A113682 -id:A113682 - cp's OEIS frontend

A097613 a(n) = binomial(2n-3,n-1) + binomial(2n-2,n-2).

1, 2, 7, 25, 91, 336, 1254, 4719, 17875, 68068, 260338, 999362, 3848222, 14858000, 57500460, 222981435, 866262915, 3370764540, 13135064250, 51250632510, 200205672810, 782920544640, 3064665881940, 12007086477750, 47081501377326, 184753963255176, 725510446350004

Offset: 1

David Callan, Sep 20 2004

nonn

a(n) is the number of Dyck (2n-1)-paths with maximum pyramid size = n. A pyramid in a Dyck path is a maximal subpath of the form k upsteps immediately followed by k downsteps and its size is k.
a(n) is the total number of runs of peaks in all Dyck (n+1)-paths. A run of peaks is a maximal subpath of the form (UD)^k with k>=1. For example, a(2)=7 because the 5 Dyck 3-paths contain a total of 7 runs of peaks (in uppercase type): uuUDdd, uUDUDd, uUDdUD, UDuUDd, UDUDUD. - David Callan, Jun 07 2006
Binomial transform of A113682. - Paul Barry, Aug 21 2007
If Y is a fixed 2-subset of a (2n+1)-set X then a(n+1) is the number of n-subsets of X intersecting Y. - Milan Janjic, Oct 21 2007
Equals the Catalan sequence, A000108, convolved with A051924 prefaced with a 1: (1, 1, 4, 14, 50, ...). - Gary W. Adamson, May 15 2009
Central terms of triangle A209561. - Reinhard Zumkeller, Dec 26 2012
Also the number of compositions of 2*(n-1) in which the odd parts appear as many times in odd as in even positions. - Alois P. Heinz, May 26 2018

			a(2) = 2 because UUDDUD and UDUUDD each have maximum pyramid size = 2.

Michael De Vlieger, Table of n, a(n) for n = 1..1664
Paul Barry, Extensions of Riordan Arrays and Their Applications, Mathematics (2025) Vol. 13, No. 2, 242. See p. 16.
Gi-Sang Cheon, Hana Kim, and Louis W. Shapiro, Mutation effects in ordered trees, arXiv preprint arXiv:1410.1249 [math.CO], 2014.
Milan Janjic, Two Enumerative Functions
Toufik Mansour and I. L. Ramirez, Enumerations of polyominoes determined by Fuss-Catalan words, Australas. J. Combin. 81 (3) (2021) 447-457, table 2.
Lin Yang and Shengliang Yang, Protected Branches in Ordered Trees, J. Math. Study (2023) Vol. 56, No. 1, 1-17.

Same as A024482 except for first term.

Cf. A000108, A000712, A026010, A051924, A113682, A209561.

Flat(List([1..30], n->Binomial(2*n-3, n-1)+Binomial(2*n-2, n-2))); # Stefano Spezia, Oct 27 2018

a097613 n = a209561 (2 * n - 1) n  -- Reinhard Zumkeller, Dec 26 2012

[((3*n-2)*Catalan(n-1)+0^(n-1))/2: n in [1..40]]; // G. C. Greubel, Apr 04 2024

Z:=(1-z-sqrt(1-4*z))/sqrt(1-4*z)/2: Zser:=series(Z, z=0, 32): seq (ceil(coeff(Zser, z, n)), n=1..22); # Zerinvary Lajos, Jan 16 2007
a := n -> `if`(n=1, 1, (2-3*n)/(4-8*n)*binomial(2*n, n)):
seq(a(n), n=1..27); # Peter Luschny, Sep 06 2014

a[1]=1; a[n_] := (3n-2)(2n-3)!/(n!(n-2)!); Array[a, 27] (* Jean-François Alcover, Oct 27 2018 *)

a(n)=binomial(2*n-3,n-1)+binomial(2*n-2,n-2) \\ Charles R Greathouse IV, Aug 05 2013

@CachedFunction
def A097613(n):
    if n < 3: return n
    return (6*n-4)*(2*n-3)*A097613(n-1)/(n*(3*n-5))
[A097613(n) for n in (1..27)] # Peter Luschny, Sep 06 2014

G.f.: (x-1)*(1 - 1/sqrt(1-4*x))/2.
a(n) = ceiling(A051924(n)/2). - Zerinvary Lajos, Jan 16 2007
Integral representation as n-th moment of a signed weight function W(x) = W_a(x) + W_c(x), where W_a(x) = Dirac(x)/2 is the discrete (atomic) part, and  W_c(x) = (1/(2*Pi))*((x-1))*sqrt(1/(x*(4-x))) is the continuous part of W(x) defined on (0,4): a(n) = Integral_{x=-eps..eps} x^n*W_a(x) + Integral_{x=0..4} x^n*W_c(x) for any eps > 0, n >= 0. W_c(0) = -infinity, W_c(1) = 0 and W_c(4) = infinity. For 0 < x < 1, W_c(x) < 0, and for 1 < x < 4, W_c(x) > 0. - Karol A. Penson, Aug 05 2013
From Peter Luschny, Sep 06 2014: (Start)
a(n) = ((2-3*n)/(4-8*n))*binomial(2*n,n) for n >= 2.
D-finite with recurrence: a(n) = (6*n-4)*(2*n-3)*a(n-1)/(n*(3*n-5)) for n >= 3. (End)
a(n) ~ 3*2^(2*n-3)/sqrt(n*Pi). - Stefano Spezia, May 09 2023
From G. C. Greubel, Apr 04 2024: (Start)
a(n) = (1/2)*( (3*n-2)*A000108(n-1) + [n=1]).
E.g.f.: (1/2)*(-1+x + exp(2*x)*((1-x)*BesselI(0,2*x) + x*BesselI(1,2*x) )). (End)

A158793 Triangle read by rows: product of A130595 and A092392 considered as infinite lower triangular arrays.

Original entry on oeis.org

1, 1, 1, 3, 1, 1, 7, 4, 1, 1, 19, 9, 5, 1, 1, 51, 26, 11, 6, 1, 1, 141, 70, 34, 13, 7, 1, 1, 393, 197, 92, 43, 15, 8, 1, 1, 1107, 553, 265, 117, 53, 17, 9, 1, 1, 3139, 1570, 751, 346, 145, 64, 19, 10, 1, 1, 8953, 4476, 2156, 991, 441, 176, 76, 21, 11, 1, 1

Offset: 0

Gary W. Adamson & Roger L. Bagula, Mar 26 2009

Riordan array (f(x), x*g(x)) where f(x) is the g.f. of A002426 and where g(x) is the g.f. of A005043. - Philippe Deléham, Dec 05 2009

Matrix product P * Q * P^(-1), where P denotes Pascal's triangle A007318 and Q denotes A061554 (formed from P by sorting the rows into descending order). Cf. A158815 and A171243. - Peter Bala, Jul 13 2021

			First rows of the triangle:
     1;
     1,    1;
     3,    1,    1;
     7,    4,    1,   1;
    19,    9,    5,   1,   1;
    51,   26,   11,   6,   1,   1;
   141,   70,   34,  13,   7,   1,  1;
   393,  197,   92,  43,  15,   8,  1,  1;
  1107,  553,  265, 117,  53,  17,  9,  1,  1;
  3139, 1570,  751, 346, 145,  64, 19, 10,  1, 1;
  8953, 4476, 2156, 991, 441, 176, 76, 21, 11, 1, 1;

T(n, 0) = A002426(n), A005773 (row sums).

Cf. A046899, A007318, A158815, A171243.

A158793 := proc (n, k)
  add((-1)^(n+j)*binomial(n, j)*binomial(2*j-k, j-k), j = k..n);
end proc:
seq(seq(A158793(n, k), k = 0..n), n = 0..10); # Peter Bala, Jul 13 2021

T[n_, k_] := (-1)^(k + n) Binomial[n, k] HypergeometricPFQ[{k/2 + 1/2, k/2 + 1, k - n}, {k + 1, k + 1}, 4];
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Peter Luschny, Jul 17 2021 *)

T(n, m) = Sum_{k=m..n-1} A130595(n,k) * A092392(k+1,m+1), with the triangular interpretation of A092392.
Conjecture: T(n,1) = A113682(n-1). - R. J. Mathar, Oct 06 2009
Sum_{k=0..n} T(n,k)*x^k = A002426(n), A005773(n+1), A000244(n), A126932(n) for x = 0,1,2,3 respectively. - Philippe Deléham, Dec 03 2009
T(n, k) = (-1)^(k + n) binomial(n, k) hypergeom([k/2 + 1/2, k/2 + 1, k - n], [k + 1, k + 1], 4). - Peter Luschny, Jul 17 2021

Simplified definition from R. J. Mathar, Oct 06 2009

A178301 Triangle T(n,k) = binomial(n,k)*binomial(n+k+1,n+1) read by rows, 0 <= k <= n.

Original entry on oeis.org

1, 1, 3, 1, 8, 10, 1, 15, 45, 35, 1, 24, 126, 224, 126, 1, 35, 280, 840, 1050, 462, 1, 48, 540, 2400, 4950, 4752, 1716, 1, 63, 945, 5775, 17325, 27027, 21021, 6435, 1, 80, 1540, 12320, 50050, 112112, 140140, 91520, 24310, 1, 99, 2376, 24024, 126126, 378378, 672672, 700128, 393822, 92378

Offset: 0

Alford Arnold, May 30 2010

Antidiagonal sums are given by A113682. - Johannes W. Meijer, Mar 24 2013
The rows seem to give (up to sign) the coefficients in the expansion of the integer-valued polynomial binomial(x+n,n)*binomial(x+n,n-1) in the basis made of the binomial(x+i,i). - F. Chapoton, Nov 01 2022
Chapoton's observation above is correct: the precise expansion is  binomial(x+n,n)*binomial(x+n,n-1) = Sum_{k = 0..n-1} (-1)^k*T(n-1,n-1-k)*binomial(x+2*n-1-k,2*n-1-k), as can be verified using the WZ algorithm. For example, n = 4 gives binomial(x+4,4)*binomial(x+4,3)  = 35*binomial(x+7,7) - 45*binomial(x+6,6) + 15*binomial(x+5,5) - binomial(x+4,4). - Peter Bala, Jun 24 2023

			n=0: 1;
n=1: 1,  3;
n=2: 1,  8,  10;
n=3: 1, 15,  45,   35;
n=4: 1, 24, 126,  224,   126;
n=5: 1, 35, 280,  840,  1050,   462;
n=6: 1, 48, 540, 2400,  4950,  4752,  1716;
n=7: 1, 63, 945, 5775, 17325, 27027, 21021, 6435;

David A. Corneth, Table of n, a(n) for n = 0..10010
Author?, Norm of a continuous function, dxdy.ru (in Russian).

Cf. A007318, A047781 (row sums), A178300, A182626, A123160, A132813.

A178301 := proc(n,k)
        binomial(n,k)*binomial(n+k+1,n+1) ;
end proc: # R. J. Mathar, Mar 24 2013
R := proc(n) add((-1)^(n+k)*(2*k+1)*orthopoly:-P(k,2*x+1)/(n+1), k=0..n) end:
for n from 0 to 6 do seq(coeff(R(n), x, k), k=0..n) od; # Peter Luschny, Aug 25 2021

Flatten[Table[Binomial[n,k]Binomial[n+k+1,n+1],{n,0,10},{k,0,n}]] (* Harvey P. Dale, Aug 23 2014 *)

create_list(binomial(n,k)*binomial(n+k+1,n+1),n,0,12,k,0,n); /* Emanuele Munarini, Dec 16 2016 */

R(n,x) = sum(k=0,n, (-1)^(n+k) * (2*k+1) * pollegendre(k,2*x+1)) / (n+1); \\ Max Alekseyev, Aug 25 2021

T(n,k) = A007318(n,k) * A178300(n+1,k+1).
From Peter Bala, Jun 18 2015: (Start)
n-th row polynomial R(n,x) = Sum_{k = 0..n} binomial(n,k)*binomial(n+k+1,n+1)*x^k = Sum_{k = 0..n} (-1)^(n+k)*binomial(n+1,k+1)*binomial(n+k+1,n+1)*(1 + x)^k.
Recurrence: (2*n - 1)*(n + 1)*R(n,x) = 2*(4*n^2*x + 2*n^2 - x - 1)*R(n-1,x) - (2*n + 1)(n - 1)*R(n-2,x) with R(0,x) = 1, R(1,x) = 1 + 3*x.
A182626(n) = -R(n-1,-2) for n >= 1. (End)
From Peter Bala, Jul 20 2015: (Start)
n-th row polynomial R(n,x) = Jacobi_P(n,0,1,2*x + 1).
(1 + x)*R(n,x) gives the row polynomials of A123160. (End)
G.f.: (1+x-sqrt(1-2*x+x^2-4*x*y))/(2*(1+y)*x*sqrt(1-2*x+x^2-4*x*y)). - Emanuele Munarini, Dec 16 2016
R(n,x) = Sum_{k=0..n} (-1)^(n+k)*(2*k+1)*P(k,2*x+1)/(n+1), where P(k,x) is the k-th Legendre polynomial (cf. A100258) and P(k,2*x+1) is the k-th shifted Legendre polynomial (cf. A063007). - Max Alekseyev, Jun 28 2018; corrected by Peter Bala, Aug 08 2021
Polynomial g(n,x) = R(n,-x)/(n+1) delivers the maximum of f(1)^2/(Integral_{x=0..1} f(x)^2 dx) over all polynomials f(x) with real coefficients and deg(f(x)) <= n. This maximum equals (n+1)^2. See dxdy.ru link. - Max Alekseyev, Jun 28 2018

A217539 Number of Dyck paths of semilength n which satisfy the condition: number of returns + number of hills < number of peaks.

Original entry on oeis.org

0, 0, 0, 1, 4, 17, 66, 252, 946, 3523, 13054, 48248, 178146, 657813, 2430962, 8995521, 33342588, 123822171, 460772982, 1718304786, 6421729878, 24051429321, 90272123682, 339522804129, 1279556832780, 4831639423695, 18278491474726, 69272752632502, 262981858878706

Offset: 0

Peter Luschny, Oct 22 2012

nonn

David Scambler observed that [1, 0, A113682(n-2)] for n>=2 count the Dyck paths of semilength n which satisfy the condition "number of peaks = number of returns + number of hills" and [1, A189912(n-1)] for n>=1 count the paths which satisfy the condition "number of peaks <= number of returns + number of hills".

			a(4) = 4 count the Dyck words
[11010100] (()()()) [11011000] (()(()))
[11100100] ((())()) [11101000] ((()())) .

Cf. A217540.

A217539 := proc(n) local k; if n = 0 then 0 else (2*n)!/(n!^2*(n+1)) - add((n-1)!/(((n-1-k)!*iquo(k,2)!^2)*(iquo(k,2)+1)), k=0..n-1) fi end: seq(A217539(i), i=0..28);

MotzkinNumber[n_] := Sum[ Binomial[n+1, k]*Binomial[n+1-k, k-1], {k, 0, Ceiling[(n+1)/2]}]/(n+1); a[0] = a[1] = 0; a[n_] := CatalanNumber[n] - (n-1)*MotzkinNumber[n-2] - MotzkinNumber[n-1]; Table[a[n], {n, 0, 28}] (* Jean-François Alcover, Jun 27 2013, from 3rd formula *)

def A217539(n):
    @CachedFunction
    def M(n): return (3*(n-1)*M(n-2)+(2*n+1)*M(n-1))/(n+2) if n>1 else 1
    @CachedFunction
    def catalan(n): return ((4*n-2)*catalan(n-1))/(n+1) if n>0 else 1
    return catalan(n) - (n-1)*M(n-2) - M(n-1) if n!=0 else 0
[A217539(i) for i in (0..28)]

a(n) = Sum_{k < 0} A217540(n, k).
a(n) = A000108(n) - A189912(n-1) for n > 0.
a(n) = C(n)-(n-1)*M(n-2)-M(n-1) for n > 0; C(n) Catalan, M(n) Motzkin numbers.
Conjecture: 2*(n+1)*(n-3)*a(n) +(-15*n^2+53*n-12)*a(n-1) +(28*n^2-157*n+165)*a(n-2) + 3*(3*n^2+2*n-26)*a(n-3) -18*(2*n-7)*(n-4)*a(n-4)=0. - R. J. Mathar, Nov 11 2012

A217540 Scambler statistic on Dyck paths. Triangle T(n, k) read by rows, n >= 0, -n <= k <= n, T(n, k) is the number of Dyck paths of semilength n and k = number of returns + number of hills - number of peaks.

Original entry on oeis.org

1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1, 2, 0, 1, 0, 0, 1, 3, 4, 2, 3, 0, 1, 0, 0, 1, 6, 10, 9, 8, 3, 4, 0, 1, 0, 0, 1, 10, 25, 30, 26, 17, 13, 4, 5, 0, 1, 0, 0, 1, 15, 56, 90, 90, 70, 49, 27, 19, 5, 6, 0, 1, 0, 0, 1, 21, 112, 245, 301, 266, 197, 128, 80, 39, 26, 6, 7, 0, 1

Offset: 0

Peter Luschny, Oct 21 2012

			[n\k] -8,-7,-6, -5,  -4,  -3,  -2,  -1,  0,   1,  2,  3,  4, 5, 6, 7, 8
-----------------------------------------------------------------------
[ 0 ]                                    1,
[ 1 ]                               0,   0,   1,
[ 2 ]                          0,   0,   1,   0,  1,
[ 3 ]                     0,   0,   1,   1,   2,  0,  1,
[ 4 ]                0,   0,   1,   3,   4,   2,  3,  0,  1,
[ 5 ]           0,   0,   1,   6,  10,   9,   8,  3,  4,  0, 1,
[ 6 ]       0,  0,   1,  10,  25,  30,  26,  17, 13,  4,  5, 0, 1,
[ 7 ]    0, 0,  1,  15,  56,  90,  90,  70,  49, 27, 19,  5, 6, 0, 1,
[ 8 ] 0, 0, 1, 21, 112, 245, 301, 266, 197, 128, 80, 39, 26, 6, 7, 0, 1
.
T(5, -2) = 6 counting the Dyck words
[1101011000] (()()(())) [1101100100] (()(())()) [1101101000] (()(()()))
[1110010100] ((())()()) [1110100100] ((()())()) [1110101000] ((()()())) .

Alois P. Heinz, Rows n = 0..100, flattened
Peter Luschny, The Scambler_statistic_on_Dyck_words.

b:= proc(x, y, t) option remember; `if`(y<0 or y>x, 0,
      `if`(x=0, z, expand(`if`(y=0, z, 1)*(b(x-1, y+1, true)
      +b(x-1, y-1, false)*`if`(t and y<>1, 1/z, 1)))))
    end:
T:= n-> `if`(n=0, 1, (p-> seq(coeff(p, z, i), i=-n..n))
                     (b(2*n-1, 1, true))):
seq(T(n), n=0..10);  # Alois P. Heinz, Jun 10 2014

b[x_, y_, t_] := b[x, y, t] = If[y<0 || y>x, 0, If[x == 0, z, Expand[If[y == 0, z, 1]*(b[x-1, y+1, True]+b[x-1, y-1, False]*If[t && y != 1, 1/z, 1])]]]; T[n_] := If[n == 0, 1, Function[p, Table[Coefficient[p, z, i], {i, -n, n}]][b[2*n-1, 1, True]]]; Table[T[n], {n, 0, 10}] // Flatten (* Jean-François Alcover, Oct 24 2016, after Alois P. Heinz *)

def A217540(n, k):
    def characteristic(d):
        count = 1
        h = d.heights()
        for i in (1..len(d)-1):
            if d[i-1]==1 and d[i]==0: count -= 1
            if h[i]==0: count +=1
            else:
                if h[i-1]==0 and h[i+1]==0: count += 1
        return count
    if n == 0: return 1
    count = 0
    for d in DyckWords(n):
        if k == characteristic(d): count += 1
    return count
for n in (0..6): [A217540(n, k) for k in (-n..n)]

T(n,-1) = A014531(n-2) = [0,0,0],1,3,10,30,90,...
T(n, 0) = A113682(n-2) = [1,0],1,1,4,9,26,70,197,...
T(n, 1) = A194588(n-1) = [0],1,0,2,2,8,17,49,128,...
Sum(k>=0,T(n,k))     = A189912(n-1) = [1],1,2,4,10,25, 66,177,..
Sum(k< 0,T(n,k))     = A217539(n)   =   0,0,0,1, 4,17, 66,252,..
Sum(-n<=k<=n,T(n,k)) = A000108(n)   =   1,1,2,5,14,42,132,429,..

A164611 Expansion of (1 + x + 2x^2 - x^3)/(1 - 2x + 3x^2 - 2x^3 + x^4).

Original entry on oeis.org

1, 3, 5, 2, -6, -11, -5, 9, 17, 8, -12, -23, -11, 15, 29, 14, -18, -35, -17, 21, 41, 20, -24, -47, -23, 27, 53, 26, -30, -59, -29, 33, 65, 32, -36, -71, -35, 39, 77, 38, -42, -83, -41, 45, 89, 44, -48, -95, -47, 51, 101

Offset: 0

Paul Barry, Aug 17 2009

Hankel transform of A113682.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,-3,2,-1).

CoefficientList[Series[(1+x+2x^2-x^3)/(1-2x+3x^2-2x^3+x^4),{x,0,80}],x] (* or *) LinearRecurrence[{2,-3,2,-1},{1,3,5,2},80] (* Harvey P. Dale, May 28 2013 *)

x='x+O('x^50); Vec((1 +x +2*x^2 -x^3)/(1 -2*x +3*x^2 -2*x^3 +x^4)) \\ G. C. Greubel, Aug 10 2017

G.f.: (1+x+2*x^2-x^3)/(1-x+x^2)^2.

a(n) = 2*a(n-1)-3*a(n-2)+2*a(n-3)-a(n-4), with a(0)=1, a(1)=3, a(2)=5, a(3)=2. - Harvey P. Dale, May 28 2013

A260056 Irregular triangle read by rows: coefficients T(n, k) of certain polynomials p(n, x) with exponents in increasing order, n >= 0 and 0 <= k <= 2*n.

Original entry on oeis.org

1, 2, 1, 1, 3, 3, 4, 2, 1, 4, 6, 10, 9, 7, 3, 1, 5, 10, 20, 25, 26, 19, 11, 4, 1, 6, 15, 35, 55, 71, 70, 56, 34, 16, 5, 1, 7, 21, 56, 105, 161, 196, 197, 160, 106, 55, 22, 6, 1, 8, 28, 84, 182, 322, 462, 554, 553, 463, 321, 183, 83, 29, 7, 1, 9, 36, 120, 294, 588, 966, 1338, 1569, 1570, 1337, 967, 587, 295, 119, 37, 8, 1, 10, 45, 165, 450, 1002, 1848, 2892, 3873, 4477, 4476, 3874

Offset: 0

Werner Schulte, Nov 08 2015

The triangle is related to the triangle of trinomial coefficients.

			The irregular triangle T(n,k) begins:
n\k:  0   1   2    3    4    5    6    7    8    9   10  11  12  13  14  ...
0     1;
1     2   1   1;
2     3   3   4    2    1;
3     4   6  10    9    7    3    1;
4     5  10  20   25   26   19   11    4    1;
5     6  15  35   55   71   70   56   34   16    5    1;
6     7  21  56  105  161  196  197  160  106   55   22   6   1;
7     8  28  84  182  322  462  554  553  463  321  183  83  29   7   1;
etc.
The polynomial corresponding to row 2 is p(2,x) = 3+3*x+4*x^2+2*x^3+x^4.

G. C. Greubel, Table of n, a(n) for the first 100 rows, flattened

Cf. A000027 (col 0), A000217 (col 1), A000292 (col 2), A001590, A002426, A004524, A005582 (col 3), A008937, A027907, A095662 (col 5), A113682, A246437.

A027907[n_, k_] := Sum[Binomial[n, j]*Binomial[j, k - j], {j, 0, n}]; Table[ Sum[A027907[j, k], {j, 0, n}], {n,0,10}, {k, 0, 2*n} ] // Flatten (* G. C. Greubel, Mar 07 2017 *)

T(n,0) = n+1, and T(n,k) = 0 for k < 0 or k > 2*n, and T(n+1,k) = T(n,k) + T(n,k-1) + T(n,k-2) for k > 0.
T(n,k) = Sum_{j=0..n} A027907(j,k) for 0 <= k <= 2*n.
T(n,k) = Sum_{j=0..k} (-1)^(k-j)*A027907(n+1,j+1) for 0 <= k <= 2*n.
T(n,k) = T(n,2*n-1-k) + (-1)^k for 0 <= k < 2*n.
p(n,x) = Sum_{k=0..2*n} T(n,k)*x^k = Sum_{k=0..n} (1+x+x^2)^k for n >= 0.
p(n,x) = ((1+x+x^2)^(n+1)-1)/(x+x^2), p(n,0) = p(n,-1) = n+1 for n >= 0.
p(n+1,x) = (1+x+x^2)*p(n,x)+1 for n >= 0.
Sum_{n>=0} p(n,x)*t^n = 1/((1-t)*(1-t*(1+x+x^2))).
T(n,2*n) = 1, and T(n,n) = A113682(n) for n >= 0.
T(n,n-1) = A246437(n+1), and T(n,n-1)+T(n,n) = A002426(n+1) for n > 0.
If d(n) is n-th antidiagonal sum of the triangle then: d(n) = A008937(n+1), and d(n+2)-d(n) = A001590(n+5) for n >= 0.
Conjecture: If a(n) is n-th antidiagonal alternating sum of the triangle then: a(n) = A004524(n+3).
Sum_{k=0..2*n} (-1)^k*T(n,k)^2 = (3^(n+1)-1)/2 for n >= 0.
Sum_{k=0..2*n} (-1)^k*(y*k+1)*T(n,k) = Sum{k=0..n} y*k+1 = (n+1)*(y*n+2)/2 for real y and n >= 0.
Conjecture of linear recurrence for column k: Sum_{m=0..k+2} (-1)^m*T(n+m,k)* binomial(k+2,m) = 0 for k >= 0 and n >= 0.

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A097613 a(n) = binomial(2n-3,n-1) + binomial(2n-2,n-2).

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A178301 Triangle T(n,k) = binomial(n,k)*binomial(n+k+1,n+1) read by rows, 0 <= k <= n.

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A217539 Number of Dyck paths of semilength n which satisfy the condition: number of returns + number of hills < number of peaks.

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A217540 Scambler statistic on Dyck paths. Triangle T(n, k) read by rows, n >= 0, -n <= k <= n, T(n, k) is the number of Dyck paths of semilength n and k = number of returns + number of hills - number of peaks.

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A164611 Expansion of (1 + x + 2*x^2 - x^3)/(1 - 2*x + 3*x^2 - 2*x^3 + x^4).

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A260056 Irregular triangle read by rows: coefficients T(n, k) of certain polynomials p(n, x) with exponents in increasing order, n >= 0 and 0 <= k <= 2*n.

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