1), ..., T(2n-1, n-1), T(2n, n). - _Daniel Forgues_, Jul 07 2018

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Index entries for triangles and arrays related to Pascal's triangle

Row sums = A000045(n+1) (Fibonacci numbers). - Michael Somos, Apr 02 1999
Cf. A000129, A054123, A052553, A000930, A030528, A053122.
All of A011973, A092865, A098925, A102426, A169803 describe essentially the same triangle in different ways.
Cf. A007318, A025581, A055087.
Cf. A000108, A049310, A055151, A098474, A124644, A134264.
Cf. A054142, A053123, A332602.

a011973 n k = a011973_tabf !! n !! k
a011973_row n = a011973_tabf !! n
a011973_tabf = zipWith (zipWith a007318) a025581_tabl a055087_tabf
-- Reinhard Zumkeller, Jul 14 2015

a := proc(n) local k; [ seq(binomial(n-k,k),k=0..floor(n/2)) ]; end;
T := proc(n, k): if k<0 or k>floor(n/2) then return(0) fi: binomial(n-k, k) end: seq(seq(T(n,k), k=0..floor(n/2)), n=0..15); # Johannes W. Meijer, Aug 26 2013

(* first: sum method *) Table[CoefficientList[Sum[Binomial[n - m + 1, m]*x^m, {m, 0, Floor[(n + 1)/2]}], x], {n, 0, 12}] (* Roger L. Bagula, Feb 20 2009 *)
(* second: polynomial recursion method *) Clear[L, p, x, n, m]; L[x, 0] = 1; L[x, 1] = 1 + x; L[x_, n_] := L[x, n - 1] + x*L[x, n - 2]; Table[ExpandAll[L[x, n]], {n, 0, 10}]; Table[CoefficientList[ExpandAll[L[x, n]], x], {n, 0, 12}]; Flatten[%] (* Roger L. Bagula, Feb 20 2009 *)
(* Center option shows falling diagonals are A224838 *) Column[Table[Binomial[n - m, m], {n, 0, 25}, {m, 0, Floor[n/2]}], Center] (* John Molokach, Jul 26 2013 *)
Table[ Select[ CoefficientList[ Fibonacci[n, x], x], Positive] // Reverse, {n, 1, 18} ] // Flatten (* Jean-François Alcover, Oct 21 2013 *)
CoefficientList[LinearRecurrence[{1, x}, {1 + x, 1 + 2 x}, {-1, 10}], x] // Flatten (* Eric W. Weisstein, Apr 07 2017 *)
CoefficientList[Table[x^((n - 1)/2) Fibonacci[n, 1/Sqrt[x]], {n, 15}], x] // Flatten (* Eric W. Weisstein, Apr 07 2017 *)

{T(n, k) = if( k<0 || 2*k>n, 0, binomial(n-k, k))};

# Prints the table; cf. A145574.
for n in (2..20): [Compositions(n, length=m, min_part=2).cardinality() for m in (1..n//2)]  # Peter Luschny, Oct 18 2012

Let F(n, x) be the n-th Fibonacci polynomial in x; the g.f. for F(n, x) is Sum_{n>=0} F(n, x)*y^n = (1 + x*y)/(1 - y - x*y^2). - Paul D. Hanna
T(m, n) = 0 for n != 0 and m <= 1 T(0, 0) = T(1, 0) = 1 T(m, n) = T(m - 1, n) + T(m-2, n-1) for m >= 2 (i.e., like the recurrence for Pascal's triangle A007318, but going up one row as well as left one column for the second summand). E.g., T(7, 2) = 10 = T(6, 2) + T(5, 1) = 6 + 4. - Rob Arthan, Sep 22 2003
G.f. for k-th column: x^(2*k-1)/(1-x)^(k+1).
Identities for the Fibonacci polynomials F(n, x):
F(m+n+1, x) = F(m+1, x)*F(n+1, x) + x*F(m, x)F(n, x).
F(n, x)^2-F(n-1, x)*F(n+1, x) = (-x)^(n-1).
The degree of F(n, x) is floor((n-1)/2) and F(2p, x) = F(p, x) times a polynomial of equal degree which is 1 mod p.
From Roger L. Bagula, Feb 20 2009: (Start)
p(x,n) = Sum_{m=0..floor((n+1)/2)} binomial(n-m+1, m)*x^m;
p(x,n) = p(x, n - 1) + x*p(x, n - 2). (End)
T(n, k) = A102541(2*n+2, 2*k+1) + A102541(2*n+1, 2*k) - A102541(2*n+3, 2*k+1), n >= 0 and 0 <= k <= floor(n/2). - Johannes W. Meijer, Aug 26 2013
G.f.: 1/(1-x-y*x^2) = R(0)/2, where R(k) = 1 + 1/(1 - (2*k+1+ x*y)*x/((2*k+2+ x*y)*x + 1/R(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Nov 09 2013
O.g.f. G(x,t) = x/(1-x-tx^2) = x + x^2 + (1+t) x^3 + (1+2t) x^4 + ... has the inverse Ginv(x,t) = -[1+x-sqrt[(1+x)^2 + 4tx^2]]/(2tx) = x - x^2 + (1-t) x^3 + (-1+3t) x^4 + ..., an o.g.f. for the signed Motzkin polynomials of A055151, consistent with A134264 with h_0 = 1, h_1 = -1, h_2 = -t, and h_n = 0 otherwise. - Tom Copeland, Jan 21 2016
O.g.f. H(x,t) = x (1+tx)/ [1-x(1+tx)] = x + (1+t) x^2 + (1+2t) x^3 + ... = -L[Cinv(-tx)/t], where L(x) = x/(1+x) with inverse Linv(x) = x/(1-x) and Cinv(x) = x (1-x) is the inverse of C(x) = (1-sqrt(1-4x))/2, the o.g.f. of the shifted Catalan numbers A000108. Then Hinv(x,t) = -C[t Linv(-x)]/t = [-1 + sqrt(1+4tx/(1+x))]/2t = x - (1+t) x^2 + (1+2t+2t^2) x^3 - (1+3t+6t^2+5t^3) x^4 + ..., which is signed A098474, reverse of A124644. - Tom Copeland, Jan 25 2016
T(n, k) = GegenbauerC(k, (n+1)/2-k, 1). - Peter Luschny, May 10 2016

A078812 Triangle read by rows: T(n, k) = binomial(n+k-1, 2*k-1).

Original entry on oeis.org

1, 2, 1, 3, 4, 1, 4, 10, 6, 1, 5, 20, 21, 8, 1, 6, 35, 56, 36, 10, 1, 7, 56, 126, 120, 55, 12, 1, 8, 84, 252, 330, 220, 78, 14, 1, 9, 120, 462, 792, 715, 364, 105, 16, 1, 10, 165, 792, 1716, 2002, 1365, 560, 136, 18, 1, 11, 220, 1287, 3432, 5005, 4368, 2380, 816, 171, 20, 1

Offset: 0

Michael Somos, Dec 05 2002

Warning: formulas and programs sometimes refer to offset 0 and sometimes to offset 1.
Apart from signs, identical to A053122.
Coefficient array for Morgan-Voyce polynomial B(n,x); see A085478 for references. - Philippe Deléham, Feb 16 2004
T(n,k) is the number of compositions of n having k parts when there are q kinds of part q (q=1,2,...). Example: T(4,2) = 10 because we have (1,3),(1,3'),(1,3"), (3,1),(3',1),(3",1),(2,2),(2,2'),(2',2) and (2',2'). - Emeric Deutsch, Apr 09 2005
T(n, k) is also the number of idempotent order-preserving full transformations (of an n-chain) of height k (height(alpha) = |Im(alpha)|). - Abdullahi Umar, Oct 02 2008
This sequence is jointly generated with A085478 as a triangular 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*v(n-1)x and v(n,x) = u(n-1,x) + (x+1)*v(n-1,x). See the Mathematica section. - Clark Kimberling, Feb 25 2012
Concerning Kimberling's recursion relations, see A102426. - Tom Copeland, Jan 19 2016
Subtriangle of the triangle T(n,k), 0 <= k <= n, read by rows, given by (0, 2, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 27 2012
From Wolfdieter Lang, Aug 30 2012: (Start)
With offset [0,0] the triangle with entries R(n,k) = T(n+1,k+1):= binomial(n+k+1, 2*k+1), n >= k >= 0, and zero otherwise, becomes the Riordan lower triangular convolution matrix R = (G(x)/x, G(x)) with G(x):=x/(1-x)^2 (o.g.f. of A000027). This means that the o.g.f. of column number k of R is (G(x)^(k+1))/x. This matrix R is the inverse of the signed Riordan lower triangular matrix A039598, called in a comment there S.
The Riordan matrix with entries R(n,k), just defined, provides the transition matrix between the sequence entry F(4*m*(n+1))/L(2*l), with m >= 0, for n=0,1,... and the sequence entries 5^k*F(2*m)^(2*k+1) for k = 0,1,...,n, with F=A000045 (Fibonacci) and L=A000032 (Lucas). Proof: from the inverse of the signed triangle Riordan matrix S used in a comment on A039598.
For the transition matrix R (T with offset [0,0]) defined above, row n=2: F(12*m) /L(2*m) = 3*5^0*F(2*m)^1 + 4*5^1*F(2*m)^3 + 1*5^2*F(2*m)^5, m >= 0. (End)
From R. Bagula's comment in A053122 (cf. Damianou link p. 10), this array gives the coefficients (mod sign) of the characteristic polynomials for the Cartan matrix of the root system A_n. - Tom Copeland, Oct 11 2014
For 1 <= k <= n, T(n,k) equals the number of (n-1)-length ternary words containing k-1 letters equal 2 and avoiding 01. - Milan Janjic, Dec 20 2016
The infinite sum (Sum_{i >= 0} (T(s+i,1+i) / 2^(s+2*i)) * zeta(s+1+2*i)) = 1 allows any zeta(s+1) to be expressed as a sum of rational multiples of zeta(s+1+2*i) having higher arguments. For example, zeta(3) can be expressed as a sum involving zeta(5), zeta(7), etc. The summation for each s >= 1 uses the s-th diagonal of the triangle. - Robert B Fowler, Feb 23 2022
The convolution triangle of the nonnegative integers. - Peter Luschny, Oct 07 2022

			Triangle begins, 1 <= k <= n:
                          1
                        2   1
                      3   4   1
                    4  10   6   1
                  5  20  21   8   1
                6  35  56  36  10   1
              7  56 126 120  55  12   1
            8  84 252 330 220  78  14   1
From _Peter Bala_, Feb 11 2025: (Start)
The array factorizes as an infinite product of lower triangular arrays:
  / 1               \    / 1              \ / 1              \ / 1             \
  | 2    1           |   | 2   1          | | 0  1           | | 0  1          |
  | 3    4   1       | = | 3   2   1      | | 0  2   1       | | 0  0  1       | ...
  | 4   10   6   1   |   | 4   3   2  1   | | 0  3   2  1    | | 0  0  2  1    |
  | 5   20  21   8  1|   | 5   4   3  2  1| | 0  4   3  2  1 | | 0  0  3  2  1 |
  |...               |   |...             | |...             | |...            |
Cf. A092276. (End)

T. D. Noe, Rows n = 0..50 of triangle, flattened
J.-P. Allouche and M. Mendès France, Stern-Brocot polynomials and power series, arXiv preprint arXiv:1202.0211 [math.NT], 2012. - From _N. J. A. Sloane_, May 10 2012
Peter Bala, Factorisations of some Riordan arrays as infinite products
Paul Barry, Symmetric Third-Order Recurring Sequences, Chebyshev Polynomials, and Riordan Arrays, JIS 12 (2009) 09.8.6.
Paul Barry, Riordan arrays, generalized Narayana triangles, and series reversion, Linear Algebra and its Applications, 491 (2016) 343-385.
Paul Barry, Notes on the Hankel transform of linear combinations of consecutive pairs of Catalan numbers, arXiv:2011.10827 [math.CO], 2020.
P. Damianou, On the characteristic polynomials of Cartan matrices and Chebyshev polynomials, arXiv preprint arXiv:1110.6620 [math.RT], 2014.
Nour-Eddine Fahssi, On the combinatorics of exclusion in Haldane fractional statistics, arXiv:1808.00045 [cond-mat.stat-mech], 2018.
Milan Janjić, Words and Linear Recurrences, J. Int. Seq. 21 (2018), #18.1.4.
A. Laradji, and A. Umar, Combinatorial results for semigroups of order-preserving full transformations, Semigroup Forum 72 (2006), 51-62.
Yidong Sun, Numerical triangles and several classical sequences, Fib. Quart. 43, no. 4, (2005) 359-370.

This triangle is formed from odd-numbered rows of triangle A011973 read in reverse order.
Row sums give A001906. With signs: A053122.
The column sequences are A000027, A000292, A000389, A000580, A000582, A001288 for k=1..6, resp. For k=7..24 they are A010966..(+2)..A011000 and for k=25..50 they are A017713..(+2)..A017763.
Cf. A128908, A053123, A049310, A119900, A085478, A029653, A106195.

Flat(List([0..12], n-> List([0..n], k-> Binomial(n+k+1, 2*k+1) ))); # G. C. Greubel, Aug 01 2019

a078812 n k = a078812_tabl !! n !! k
a078812_row n = a078812_tabl !! n
a078812_tabl = [1] : [2, 1] : f [1] [2, 1] where
   f us vs = ws : f vs ws where
     ws = zipWith (-) (zipWith (+) ([0] ++ vs) (map (* 2) vs ++ [0]))
                      (us ++ [0, 0])
-- Reinhard Zumkeller, Dec 16 2013

/* As triangle */ [[Binomial(n+k-1, 2*k-1): k in [1..n]]: n in [1.. 15]]; // Vincenzo Librandi, Jun 01 2018

for n from 1 to 11 do seq(binomial(n+k-1,2*k-1),k=1..n) od; # yields sequence in triangular form; Emeric Deutsch, Apr 09 2005
# Uses function PMatrix from A357368. Adds a row and column above and to the left.
PMatrix(10, n -> n); # Peter Luschny, Oct 07 2022

(* First program *)
u[1, x_]:= 1; v[1, x_]:= 1; z = 13;
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];
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[%] (* A085478 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%] (* A078812 *) (* Clark Kimberling, Feb 25 2012 *)
(* Second program *)
Table[Binomial[n+k+1, 2*k+1], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Aug 01 2019 *)

T(n,m):=sum(binomial(2*k,n-m)*binomial(m+k,k)*(-1)^(n-m+k)*binomial(n+1,m+k+1),k,0,n-m); /* Vladimir Kruchinin, Apr 13 2016 */

{T(n, k) = if( n<0, 0, binomial(n+k-1, 2*k-1))};

{T(n, k) = polcoeff( polcoeff( x*y / (1 - (2 + y) * x + x^2) + x * O(x^n), n), k)};

@cached_function
def T(k,n):
    if k==n: return 1
    if k==0: return 0
    return sum(i*T(k-1,n-i) for i in (1..n-k+1))
A078812 = lambda n,k: T(k,n)
[[A078812(n,k) for k in (1..n)] for n in (1..8)] # Peter Luschny, Mar 12 2016

[[binomial(n+k+1, 2*k+1) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Aug 01 2019

G.f.: x*y / (1 - (2 + y)*x + x^2). To get row n, expand this in powers of x then expand the coefficient of x^n in increasing powers of y.
From Philippe Deléham, Feb 16 2004: (Start)
If indexing begins at 0 we have
T(n,k) = (n+k+1)!/((n-k)!*(2k+1))!.
T(n,k) = Sum_{j>=0} T(n-1-j, k-1)*(j+1) with T(n, 0) = n+1, T(n, k) = 0 if n < k.
T(n,k) = T(n-1, k-1) + T(n-1, k) + Sum_{j>=0} (-1)^j*T(n-1, k+j)*A000108(j) with T(n,k) = 0 if k < 0, T(0, 0)=1 and T(0, k) = 0 for k > 0.
G.f. for the column k: Sum_{n>=0} T(n, k)*x^n = (x^k)/(1-x)^(2k+2).
Row sums: Sum_{k>=0} T(n, k) = A001906(n+1). (End)
Antidiagonal sums are A000079(n) = Sum_{k=0..floor(n/2)} binomial(n+k+1, n-k). - Paul Barry, Jun 21 2004
Riordan array (1/(1-x)^2, x/(1-x)^2). - Paul Barry, Oct 22 2006
T(0,0) = 1, T(n,k) = 0 if k < 0 or if k > n, T(n,k) = T(n-1,k-1) + 2*T(n-1,k) - T(n-2,k). - Philippe Deléham, Jan 26 2010
For another version see A128908. - Philippe Deléham, Mar 27 2012
T(n,m) = Sum_{k=0..n-m} (binomial(2*k,n-m)*binomial(m+k,k)*(-1)^(n-m+k)* binomial(n+1,m+k+1)). - Vladimir Kruchinin, Apr 13 2016
T(n, k) = T(n-1, k) + (T(n-1, k-1) + T(n-2, k-1) + T(n-3, k-1) + ...) for k >= 2 with T(n, 1) = n. - Peter Bala, Feb 11 2025
From Peter Bala, May 04 2025: (Start)
With the column offset starting at 0, the n-th row polynomial B(n, x) = 1/sqrt(x + 4) * Chebyshev_U(2*n+1, (1/2)*sqrt(x + 4)) = (-1)^n * Chebyshev_U(n, -(1/2)*(x + 2)).
B(n, x) / Product_{k = 1..2*n} (1 + 1/B(k, x)) = b(n, x), the n-th row polynomial of A085478. (End)

Edited by N. J. A. Sloane, Apr 28 2008

A053122 Triangle of coefficients of Chebyshev's S(n,x-2) = U(n,x/2-1) polynomials (exponents of x in increasing order).

Original entry on oeis.org

1, -2, 1, 3, -4, 1, -4, 10, -6, 1, 5, -20, 21, -8, 1, -6, 35, -56, 36, -10, 1, 7, -56, 126, -120, 55, -12, 1, -8, 84, -252, 330, -220, 78, -14, 1, 9, -120, 462, -792, 715, -364, 105, -16, 1, -10, 165, -792, 1716, -2002, 1365, -560, 136, -18, 1, 11, -220, 1287, -3432, 5005, -4368, 2380, -816, 171, -20

Offset: 0

Wolfdieter Lang

Apart from signs, identical to A078812.
Another version with row-leading 0's and differing signs is given by A285072.
G.f. for row polynomials S(n,x-2) (signed triangle): 1/(1+(2-x)*z+z^2). Unsigned triangle |a(n,m)| has g.f. 1/(1-(2+x)*z+z^2) for row polynomials.
Row sums (signed triangle) A049347(n) (periodic(1,-1,0)). Row sums (unsigned triangle) A001906(n+1)=F(2*(n+1)) (even-indexed Fibonacci).
In the language of Shapiro et al. (see A053121 for the reference) such a lower triangular (ordinary) convolution array, considered as a matrix, belongs to the Bell-subgroup of the Riordan-group.
The (unsigned) column sequences are A000027, A000292, A000389, A000580, A000582, A001288 for m=0..5, resp. For m=6..23 they are A010966..(+2)..A011000 and for m=24..49 they are A017713..(+2)..A017763.
Riordan array (1/(1+x)^2,x/(1+x)^2). Inverse array is A039598. Diagonal sums have g.f. 1/(1+x^2). - Paul Barry, Mar 17 2005. Corrected by Wolfdieter Lang, Nov 13 2012.
Unsigned version is in A078812. - Philippe Deléham, Nov 05 2006
Also row n gives (except for an overall sign) coefficients of characteristic polynomial of the Cartan matrix for the root system A_n. - Roger L. Bagula, May 23 2007
From Wolfdieter Lang, Nov 13 2012: (Start)
The A-sequence for this Riordan triangle is A115141, and the Z-sequence is A115141(n+1), n>=0. For A- and Z-sequences for Riordan matrices see the W. Lang link under A006232 with details and references.
S(n,x^2-2) = sum(r(j,x^2),j=0..n) with Chebyshev's S-polynomials and r(j,x^2) := R(2*j+1,x)/x, where R(n,x) are the monic integer Chebyshv T-polynomials with coefficients given in A127672. Proof from comparing the o.g.f. of the partial sum of the r(j,x^2) polynomials (see a comment on the signed Riordan triangle A111125) with the present Riordan type o.g.f. for the row polynomials with x -> x^2.  (End)
S(n,x^2-2) = S(2*n+1,x)/x, n >= 0, from the odd part of the bisection of the o.g.f. - Wolfdieter Lang, Dec 17 2012
For a relation to a generator for the Narayana numbers A001263, see A119900, whose columns are unsigned shifted rows (or antidiagonals) of this array, referring to the tables in the example sections. - Tom Copeland, Oct 29 2014
The unsigned rows of this array are alternating rows of a mirrored A011973 and alternating shifted rows of A030528 for the Fibonacci polynomials. - Tom Copeland, Nov 04 2014
Boas-Buck type recurrence for column k >= 0 (see Aug 10 2017 comment in A046521 with references): a(n, m) =  (2*(m + 1)/(n - m))*Sum_{k = m..n-1} (-1)^(n-k)*a(k, m), with input a(n, n) = 1, and a(n,k) = 0 for n < k. - Wolfdieter Lang, Jun 03 2020
Row n gives the characteristic polynomial of the (n X n)-matrix M where M[i,j] = 2 if i = j, -1 if |i-j| = 1 and 0 otherwise. The matrix M is positive definite and has 2-condition number (cot(Pi/(2*n+2)))^2. - Jianing Song, Jun 21 2022
Also the convolution triangle of (-1)^(n+1)*n. - Peter Luschny, Oct 07 2022

			The triangle a(n,m) begins:
n\m   0    1    2     3     4     5     6    7    8  9
0:    1
1:   -2    1
2:    3   -4    1
3:   -4   10   -6     1
4:    5  -20   21    -8     1
5:   -6   35  -56    36   -10     1
6:    7  -56  126  -120    55   -12     1
7:   -8   84 -252   330  -220    78   -14    1
8:    9 -120  462  -792   715  -364   105  -16    1
9:  -10  165 -792  1716 -2002  1365  -560  136  -18  1
... Reformatted and extended by _Wolfdieter Lang_, Nov 13 2012
E.g., fourth row (n=3) {-4,10,-6,1} corresponds to the polynomial S(3,x-2) = -4+10*x-6*x^2+x^3.
From _Wolfdieter Lang_, Nov 13 2012: (Start)
Recurrence: a(5,1) = 35 = 1*5 + (-2)*(-20) -1*(10).
Recurrence from Z-sequence [-2,-1,-2,-5,...]: a(5,0) = -6 = (-2)*5 + (-1)*(-20) + (-2)*21 + (-5)*(-8) + (-14)*1.
Recurrence from A-sequence [1,-2,-1,-2,-5,...]: a(5,1) = 35 = 1*5  + (-2)*(-20) + (-1)*21 + (-2)*(-8) + (-5)*1.
(End)
E.g., the fourth row (n=3) {-4,10,-6,1} corresponds also to the polynomial S(7,x)/x = -4 + 10*x^2 - 6*x^4 + x^6. - _Wolfdieter Lang_, Dec 17 2012
Boas-Buck type recurrence: -56 = a(5, 2) = 2*(-1*1 + 1*(-6) - 1*21) = -2*28 = -56. - _Wolfdieter Lang_, Jun 03 2020

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 795.
Theodore J. Rivlin, Chebyshev polynomials: from approximation theory to algebra and number theory, 2. ed., Wiley, New York, 1990.
R. N. Cahn, Semi-Simple Lie Algebras and Their Representations, Dover, NY, 2006, ISBN 0-486-44999-8, p. 62.
Sigurdur Helgasson, Differential Geometry, Lie Groups and Symmetric Spaces, Graduate Studies in Mathematics, volume 34. A. M. S.: ISBN 0-8218-2848-7, 1978, p. 463.

T. D. Noe, Rows n=0..50 of triangle, flattened
Wolfdieter Lang, First rows of the triangle.
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].
P. Barry, Symmetric Third-Order Recurring Sequences, Chebyshev Polynomials, and Riordan Arrays, JIS 12 (2009) 09.8.6.
Naiomi T. Cameron and Asamoah Nkwanta, On some (pseudo) involutions in the Riordan group, J. of Integer Sequences, 8(2005), 1-16.
P. Damianou, On the characteristic polynomials of Cartan matrices and Chebyshev polynomials, arXiv preprint arXiv:1110.6620 [math.RT], 2014 (p. 10). - _Tom Copeland_, Oct 11 2014
Pentti Haukkanen, Jorma Merikoski, Seppo Mustonen, Some polynomials associated with regular polygons, Acta Univ. Sapientiae, Mathematica, 6, 2 (2014) 178-193.
Eric Weisstein's World of Mathematics, Cartan Matrix
Eric Weisstein's World of Mathematics, Dynkin Diagram
Index entries for sequences related to Chebyshev polynomials.

Cf. A005248, A127677, A053123, A049310.
Cf. A011973, A030528, A034867, A119900.
Cf. A285072 (version with row-leading 0's and differing signs). - Eric W. Weisstein, Apr 09 2017

seq(seq((-1)^(n+m)*binomial(n+m+1,2*m+1),m=0..n),n=0..10); # Robert Israel, Oct 15 2014
# Uses function PMatrix from A357368. Adds a row above and a column to the left.
PMatrix(10, n -> -(-1)^n*n); # Peter Luschny, Oct 07 2022

T[n_, m_, d_] := If[ n == m, 2, If[n == m - 1 || n == m + 1, -1, 0]]; M[d_] := Table[T[n, m, d], {n, 1, d}, {m, 1, d}]; a = Join[M[1], Table[CoefficientList[Det[M[d] - x*IdentityMatrix[d]], x], {d, 1, 10}]]; Flatten[a] (* Roger L. Bagula, May 23 2007 *)
(* Alternative code for the matrices from MathWorld: *)
sln[n_] := 2IdentityMatrix[n] - PadLeft[PadRight[IdentityMatrix[n - 1], {n, n - 1}], {n, n}] - PadLeft[PadRight[IdentityMatrix[n - 1], {n - 1, n}], {n, n}] (* Roger L. Bagula, May 23 2007 *)

@CachedFunction
def A053122(n,k):
    if n< 0: return 0
    if n==0: return 1 if k == 0 else 0
    return A053122(n-1,k-1)-A053122(n-2,k)-2*A053122(n-1,k)
for n in (0..9): [A053122(n,k) for k in (0..n)] # Peter Luschny, Nov 20 2012

a(n, m) := 0 if n

a(n, m) = -2*a(n-1, m) + a(n-1, m-1) - a(n-2, m), a(n, -1) := 0 =: a(-1, m), a(0, 0)=1, a(n, m) := 0 if n

O.g.f. for m-th column (signed triangle): ((x/(1+x)^2)^m)/(1+x)^2.

From Jianing Song, Jun 21 2022: (Start)

T(n,k) = [x^k]f_n(x), where f_{-1}(x) = 0, f_0(x) = 1, f_n(x) = (x-2)*f_{n-1}(x) - f_{n-2}(x) for n >= 2.

f_n(x) = (((x-2+sqrt(x^2-4*x))/2)^(n+1) - ((x-2-sqrt(x^2-4*x))/2)^(n+1))/sqrt(x^2-4x).

The roots of f_n(x) are 2 + 2*cos(k*Pi/(n+1)) = 4*(cos(k*Pi/(2*n+2)))^2 for 1 <= k <= n. (End)

A054142 Triangular array binomial(2*n-k, k), k=0..n, n >= 0.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 5, 6, 1, 1, 7, 15, 10, 1, 1, 9, 28, 35, 15, 1, 1, 11, 45, 84, 70, 21, 1, 1, 13, 66, 165, 210, 126, 28, 1, 1, 15, 91, 286, 495, 462, 210, 36, 1, 1, 17, 120, 455, 1001, 1287, 924, 330, 45, 1, 1, 19, 153, 680, 1820, 3003, 3003, 1716, 495, 55, 1

Offset: 0

Clark Kimberling

Row sums are odd-indexed Fibonacci numbers.
T(n,k) is the number of nondecreasing Dyck paths of semilength n+1, having k double rises. Mirror image of A085478. - Emeric Deutsch, May 31 2004
Diagonal sums are A052535. - Paul Barry, Jan 21 2005
Matrix inverse is the triangle of Salie numbers A098435. - Paul Barry, Jan 21 2005
Coefficients of Morgan-Voyce polynomial b(n,x); e.g., b(3,x)=x^3+5x^2+6x+1. See A172431 for coefficients of Morgan-Voyce polynomial B(n,x). - Clark Kimberling, Feb 13 2010
T(n,k) is the number of stack polyominoes of perimeter 2n+4 with k+1 columns. - Emanuele Munarini, Apr 07 2011
Roots of signed n-th polynomials are chaotic with respect to the operation (-2, x^2), with cycle lengths A003558(n). Example: starting with a root to x^3 - 5x^2 + 6x - 1 = 0; (2 + 2*cos(2*Pi/N) = 3.24697... = A116415; we obtain the trajectory (3.24697...-> 1.55495...-> 0.198062...; the 3 roots to the polynomial with cycle length 3 matching A003558(3) = 3. The operation (-2, x^2) is the reversal of the well known chaotic operation (x^2 - 2) [Kappraff, Adamson, 2004] starting with seed 2*cos(2*Pi/N). Check: given 2*cos(2*Pi/7) = 1.24697..., we obtain the 3-cycle using (x^2 - 2): (1.24697...-> -0.445041...-> 1.801937...; where the terms in either set are intermediate terms in the other, irrespective of sign. - Gary W. Adamson, Sep 22 2011
A054142 is jointly generated with A172431 as an array of coefficients of polynomials u(n,x): initially, u(1,x)=v(1,x)=1; for n>1, u(n,x)=x*u(n-1,x)+v(n-1,x) and v(n,x)=x*u(n-1,x)+(x+1)*v(n-1,x). See the Mathematica section of A172431. - Clark Kimberling, Mar 09 2012
Subtriangle of the triangle given by (0, 1, 0, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Apr 01 2012
The o.g.f. for row n of the array A(n, k) = binomial(2*n-k,k), k >= 0, n >= 0 is G(n,x) = Sum_{k=0..n} T(n, k)*x^k + (-x)^(2*n+1) * c(-x)^(2*n+1) / sqrt(1-4*(-x)), for n >= 0. Here c(x) is the o.g.f. of A000108 (Catalan). For powers of c(x) see the W. Lang link in A115139. For the alternating sign case replace x by -x. - Wolfdieter Lang, Sep 12 2016
Multiplying the n-th diagonal by A001147(n) generates A001497. - Tom Copeland, Oct 04 2016

			Triangle begins:
  1;
  1,  1;
  1,  3,  1;
  1,  5,  6,   1;
  1,  7, 15,  10,   1;
  1,  9, 28,  35,  15,   1;
  1, 11, 45,  84,  70,  21,   1;
  1, 13, 66, 165, 210, 126,  28,  1;
  1, 15, 91, 286, 495, 462, 210, 36, 1; ...
...
(0, 1, 0, 0, 0, 0, ...) DELTA (1, 0, 1, 0, 0, 0, ...) begins:
  1;
  0, 1;
  0, 1, 1;
  0, 1, 3,  1;
  0, 1, 5,  6,  1;
  0, 1, 7, 15, 10,  1;
  0, 1, 9, 28, 35, 15, 1. _Philippe Deléham_, Apr 01 2012

Vincenzo Librandi, Table of n, a(n) for n = 0..495
E. Barcucci, A. Del Lungo, S. Fezzi and R. Pinzani, Nondecreasing Dyck paths and q-Fibonacci numbers, Discrete Math., 170, 1997, 211-217.
D. Dumont and J. Zeng, Polynomes d'Euler et les fractions continues de Stieltjes-Rogers, Ramanujan J. 2 (1998) 3, 387-410.
Molly Fenn and Eric Sommers, A transitivity result for ad-nilpotent ideals in type A, arXiv:2101.04091 [math.RT], 2021.
J. L. Jacobsen, and J. Salas, Transfer Matrices and Partition-Function Zeros for Antiferromagnetic Potts Models IV. Chromatic polynomial with cyclic boundary conditions, J. Stat. Phys. 122 (2006) 705-760, arXiv:cond-mat/0407444 See Eq. 2.27. Mentions this sequence. - _N. J. A. Sloane_, Mar 14 2014
Jay Kappraff and Gary W. Adamson, Polygons and Chaos, 5th Interdispl Symm. Congress and Exh. Jul 8-14, Sydney, 2001 - [with commercial pop-ups].
Jay Kappraff and Gary W. Adamson, Polygons and Chaos, Journal of Dynamical Systems and Geometric Theories, Vol. 2 pp. 79-94, (Nov 2004).

These are the even-indexed rows of A011973, the odd-indexed rows form A053123.
Cf. A000108, A003558, A027989, A052535, A054142, A076756, A084938, A085478, A098435, A115139, A172431, A172991, A188648.
Cf. A001147, A001497.

Flat(List([0..12], n-> List([0..n], k-> Binomial(2*n-k,k) ))); # G. C. Greubel, Aug 01 2019

[Binomial(2*n-k,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Aug 01 2019

T:=(n,k)->binomial(2*n-k,k): seq(seq(T(n,k), k=0..n), n=0..11);

Flatten[Table[Binomial[2n - k, k], {n, 0, 11}, {k, 0, n}]] (* Emanuele Munarini, Apr 07 2011 *)

create_list(binomial(2*n-k,k),n,0,10,k,0,n); /* Emanuele Munarini, Apr 07 2011 */

T(n,k)=if(n<0,0,polcoeff(charpoly(matrix(n,n,i,j,-min(i,j))),k))

[[binomial(2*n-k,k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Aug 01 2019

G.f.: (1-t*z)/((1-t*z)^2-z). - Emeric Deutsch, May 31 2004
Column k has g.f.: (Sum_{j=0..k+1} binomial(k+1, 2j)*x^j)*x^k/(1-x)^(k+1). - Paul Barry, Jun 22 2005
Recurrence: T(n+2,k+2) = T(n+1,k+2) + 2*T(n+1,k+1) - T(n,k). - Emanuele Munarini, Apr 07 2011
T(n, k) = binomial(2*n-k, k) = A085478(n, n-k), for n >= 0, k = 0..n. - Wolfdieter Lang, Mar 25 2020

A053125 Triangle of coefficients of Chebyshev's U(n,2*x-1) polynomials (exponents of x in decreasing order).

Original entry on oeis.org

1, 4, -2, 16, -16, 3, 64, -96, 40, -4, 256, -512, 336, -80, 5, 1024, -2560, 2304, -896, 140, -6, 4096, -12288, 14080, -7680, 2016, -224, 7, 16384, -57344, 79872, -56320, 21120, -4032, 336, -8, 65536, -262144, 430080, -372736, 183040, -50688, 7392, -480, 9, 262144, -1179648, 2228224, -2293760, 1397760

Offset: 0

Wolfdieter Lang

A000302 (powers of 4), A002699, A002700 unsigned column sequences for m=0..2.
G.f. for row polynomials U(n,2*x-1) and row sums same as for A053124.
With offset 1 this is also the coefficient triangle of 2* U(2*n-1,x) expanded in decreasing powers of x. W. Lang, Mar 07 2007.

			{1}; {4,-2}; {16,-16,3}; {64,-96,40,-4}; {256,-512,336,-80,5};... E.g. fourth row (n=3) corresponds to polynomial U^{*}(3,m)=U(3,2*x-1)= 64*x^3-96*x^2+40*x-4.

C. Lanczos, Applied Analysis. Prentice-Hall, Englewood Cliffs, NJ, 1956, p. 518.
Theodore J. Rivlin, Chebyshev polynomials: from approximation theory to algebra and number theory, 2. ed., Wiley, New York, 1990.

T. D. Noe, Rows n=0..50 of triangle, flattened
W. Lang, First rows and related triangles .
Eric Weisstein's World of Mathematics, Chebyshev Polynomial of the Second Kind
Index entries for sequences related to Chebyshev polynomials.

Cf. A053124, A053123.

Reverse /@ CoefficientList[Table[ChebyshevU[n, 2 x - 1], {n, 0, 10}], x] // Flatten (* Eric W. Weisstein, Apr 04 2018 *)
Reverse /@ CoefficientList[ChebyshevU[Range[0, 10], 2 x - 1], x] // Flatten (* Eric W. Weisstein, Apr 04 2018 *)

a(n, m) = A053124(n, n-m)= (4^(n-m))*A053123(n, m)= (4^(n-m))*((-1)^m)*binomial(2*n+1-m, m) if n >= m, else 0.

a(n, m) := -2*a(n-1, m-1)+4*a(n-1, m)-a(n-2, m-2), a(-2, m) := 0=: a(n, -2), a(-1, m) := 0=: a(n, -1), a(0, 0)=1, a(n, m)=0 if n

G.f. for m-th column (signed triangle): ((-x)^m)*Po(m+1, 4*x)/(1-4*x)^(m+1), with Po(k, x) := sum('binomial(k, 2*j+1)*x^j', 'j'=0..floor(k/2)).

A053126 Binomial coefficients binomial(2*n-3,4).

Original entry on oeis.org

5, 35, 126, 330, 715, 1365, 2380, 3876, 5985, 8855, 12650, 17550, 23751, 31465, 40920, 52360, 66045, 82251, 101270, 123410, 148995, 178365, 211876, 249900, 292825, 341055, 395010, 455126, 521855, 595665, 677040, 766480

Offset: 4

Wolfdieter Lang

Number of intersections of diagonals in the interior of regular (2n-3)-gon. - Philippe Deléham, Jun 07 2013

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

Vincenzo Librandi, Table of n, a(n) for n = 4..200
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].
Milan Janjic, Two Enumerative Functions University of Banja Luka (Bosnia and Herzegovina, 2017).
Ângela Mestre and José Agapito, Square Matrices Generated by Sequences of Riordan Arrays, J. Int. Seq., Vol. 22 (2019), Article 19.8.4.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A053123, A002492.

[Binomial(2*n-3,4): n in [4..40]]; // Vincenzo Librandi, Oct 07 2011

Table[Binomial[2*n-3,4], {n,4,50}] (* G. C. Greubel, Aug 26 2018 *)

for(n=4,50, print1(binomial(2*n-3,4), ", ")) \\ G. C. Greubel, Aug 26 2018

a(n) = binomial(2*n-3, 4) if n >= 4 else 0;
G.f.: (5+10*x+x^2)/(1-x)^5.
a(n) = A053123(n,4), n >= 4; a(n) = 0, n=0..3 (fifth column of shifted Chebyshev's S-triangle, decreasing order).
a(n) = A006561(2n-3). - Philippe Deléham, Jun 07 2013
E.g.f.: (90 - 84*x + 39*x^2 - 12*x^3 + 4*x^4)*exp(x)/6. - G. C. Greubel, Aug 26 2018
From Amiram Eldar, Jan 04 2022: (Start)
Sum_{n>=4} 1/a(n) = 34/3 - 16*log(2).
Sum_{n>=4} (-1)^n/a(n) = 2*Pi - 4*log(2) - 10/3. (End)

A172431 Even row Pascal-square read by antidiagonals.

Original entry on oeis.org

1, 1, 2, 1, 4, 3, 1, 6, 10, 4, 1, 8, 21, 20, 5, 1, 10, 36, 56, 35, 6, 1, 12, 55, 120, 126, 56, 7, 1, 14, 78, 220, 330, 252, 84, 8, 1, 16, 105, 364, 715, 792, 462, 120, 9, 1, 18, 136, 560, 1365, 2002, 1716, 792, 165, 10

Offset: 1

Mark Dols, Feb 02 2010

Apart from signs identical to A053123. Mirror of A078812.
As a triangle, row n consists of the coefficients of Morgan-Voyce polynomial B(n,x); e.g., B(3,x)=x^3+6x^2+10x+4. As a triangle, rows 0 to 4 are as follows: 1 1...2 1...4...3 1...6...10...4 1...8...21...20...5 See A054142 for coefficients of Morgan-Voyce polynomial b(n,x).
Scaled version of A119900. - Philippe Deléham, Feb 24 2012
A172431 is jointly generated with A054142 as an array of coefficients of polynomials v(n,x):  initially, u(1,x)=v(1,x)=1; for n>1, u(n,x)=x*u(n-1,x)+v(n-1,x) and v(n,x)=x*u(n-1,x)+(x+1)*v(n-1,x). See the Mathematica section. - Clark Kimberling, Mar 09 2012
Subtriangle of the triangle given by (1, 0, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 2, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 22 2012

			Array begins:
  1,  2,  3,  4,  5,  6, ...
  1,  4, 10, 20, 35, ...
  1,  6, 21, 56, ...
  1,  8, 36, ...
  1, 10, ...
  1, ...
  ...
Example:
Starting with 1, every entry is twice the one to the left minus the second one to the left, plus the one above.
For n = 9 the a(9) = 10 solution is 2*4 - 1 + 3.
From _Philippe Deléham_, Feb 24 2012: (Start)
Triangle T(n,k) begins:
  1;
  1,   2;
  1,   4,   3;
  1,   6,  10,   4;
  1,   8,  21,  20,   5;
  1,  10,  36,  56,  35,   6;
  1,  12,  55, 120, 126,  56,   7; (End)
From _Philippe Deléham_, Mar 22 2012: (Start)
(1, 0, 0, 0, 0, 0, ...) DELTA (0, 2, -1/2, 1/2, 0, 0, ...) begins:
  1;
  1,   0;
  1,   2,   0;
  1,   4,   3,   0;
  1,   6,  10,   4,   0;
  1,   8,  21,  20,   5,   0;
  1,  10,  36,  56,  35,   6,   0;
  1,  12,  55, 120, 126,  56,   7,   0; (End)

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

Cf. A078812, A053123, A007318, A001906 (antidiagonals sums), A007685.

Cf. also A054142, A082985.

F:=Factorial;; Flat(List([1..15], n-> List([1..n], k-> Sum([0..Int((k-1)/2)], j-> (-1)^j*F(n-j-1)*2^(k-2*j-1)/(F(j)*F(n-k)*F(k-2*j-1)) )))); # G. C. Greubel, Dec 15 2019

F:=Factorial; [ &+[(-1)^j*F(n-j-1)*2^(k-2*j-1)/(F(j)*F(n-k)*F(k-2*j-1)): j in [0..Floor((k-1)/2)]]: k in [1..n], n in [1..15]]; // G. C. Greubel, Dec 15 2019

T := (n, k) -> simplify(GegenbauerC(k, n-k, 1)):
for n from 0 to 10 do seq(T(n,k), k=0..n-1) od; # Peter Luschny, May 10 2016

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := x*u[n - 1, x] + v[n - 1, x];
v[n_, x_] := x*u[n - 1, x] + (x + 1)*v[n - 1, x];
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[%]    (* A054142 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]    (* A172431 *)
(* Clark Kimberling, Mar 09 2012 *)
Table[GegenbauerC[k-1, n-k+1, 1], {n, 15}, {k, n}]//Flatten (* G. C. Greubel, Dec 15 2019 *)

T(n,k) = sum(j=0, (k-1)\2, (-1)^j*(n-j-1)!*2^(k-2*j-1)/(j!*(n-k)!*(k-2*j-1)!) );
for(n=1, 10, for(k=1, n, print1(T(n,k), ", "))) \\ G. C. Greubel, Dec 15 2019

[[gegenbauer(k-1, n-k+1, 1) for k in (1..n)] for n in (1..15)] # G. C. Greubel, Dec 15 2019

As a decimal sequence: a(n)= 12*a(n-1)- a(n-2) with a(1)=1. [I interpret this remark as: 1, 12=1,2, 143=1,4,3, 1704=1,6,10,4,... taken from A004191 are decimals on the diagonal. - R. J. Mathar, Sep 08 2013]
As triangle T(n,k): T(n,k) = T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k-2). - Philippe Deléham, Feb 24 2012
As DELTA-triangle T(n,k) with 0<=k<=n: G.f.: (1-y*x)^2/((1-y*x)^2-x). - Philippe Deléham, Mar 22 2012
T(n, k) = GegenbauerC(k, n-k, 1). - Peter Luschny, May 10 2016
As triangle T(n,k): Product_{k=1..n} T(n,k) = Product_{k=0..n-1} binomial(2*k,k) = A007685(n-1)  for n >= 1. - Werner Schulte, Apr 26 2017
As triangle T(n,k) with 1 <= k <= n: T(n,k) = binomial(2*n-k, k-1). - Paul Weisenhorn, Nov 25 2019

A332602 Tridiagonal matrix M read by antidiagonals: main diagonal is 1,2,2,2,2,..., two adjacent diagonals are 1,1,1,1,1,...

Original entry on oeis.org

1, 1, 1, 0, 2, 0, 0, 1, 1, 0, 0, 0, 2, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0

Offset: 0

N. J. A. Sloane, Mar 06 2020, following a suggestion from Gary W. Adamson

From Gary W. Adamson, Mar 11 2020: (Start)
The upper left entry of M^n gives the Catalan numbers A000108. Extracting 2 X 2, 3 X 3, and 4 X 4 submatrices from M; then generating sequences from the upper left entries of M^n, we obtain the following sequences:
1, 1, 2, 5, 13, ... = A001519 and the convergent is 2.61803... = 2 + 2*cos(2*Pi/5) = (2*cos(Pi/5))^2.
1, 1, 2, 5, 14, 42, 131, ... = A080937 and the convergent is 3.24697... = 2 + 2*cos(2*Pi/7) = (2*cos(Pi/7))^2.
1, 1, 2, 5, 14, 42, 132, 429, 1429, ... = A080938 and the convergent is 3.53208... = 2 + 2*cos(2*Pi/9) = (2*cos(Pi/9))^2. (End)
The characteristic polynomial for the N X N main submatrix M_N is Phi(N, x) = S(N, 2-x) - S(N-1, 2-x), with Chebyshev's S polynomial (see A049310) evaluated at 2-x. Proof by determinant expansion, to obtain the recurrence Phi(N, x) - (x-2)*Phi(N-1, x) - Phi(N-2, x), for N >= 2, and Phi(0, x) = 1 and Phi(1, x) = 1 - x, that is Phi(-1, x) = 1. The trace is tr(M_N) = 1 + 2^(N-1) = A000051(N-1), and Det(M_N) = 1. - Wolfdieter Lang, Mar 13 2020
The explicit form of the characteristic polynomial for the N X N main submatrix M_N is Phi(N, x) := Det(M_N - x*1_N) = Sum_{k=0..N} binomial(N+k, 2*k)*(-x)^k = Sum_{k=0..N} A085478(N, k)*(-x)^k, for N >= 0, with Phi(0, x) := 1. Proof from the recurrence given in the preceding comment. - Wolfdieter Lang, Mar 25 2020
For the proofs of the 2 X 2, 3 X 3 and 4 X 4 conjectures, see the comments in the respective A-numbers A001519, A080937 and A080938. - Wolfdieter Lang, Mar 30 2020
Replace the main diagonal 1,2,2,2,... of the matrix M with 1,0,0,0,...; 1,1,1,1,...; 1,3,3,3,...; 1,2,1,2,...; 1,2,3,4,...; 1,0,1,0...; and 1,1,0,0,1,1,0,0,.... Take powers of M and extract the upper left terms, resulting in respectively: A001405, A001006, A033321, A176677, A006789, A090344, and A007902. - Gary W. Adamson, Apr 12 2022
The statement that the upper left entry of M^n is a Catalan number is equivalent to Exercise 41 of R. Stanley, "Catalan Numbers." - Richard Stanley, Feb 28 2023
If the upper left 1 in matrix M is replaced with 3, taking powers of the resulting matrix and extracting the upper left terms apparently results in sequence A001700. - Gary W. Adamson, Apr 03 2023

			The matrix begins:
  1, 1, 0, 0, 0, ...
  1, 2, 1, 0, 0, ...
  0, 1, 2, 1, 0, ...
  0, 0, 1, 2, 1, ...
  0, 0, 0, 1, 2, ...
  ...
The first few antidiagonals are:
  1;
  1, 1;
  0, 2, 0;
  0, 1, 1, 0;
  0, 0, 2, 0, 0;
  0, 0, 1, 1, 0, 0;
  0, 0, 0, 2, 0, 0, 0;
  0, 0, 0, 1, 1, 0, 0, 0;
  0, 0, 0, 0, 2, 0, 0, 0, 0;
  0, 0, 0, 0, 1, 1, 0, 0, 0, 0;
  ...
Characteristic polynomial of the 3 X 3 matrix M_3: Phi(3, x) = 1 - 6*x + 5*x^2 - x^3, from {A085478(3, k)}_{k=0..3} = {1, 6, 5, 1}. - _Wolfdieter Lang_, Mar 25 2020

Richard P. Stanley, "Catalan Numbers", Cambridge University Press, 2015.

Cf. A000108, A000051, A001519, A049310, A080937, A080938, A085478.
Cf. A001006, A001405, A006789, A033321, A176677, A090344, A007902.
Cf. A001333 (permanent of the matrix M).
Cf. A054142, A053123, A011973 (characteristic polynomials of submatrices of M).
Cf. A001700.

A053127 Binomial coefficients C(2*n-4,5).

Original entry on oeis.org

6, 56, 252, 792, 2002, 4368, 8568, 15504, 26334, 42504, 65780, 98280, 142506, 201376, 278256, 376992, 501942, 658008, 850668, 1086008, 1370754, 1712304, 2118760, 2598960, 3162510, 3819816, 4582116, 5461512, 6471002, 7624512

Offset: 5

Wolfdieter Lang

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

Vincenzo Librandi, Table of n, a(n) for n = 5..200
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].
Milan Janjic, Two Enumerative Functions, University of Banja Luka (Bosnia and Herzegovina, 2017).
Ângela Mestre and José Agapito, Square Matrices Generated by Sequences of Riordan Arrays, J. Int. Seq., Vol. 22 (2019), Article 19.8.4.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A053123, A053132, A053126.

a053127 = (* 2) . a053132  -- Reinhard Zumkeller, Mar 03 2015

[Binomial(2*n-4,5): n in [5..40]]; // Vincenzo Librandi, Oct 07 2011

Binomial[2Range[5,40]-4,5] (* or *) LinearRecurrence[{6,-15,20,-15,6,-1},{6,56,252,792,2002,4368},30] (* Harvey P. Dale, Jun 03 2013 *)

for(n=5,50, print1(binomial(2*n-4,5), ", ")) \\ G. C. Greubel, Aug 26 2018

a(n) = binomial(2*n-4, 5) if n >= 5 else 0.
a(n) = -A053123(n,5), n >= 5; a(n) := 0, n=0..4 (sixth column of shifted Chebyshev's S-triangle, decreasing order).
G.f.: (6+20*x+6*x^2)/(1-x)^6.
a(n) = 6*a(n-1) - 15*a(n-2) +20*a(n-3) -15*a(n-4) +6*a(n-5) -a(n-6). - Harvey P. Dale, Jun 03 2013
E.g.f.: (-840 + 750*x - 330*x^2 + 95*x^3 - 20*x^4 + 4*x^5)*exp(x)/15. - G. C. Greubel, Aug 26 2018
a(n) = (2*n-8)*(2*n-7)*(2*n-6)*(2*n-5)*(2*n-4)/120. - Wesley Ivan Hurt, Mar 25 2020
From Amiram Eldar, Jan 03 2022: (Start)
Sum_{n>=5} 1/a(n) = 335/12 - 40*log(2).
Sum_{n>=5} (-1)^(n+1)/a(n) = 85/12 - 10*log(2). (End)

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A125175 Triangle T(n,k) = |A053123(n/2+k/2,k)| for even n+k, T(n,k)= A082985((n+k-1)/2,k) for odd n+k; read by rows, 0<=k<=n.

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A011973 Irregular triangle read by rows: T(n,k) = binomial(n-k, k), n >= 0, 0 <= k <= floor(n/2); or, coefficients of (one version of) Fibonacci polynomials.

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A078812 Triangle read by rows: T(n, k) = binomial(n+k-1, 2*k-1).

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A053122 Triangle of coefficients of Chebyshev's S(n,x-2) = U(n,x/2-1) polynomials (exponents of x in increasing order).

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A054142 Triangular array binomial(2*n-k, k), k=0..n, n >= 0.

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