A124644 -id:A124644 - 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

A030528 Triangle read by rows: a(n,k) = binomial(k,n-k).

Original entry on oeis.org

1, 1, 1, 0, 2, 1, 0, 1, 3, 1, 0, 0, 3, 4, 1, 0, 0, 1, 6, 5, 1, 0, 0, 0, 4, 10, 6, 1, 0, 0, 0, 1, 10, 15, 7, 1, 0, 0, 0, 0, 5, 20, 21, 8, 1, 0, 0, 0, 0, 1, 15, 35, 28, 9, 1, 0, 0, 0, 0, 0, 6, 35, 56, 36, 10, 1, 0, 0, 0, 0, 0, 1, 21, 70, 84, 45, 11, 1, 0, 0, 0, 0, 0, 0, 7, 56, 126, 120, 55, 12, 1

Offset: 1

Wolfdieter Lang

A convolution triangle of numbers obtained from A019590.
a(n,m) := s1(-1; n,m), a member of a sequence of triangles including s1(0; n,m)= A023531(n,m) (unit matrix) and s1(2; n,m)= A007318(n-1,m-1) (Pascal's triangle).
The signed triangular matrix a(n,m)*(-1)^(n-m) is the inverse matrix of the triangular Catalan convolution matrix A033184(n+1,m+1), n >= m >= 0, with A033184(n,m) := 0 if nRiordan array (1+x, x(1+x)). The signed triangle is the Riordan array (1-x,x(1-x)), inverse to (c(x),xc(x)) with c(x) g.f. for A000108. - Paul Barry, Feb 02 2005 [with offset 0]
Also, a(n,k)=number of compositions of n into k parts of 1's and 2's. Example: a(6,4)=6 because we have 2211, 2121, 2112, 1221, 1212 and 1122. - Emeric Deutsch, Apr 05 2005 [see MacMahon and Riordan. - Wolfdieter Lang, Jul 27 2023]
Subtriangle of A026729. - Philippe Deléham, Aug 31 2006
a(n,k) is the number of length n-1 binary sequences having no two consecutive 0's with exactly k-1 1's. Example: a(6,4)=6 because we have 01011, 01101, 01110, 10101, 10110, 11010. - Geoffrey Critzer, Jul 22 2013
Mirrored, shifted Fibonacci polynomials of A011973. The polynomials (illustrated below) of this entry have the property that p(n,t) = t * [p(n-1,t) + p(n-2,t)]. The additive properties of Pascal's triangle (A007318) are reflected in those of these polynomials, as can be seen in the Example Section below and also when the o.g.f. G(x,t) below is expanded as the series x*(1+x) + t * [x*(1+x)]^2 + t^2 * [x*(1+x)]^3 + ... . See also A053122 for a relation to Cartan matrices. - Tom Copeland, Nov 04 2014
Rows of this entry appear as columns of an array for an infinitesimal generator presented in the Copeland link. - Tom Copeland, Dec 23 2015
For n >= 2, the n-th row is also the coefficients of the vertex cover polynomial of the (n-1)-path graph P_{n-1}. - Eric W. Weisstein, Apr 10 2017
With an additional initial matrix element a_(0,0) = 1 and column of zeros a_(n,0) = 0 for n > 0, these are antidiagonals read from bottom to top 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. Cf. A011973. And A169803. - Tom Copeland, Jul 02 2018

			Triangle starts:
  [ 1]  1
  [ 2]  1   1
  [ 3]  0   2   1
  [ 4]  0   1   3   1
  [ 5]  0   0   3   4   1
  [ 6]  0   0   1   6   5   1
  [ 7]  0   0   0   4  10   6   1
  [ 8]  0   0   0   1  10  15   7   1
  [ 9]  0   0   0   0   5  20  21   8   1
  [10]  0   0   0   0   1  15  35  28   9   1
  [11]  0   0   0   0   0   6  35  56  36  10   1
  [12]  0   0   0   0   0   1  21  70  84  45  11   1
  [13]  0   0   0   0   0   0   7  56 126 120  55  12   1
  ...
From _Tom Copeland_, Nov 04 2014: (Start)
For quick comparison to other polynomials:
  p(1,t) = 1
  p(2,t) = 1 + 1 t
  p(3,t) = 0 + 2 t + 1 t^2
  p(4,t) = 0 + 1 t + 3 t^2 + 1 t^3
  p(5,t) = 0 + 0   + 3 t^2 + 4 t^3 +  1 t^4
  p(6,t) = 0 + 0   + 1 t^2 + 6 t^3 +  5 t^4 +  1 t^5
  p(7,t) = 0 + 0   + 0     + 4 t^3 + 10 t^4 +  6 t^5 + 1 t^6
  p(8,t) = 0 + 0   + 0     + 1 t^3 + 10 t^4 + 15 t^5 + 7 t^6 + 1 t^7
  ...
Reading along columns gives rows for Pascal's triangle. (End)

P. A. MacMahon, Combinatory Analysis, Two volumes (bound as one), Chelsea Publishing Company, New York, 1960, Vol. I, nr. 124, p. 151.
John Riordan, An Introduction to Combinatorial Analysis, John Wiley & Sons, London, 1958. eq. (35), p.124, 11. p. 154.

Indranil Ghosh, Rows 1..125 of triangle, flattened
Jean Luc Baril, Rigoberto Flórez, and José L. Ramirez, Generalized Narayana arrays, restricted Dyck paths, and related bijections, Univ. Bourgogne (France, 2025). See p. 18.
Tom Copeland, Generators, Inversion, and Matrix, Binomial, and Integral Transforms, Addendum to Elliptic Lie Triad
Loïc Foissy, Cointeraction on noncrossing partitions and related polynomial invariants, arXiv:2501.18212 [math.CO], 2025. See p. 21.
Wolfdieter Lang, On generalizations of Stirling number triangles, J. Integer Seqs., Vol. 3 (2000), #00.2.4.
Donatella Merlini, Renzo Sprugnoli, and M. Cecilia Verri, An algebra for proper generating trees, Colloquium on Mathematics and Computer Science, Versailles, September 2000.
Donatella Merlini, Renzo Sprugnoli, and M. Cecilia Verri, An algebra for proper generating trees, Mathematics and Computer Science, Part of the series Trends in Mathematics pp 127-139.
Peter J. Olver, The canonical contact form.
Eric Weisstein's World of Mathematics, Path Graph
Eric Weisstein's World of Mathematics, Vertex Cover Polynomial

Row sums A000045(n+1) (Fibonacci). a(n, 1)= A019590(n) (Fermat's last theorem). Cf. A049403.

Cf. A104597, A146559, A155020, A125145, A057078, A039717, A001792, A057862, A011973, A115139.

/* As triangle */ [[Binomial(k, n-k): k in [1..n]]: n in [1.. 15]]; // Vincenzo Librandi, Nov 05 2014

for n from 1 to 12 do seq(binomial(k,n-k),k=1..n) od; # yields sequence in triangular form - Emeric Deutsch, Apr 05 2005

nn=10;CoefficientList[Series[(1+x)/(1-y x - y x^2),{x,0,nn}],{x,y}]//Grid (* Geoffrey Critzer, Jul 22 2013 *)
Table[Binomial[k, n - k], {n, 13}, {k, n}] // Flatten (* Michael De Vlieger, Dec 23 2015 *)
CoefficientList[Table[x^(n/2 - 1) Fibonacci[n + 1, Sqrt[x]], {n, 10}],
   x] // Flatten (* Eric W. Weisstein, Apr 10 2017 *)

a(n, m) = 2*(2*m-n+1)*a(n-1, m)/n + m*a(n-1, m-1)/n, n >= m >= 1; a(n, m) := 0, n

G.f. for m-th column: (x*(1+x))^m.

As a number triangle with offset 0, this is T(n, k) = Sum_{i=0..n} (-1)^(n+i)*binomial(n, i)*binomial(i+k+1, 2k+1). The antidiagonal sums give the Padovan sequence A000931(n+5). Inverse binomial transform of A078812 (product of lower triangular matrices). - Paul Barry, Jun 21 2004

G.f.: (1 + x)/(1 - y*x - y*x^2). - Geoffrey Critzer, Jul 22 2013 [offset 0] [with offset 1: g.f. of row polynomials in y: x*(1+x)*y/(1 - x*(1+x)*y). - Wolfdieter Lang, Jul 27 2023]

From Tom Copeland, Nov 04 2014: (Start)

O.g.f: G(x,t) = x*(1+x) / [1 - t*x*(1+x)] = -P[Cinv(-x),t], where P(x,t)= x / (1 + t*x) and Cinv(x)= x*(1-x) are the compositional inverses in x of Pinv(x,t) = -P(-x,t) = x / (1 - t*x) and C(x) = [1-sqrt(1-4*x)]/2, an o.g.f. for the shifted Catalan numbers A000108.

Therefore, Ginv(x,t) = -C[Pinv(-x,t)] = {-1 + sqrt[1 + 4*x/(1+t*x)]}/2, which is -A124644(-x,t).

This places this array in a family of arrays related by composition of P and C and their inverses and interpolation by t, such as A091867 and A104597, and associated to the Catalan, Motzkin, Fine, and Fibonacci numbers. Cf. A104597 (polynomials shifted in t) A125145, A146559, A057078, A000045, A155020, A125145, A039717, A001792, A057862, A011973, A115139. (End)

More terms from Emeric Deutsch, Apr 05 2005

A091867 Triangle read by rows: T(n,k) = number of Dyck paths of semilength n having k peaks at odd height.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 1, 3, 0, 1, 3, 4, 6, 0, 1, 6, 15, 10, 10, 0, 1, 15, 36, 45, 20, 15, 0, 1, 36, 105, 126, 105, 35, 21, 0, 1, 91, 288, 420, 336, 210, 56, 28, 0, 1, 232, 819, 1296, 1260, 756, 378, 84, 36, 0, 1, 603, 2320, 4095, 4320, 3150, 1512, 630, 120, 45, 0, 1, 1585, 6633, 12760, 15015, 11880, 6930, 2772, 990, 165, 55, 0, 1

Offset: 0

Emeric Deutsch, Mar 10 2004

Number of ordered trees with n edges having k leaves at odd height. Row sums are the Catalan numbers (A000108). T(n,0)=A005043(n). Sum_{k=0..n} k*T(n,k) = binomial(2n-2,n-1).
T(n,k)=number of Dyck paths of semilength n and having k ascents of length 1 (an ascent is a maximal string of consecutive up steps). Example: T(4,2)=6 because we have UdUduud, UduuddUd, uuddUdUd, uudUdUdd, UduudUdd and uudUddUd (the ascents of length 1 are indicated by U instead of u).
T(n,k) is the number of Łukasiewicz paths of length n having k level steps (i.e., (1,0)). A Łukasiewicz path of length n is a path in the first quadrant from (0,0) to (n,0) using rise steps (1,k) for any positive integer k, level steps (1,0) and fall steps (1,-1) (see R. P. Stanley, Enumerative Combinatorics, Vol. 2, Cambridge Univ. Press, Cambridge, 1999, p. 223, Exercise 6.19w; the integers are the slopes of the steps). Example: T(4,1)=4 because we have HU(2)DD, U(2)HDD, U(2)DHD and U(2)DDH, where H=(1,0), U(1,1), U(2)=(1,2) and D=(1,-1). - Emeric Deutsch, Jan 06 2005
T(n,k) = number of noncrossing partitions of [n] containing k singleton blocks. Also, T(n,k) = number of noncrossing partitions of [n] containing k adjacencies. An adjacency is an occurrence of 2 consecutive integers in the same block (here 1 and n are considered consecutive). In fact, the statistics # singletons and # adjacencies have a symmetric joint distribution.
Exponential Riordan array [e^x*(Bessel_I(0,2x)-Bessel_I(1,2x)),x]. - Paul Barry, Mar 03 2011
T(n,k) is the number of ordered trees having n edges and exactly k nodes with one child. - Geoffrey Critzer, Feb 25 2013
From Tom Copeland, Nov 04 2014: (Start)
Summing the coeff. of the partitions in A134264 for a Lagrange inversion formula (see also A249548) containing (h_1)^k = (1')^k gives this triangle, so this array's o.g.f. H(x,t) = x + t * x^2 + (1 + t^2) * x^3 ... is the inverse of the o.g.f. of A104597 with a sign change, i.e., H^(-1)(x,t) = (x-x^2) / [1 + (t-1)(x-x^2)] = Cinv(x)/[1 + (t-1)Cinv(x)] = P[Cinv(x),t-1] where Cinv(x)= x * (1-x) is the inverse of C(x) = [1-sqrt(1-4*x)]/2, an o.g.f. for the Catalan numbers A000108, and P(x,t) = x/(1+t*x) with inverse Pinv(x,t) = -P(-x,t) = x/(1-t*x). Therefore,
O.g.f.: H(x,t) = C[Pinv(x,t-1)] = C[P(x,1-t)] = C[x/(1-(t-1)x)] = {1-sqrt[1-4*x/(1-(t-1)x)]}/2 (for A091867). Reprising,
Inverse O.g.f.: H^(-1)(x,t) = x*(1-x) / [1 + (t-1)x(1-x)] = P[Cinv(x),t-1].
From general arguments in A134264, the row polynomials are an Appell sequence with lowering operator d/dt, having the umbral property (p(.,t)+a)^n=p(n,t+a) with e.g.f. = e^(x*t)/w(x), where 1/w(x)= e.g.f. of first column for the Motzkin numbers in A005043. (Mislabeled argument corrected on Jan 31 2016.)
Cf. A124644 (t-shifted polynomials), A026378 (t=-4), A001700 (t=-3), A005773 (t=-2), A126930 (t=-1) and A210736 (t=-1, a(0)=0, unsigned), A005043 (t=0), A000108 (t=1), A007317 (t=2), A064613 (t=3), A104455 (t=4), A030528 (for inverses).
(End)
The sequence of binomial transforms A126930, A005043, A000108, ... in the above comment appears in A126930 and the link therein to a paper by F. Fite et al. on page 42. - Tom Copeland, Jul 23 2016

			T(4,2)=6 because we have (ud)uu(ud)dd, uu(ud)dd(ud), uu(ud)(ud)dd, (ud)(ud)uudd, (ud)uudd(ud) and uudd(ud)(ud) (here u=(1,1), d=(1,-1) and the peaks at odd height are shown between parentheses).
Triangle begins:
   1;
   0,   1;
   1,   0,   1;
   1,   3,   0,   1;
   3,   4,   6,   0,  1;
   6,  15,  10,  10,  0,  1;
  15,  36,  45,  20, 15,  0, 1;
  36, 105, 126, 105, 35, 21, 0, 1;
  ...

R. Sedgewick and P. Flajolet, Analysis of Algorithms, Addison and Wesley, 1996, page 254 (first edition)

Alois P. Heinz, Rows n = 0..200, flattened
David Callan, On conjugates for set partitions and integer compositions , arXiv:math/0508052 [math.CO], 2005.
A. Sapounakis, I. Tasoulas and P. Tsikouras, Counting strings in Dyck paths, Discrete Math., 307 (2007), 2909-2924.
F. Yano and H. Yoshida, Some set partition statistics in non-crossing partitions and generating functions, Discr. Math., 307 (2007), 3147-3160.

Cf. A000108, A001700, A005043, A005773, A007317, A026378, A030528, A064613, A104455, A104597, A124644, A126930, A210736.

T := proc(n,k) if k>n then 0 elif k=n then 1 else (binomial(n+1,k)/(n+1))*sum(binomial(n+1-k,j)*binomial(n-k-j-1,j-1),j=1..floor((n-k)/2)) fi end: seq(seq(T(n,k),k=0..n),n=0..12);
T := (n,k) -> (-1)^(n+k)*binomial(n,k)*hypergeom([-n+k,1/2],[2],4): seq(seq(simplify(T(n, k)), k=0..n), n=0..10); # Peter Luschny, Jul 27 2016
# alternative Maple program:
b:= proc(x, y, t) option remember; expand(`if`(x=0, 1,
      `if`(y>0, b(x-1, y-1, 0)*z^irem(t*y, 2), 0)+
      `if`(y (p-> seq(coeff(p, z, i), i=0..n))(b(2*n, 0$2)):
seq(T(n), n=0..16);  # Alois P. Heinz, May 12 2017

nn=10;cy = ( 1 + x - x y - ( -4x(1+x-x y) + (-1 -x + x y)^2)^(1/2))/(2(1+x-x y)); Drop[CoefficientList[Series[cy,{x,0,nn}],{x,y}],1]//Grid  (* Geoffrey Critzer, Feb 25 2013 *)
Table[Which[k == n, 1, k > n, 0, True, (Binomial[n + 1, k]/(n + 1)) Sum[Binomial[n + 1 - k, j] Binomial[n - k - j - 1, j - 1], {j, Floor[(n - k)/2]}]], {n, 0, 11}, {k, 0, n}] // Flatten (* Michael De Vlieger, Jul 25 2016 *)

T(n, k) = [binomial(n+1, k)/(n+1)]*Sum_{j=1..floor((n-k)/2)} binomial(n+1-k, j)*binomial(n-k-j-1, j-1) for kn. G.f.=G=G(t, z) satisfies z(1+z-tz)G^2-(1+z-tz)G+1=0. T(n, k)=r(n-k)*binomial(n, k), where r(n)=A005043(n) are the Riordan numbers.
G.f.: 1/(1-xy-x^2/(1-x-xy-x^2/(1-x-xy-x^2/(1-x-xy-x^2/(1-... (continued fraction). - Paul Barry, Aug 03 2009
Sum_{k=0..n} T(n,k)*x^k = A126930(n), A005043(n), A000108(n), A007317(n), A064613(n), A104455(n) for x = -1,0,1,2,3,4 respectively. - Philippe Deléham, Dec 03 2009
Sum_{k=0..n} (-1)^(n-k)*T(n,k)*x^k = A168491(n), A099323(n+1), A001405(n), A005773(n+1), A001700(n), A026378(n+1), A005573(n), A122898(n) for x = -1, 0, 1, 2, 3, 4, 5, 6 respectively. - Philippe Deléham, Dec 03 2009
E.g.f.: e^(x+xy)*(Bessel_I(0,2x)-Bessel_I(1,2x)). - Paul Barry, Mar 10 2010
From Tom Copeland, Nov 06 2014: (Start)
O.g.f.: H(x,t) = {1-sqrt[1-4x/(1-(t-1)x)]}/2 (shifted index, as given in Copeland's comment, see comp. inverse there).
H(x,t)= x / [1-(C.+(t-1))x] = Sum_{n>=1} (C.+ (t-1))^(n-1)*x^n umbrally, e.g., (a.+b.)^2 = a_0*b_2 + 2 a_1*b1_+ a_0*b_2, where (C.)^n = C_n are the Catalan numbers (1,1,2,5,14,..) of A000108.
This shows directly that the lowering operator for the polynomials is D=d/dt, i.e., D p(n,t)= D(C. + (t-1))^n = n * (C. + (t-1))^(n-1) = n*p(n-1,t), so that the polynomials form an Appell sequence, and that p(n,0) gives a Motzkin sum, or Riordan, number A005043.
(End)
T(n,k) = (-1)^(n+k)*binomial(n,k)*hypergeom([k-n,1/2],[2],4). - Peter Luschny, Jul 27 2016

A098474 Triangle read by rows, T(n,k) = C(n,k)C(2k,k)/(k+1), n >= 0, 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 1, 3, 6, 5, 1, 4, 12, 20, 14, 1, 5, 20, 50, 70, 42, 1, 6, 30, 100, 210, 252, 132, 1, 7, 42, 175, 490, 882, 924, 429, 1, 8, 56, 280, 980, 2352, 3696, 3432, 1430, 1, 9, 72, 420, 1764, 5292, 11088, 15444, 12870, 4862, 1, 10, 90, 600, 2940, 10584, 27720

Offset: 0

Paul Barry, Sep 09 2004

A Catalan scaled binomial matrix.
From Philippe Deléham, Sep 01 2005: (Start)
Table U(n,k), k >= 0, n >= 0, read by antidiagonals, begins:
  row k = 0: 1, 1,  2,   5,   14, ... is A000108
  row k = 1: 1, 2,  6,  20,   70, ... is A000984
  row k = 2: 1, 3, 12,  50,  280, ... is A007854
  row k = 3: 1, 4, 20, 104,  548, ... is A076035
  row k = 4: 1, 5, 30, 185, 1150, ... is A076036
G.f. for row k: 1/(1-(k+1)*x*C(x)) where C(x) is the g.f. = for Catalan numbers A000108.
U(n,k) = Sum_{j=0..n} A106566(n,j)*(k+1)^j. (End)
This sequence gives the coefficients (increasing powers of x) of the Jensen polynomials for the Catalan sequence A000108 of degree n and shift 0. For the definition of Jensen polynomials for a sequence see a comment in A094436. - Wolfdieter Lang, Jun 25 2019

			Rows begin:
  1;
  1, 1;
  1, 2,  2;
  1, 3,  6,   5;
  1, 4, 12,  20,  14;
  1, 5, 20,  50,  70,  42;
  1, 6, 30, 100, 210, 252, 132;
  ...
Row 3: t*(1 - 3*t + 6*t^2 - 5*t^3)/(1 - 4*t)^(9/2) = 1/2*Sum_{k >= 1} k*(k+1)*(k+2)*(k+3)/4!*binomial(2*k,k)*t^k. - _Peter Bala_, Jun 13 2016

Indranil Ghosh, Rows 0..125, flattened

Row sums are A007317.
Antidiagonal sums are A090344.
Principal diagonal is A000108.
Mirror image of A124644.
Cf. A000984, A007854, A011973, A076035, A076036, A098443, A098473, A106566, A267633.

p := proc(n) option remember; if n = 0 then 1 else normal((x*(1 + 4*x)*diff(p(n-1, x), x) + (2*x + n + 1)*p(n-1, x))/(n + 1)) fi end:
row := n -> local k; seq(coeff(p(n), x, k), k = 0..n):
for n from 0 to 6 do row(n) od;  # Peter Luschny, Jun 21 2023

Table[Binomial[n, k] Binomial[2 k, k]/(k + 1), {n, 0, 10}, {k, 0, n}] // Flatten (* or *)
Table[(-1)^k*CatalanNumber[k] Pochhammer[-n, k]/k!, {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Feb 17 2017 *)

from functools import cache
@cache
def A098474row(n: int) -> list[int]:
    if n == 0: return [1]
    a = A098474row(n - 1) + [0]
    row = [0] * (n + 1)
    row[0] = 1; row[1] = n
    for k in range(2, n + 1):
        row[k] = (a[k] * (n + k + 1) + a[k - 1] * (4 * k - 2)) // (n + 1)
    return row  # Peter Luschny, Jun 22 2023

def A098474(n,k):
    return (-1)^k*catalan_number(k)*rising_factorial(-n,k)/factorial(k)
for n in range(7): [A098474(n,k) for k in (0..n)] # Peter Luschny, Feb 05 2015

G.f.: 2/(1-x+(1-x-4*x*y)^(1/2)). - Vladeta Jovovic, Sep 11 2004
E.g.f.: exp(x*(1+2*y))*(BesselI(0, 2*x*y)-BesselI(1, 2*x*y)). - Vladeta Jovovic, Sep 11 2004
G.f.: 1/(1-x-xy/(1-xy/(1-x-xy/(1-xy/(1-x-xy/(1-xy/(1-x-xy/(1-xy/(1-... (continued fraction). - Paul Barry, Feb 11 2009
Sum_{k=0..n} T(n,k)*x^(n-k) = A126930(n), A005043(n), A000108(n), A007317(n+1), A064613(n), A104455(n) for x = -2, -1, 0, 1, 2, 3 respectively. - Philippe Deléham, Dec 12 2009
T(n,k) = (-1)^k*Catalan(k)*Pochhammer(-n,k)/k!. - Peter Luschny, Feb 05 2015
O.g.f.: [1 - sqrt(1-4tx/(1-x))]/(2tx) = 1 + (1+t) x + (1+2t+2t^2) x^2 + (1+3t+6t^2+5t^3) x^3 + ... , generating the polynomials of this entry, reverse of A124644. See A011973 for a derivation and the inverse o.g.f., connected to the Fibonacci, Chebyshev, and Motzkin polynomials. See also A267633. - Tom Copeland, Jan 25 2016
From Peter Bala, Jun 13 2016: (Start)
The o.g.f. F(x,t) = ( 1 - sqrt(1 - 4*t*x/(1 - x)) )/(2*t*x) satisfies the partial differential equation d/dx(x*(1 - x)*F) - x*t*(1 + 4*t)*dF/dt - 2*x*t*F = 1. This gives a recurrence for the row polynomials: (n + 2)*R(n+1,t) = t*(1 + 4*t)*R'(n,t) + (2*t + n + 2)*R(n,t), where the prime ' indicates differentiation with respect to t.
Equivalently, setting Q(n,t) = t^(n+2)*R(n,-t)/(1 - 4*t)^(n + 3/2) we have t^2*d/dt(Q(n,t)) = (n + 2)*Q(n+1,t).
This leads to the following expansions:
Q(0,t) = (1/2)*Sum_{k >= 1} k*binomial(2*k,k)*t^(k+1)
Q(1,t) = (1/2)*Sum_{k >= 1} k*(k+1)/2!*binomial(2*k,k)*t^(k+2)
Q(2,t) = (1/2)*Sum_{k >= 1} k*(k+1)*(k+2)/3!*binomial(2*k,k) *t^(k+3) and so on. (End)
Sum_{k=0..n} T(n,k)*x^k = A007317(n+1), A162326(n+1), A337167(n) for x = 1, 2, 3 respectively. - Sergii Voloshyn, Mar 31 2022

New name using a formula of Paul Barry by Peter Luschny, Feb 05 2015

A267633 Expansion of (1 - 4t)/(1 - x + t x^2): a Fibonacci-type sequence of polynomials.

Original entry on oeis.org

1, -4, 1, -4, 1, -5, 4, 1, -6, 8, 1, -7, 13, -4, 1, -8, 19, -12, 1, -9, 26, -25, 4, 1, -10, 34, -44, 16, 1, -11, 43, -70, 41, -4, 1, -12, 53, -104, 85, -20, 1, -13, 64, -147, 155, -61, 4, 1, -14, 76, -200, 259, -146, 24

Offset: 0

Tom Copeland, Jan 18 2016

			Row polynomials:
P(0,t) = 1 - 4t
P(1,t) = 1 - 4t = [-t(0) + (1-4t)] = -t(0) + P(0,t)
P(2,t) = 1 - 5t + 4t^2 = [-t(1-4t) + (1-4t)] = -t P(0,t) + P(1,t)
P(3,t) = 1 - 6t + 8t^2 = [-t(1-4t) + (1-5t+4t^2)] = -t P(1,t) + P(2,t)
P(4,t) = 1 - 7t + 13t^2 - 4t^3 = [-t(1-5t+4t^2) + (1-6t+8t^2)]
P(5,t) = 1 - 8t + 19t^2 - 12t^3 = [-t(1-6t+8t^2) + (1-7t+13t^2)]
P(6,t) = 1 - 9t + 26t^2 - 25t^3 + 4t^4
P(7,t) = 1 - 10t + 34t^2 - 44t^3 + 16t^4
P(8,t) = 1 - 11t + 43t^2 - 70t^3 + 41t^4 - 4t^5
P(9,t) = 1 - 12t + 53t^2 - 104t^3 + 85t^4 - 20t^5
P(10,t) = 1 - 13t + 64t^2 - 147t^3 + 155t^4 - 61t^5 + 4t^6
P(11,t) = 1 - 14t + 76t^2 - 200t^3 + 259t^4 - 146t^5 + 24t^6
...
Apparently: The odd rows for n>1 are reversed rows of A140882 (mod signs), and the even rows for n>0, the 9th, 15th, 21st, 27th, etc. rows of A228785 (mod signs). The diagonals are reverse rows of A202241.

Tom Copeland, Addendum to Elliptic Lie Triad

Cf. A000108, A000297, A008574, A008586, A011973, A022088 A034856, A084103, A098474, A124644, A140882, A202241, A228785.

p = (1 - 4 t) / (1 - x + t x^2) + O[x]^12 // CoefficientList[#, x] &;
CoefficientList[#, t] & /@ p // Flatten (* Andrey Zabolotskiy, Mar 07 2024 *)

O.g.f. G(x,t) = (1 - 4t)/(1 - x + t x^2) = a /  [t (x - (1+sqrt(a))/(2t))(x - (1-sqrt(a))/(2t))] with a = 1-4t.
Recursion P(n,t) = -t P(n-2,t) + P(n-1,t) with P(-1,t)=0 and P(0,t) = 1-4t.
Convolution of the Fibonacci polynomials of signed A011973 Fb(n,-t) with coefficients of (1-4t), e.g., (1-4t)Fb(5,-t) = (1-4t)(1-3t+t^2) = 1-7t+13t^2-4t^3, so, for n>=1 and k<=(n-1), T(n,k) = (-1)^k [-4*binomial(n-(k-1),k-1) - binomial(n-k,k)] with the convention that 1/(-m)! = 0 for m>=1, i.e., let binomial(n,k) = nint[n!/((k+c)!(n-k+c)!)] for c sufficiently small in magnitude.
Third column is A034856, and the fourth, A000297. Embedded in the coefficients of the highest order term of the polynomials is A008586 (cf. also A008574).
With P(0,t)=0, the o.g.f. is H(x,t) = (1-4t) x(1-tx)/[1-x(1-tx)] = (1-4t) Linv(Cinv(tx)/t), where Linv(x) = x/(1-x) with inverse L(x) = x/(1+x) and Cinv(x) = x (1-x) is the inverse of the o.g.f. of the shifted Catalan numbers A000108, C(x) = [1-sqrt(1-4x)]/2. Then Hinv(x,t) = C[t L(x/(1-4t))]/t = {1 - sqrt[1-4t(x/(1-4t))/[1+x/(1-4t)]]}/2t = {1-sqrt[1-4tx/(1-4t+x)]}/2t = 1/(1-4t) + (-1+t)/(1-4t)^2 x + (1-2t+2t^2)/(1-4t)^3 x^ + (-1+3t-6t^2+5t^3)/(1-4t)^4 + ..., where the numerators are the signed polynomials of A098474, reverse of A124644.
Row sums (t=1) are periodic -3,-3,0,3,3,0, repeat the six terms ... with o.g.f. -3 - 3x (1-x) / [1-x(1-x)]. Cf. A084103.
Unsigned row sums (t=-1) are shifted A022088 with o.g.f. 5 + 5x(1+x) / [x(1+x)].

Data corrected by Andrey Zabolotskiy, Mar 07 2024

A155839 A ratio of two Catalan arrays.

Original entry on oeis.org

1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 2, 3, 0, 1, 0, 4, 7, 6, 0, 1, 0, 8, 18, 16, 10, 0, 1, 0, 16, 45, 51, 30, 15, 0, 1, 0, 32, 110, 152, 115, 50, 21, 0, 1, 0, 64, 264, 436, 396, 225, 77, 28, 0, 1, 0, 128, 624, 1212, 1300, 876, 399, 112, 36, 0, 1

Offset: 0

Paul Barry, Jan 28 2009

			Triangle begins
  1;
  0,  1;
  0,  0,   1;
  0,  1,   0,   1;
  0,  2,   3,   0,   1;
  0,  4,   7,   6,   0,  1;
  0,  8,  18,  16,  10,  0,  1;
  0, 16,  45,  51,  30, 15,  0, 1;
  0, 32, 110, 152, 115, 50, 21, 0, 1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A000108, A033184, A120010 (row sums), A124644.

A155839:= func< n,k | (&+[(-1)^(n-j)*Binomial(j+1, n-j)*Binomial(j, k)*Catalan(j-k) : j in [k..n]]) >;
[A155839(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jun 04 2021

T[n_, k_] = Sum[(-1)^j*Binomial[n-j, k]*Binomial[n-j+1, j]*CatalanNumber[n-k-j], {j, 0, n-k}];
Table[T[n, k], {n, 0, 10}, {k, 0, n}]//Flatten (* G. C. Greubel, Jun 04 2021 *)

def A155839(n,k): return sum( (-1)^j*binomial(n-j,k)*binomial(n-j+1,j)*catalan_number(n-k-j) for j in (0..n-k))
flatten([[A155839(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Jun 04 2021

T(n, k) = Sum_{j=k..n} (-1)^(n-j)*binomial(j+1, n-j)*binomial(j, k)*A000108(j-k).
Sum_{k=0..n} T(n, k) = A120010(n+1).
Equals A033184^{-1}*A124644.

Showing 1-7 of 7 results.

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.

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A109466 Riordan array (1, x(1-x)).

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A030528 Triangle read by rows: a(n,k) = binomial(k,n-k).

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A091867 Triangle read by rows: T(n,k) = number of Dyck paths of semilength n having k peaks at odd height.

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A098474 Triangle read by rows, T(n,k) = C(n,k)*C(2*k,k)/(k+1), n >= 0, 0 <= k <= n.

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A267633 Expansion of (1 - 4t)/(1 - x + t x^2): a Fibonacci-type sequence of polynomials.

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A098474 Triangle read by rows, T(n,k) = C(n,k)C(2k,k)/(k+1), n >= 0, 0 <= k <= n.