A156894 -id:A156894 - cp's OEIS frontend

A103885 a(n) = [x^(2*n)] ((1 + x)/(1 - x))^n.

1, 2, 16, 146, 1408, 14002, 142000, 1459810, 15158272, 158611106, 1669752016, 17664712562, 187641279616, 2000029880786, 21380213588848, 229129634462146, 2460955893981184, 26482855453375042, 285475524009208720, 3082024598888203090, 33319523640218177408

Offset: 0

Ralf Stephan, Feb 20 2005

From Peter Bala, Mar 01 2020: (Start)
The recurrence given below can be rewritten in the form
(2*n+1)*(2*n+2)*P(2,n)*a(n+1) - (2*n-1)*(2*n-2)*P(2,-n)*a(n-1) = Q(2,n^2)*a(n), where the polynomial Q(2,n) = 4*(55*n^2 - 34*n + 3) and the polynomial P(2,n) = 5*n^2 - 5*n + 1 satisfies the symmetry condition P(2,n) = P(2,1-n) and has real zeros.
More generally, for fixed m = 1,2,3,..., we conjecture that the sequence b(n) := a(m*n) satisfies a recurrence of the form ( Product_{k = 1..2*m} (2*m*n + k) ) * P(2*m,n)*b(n+1) + (-1)^m*( Product_{k = 1..2*m} (2*m*n - k) ) * P(2*m,-n)*b(n-1) = Q(2*m,n^2)*b(n), where the polynomials P(2*m,n) and Q(2*m,n) have degree 2*m. Conjecturally, the polynomial P(2*m,n) = P(2*m,1-n) and has real zeros in the interval [0, 1]. The 4*m zeros of the polynomial Q(2*m,n^2) seem to belong to the interval [-1, 1] and 4*m - 2 of these zeros appear to be approximated by the rational numbers +- k/(3*m), where 1 <= k <= 3*m - 2, k not a multiple of 3. (End)

G. C. Greubel, Table of n, a(n) for n = 0..950 [a(0) = 1 inserted by _Georg Fischer_, Apr 03 2020]
Peter Bala, Notes on A103885
V. V. Kruchinin and D. V. Kruchinin, A Generating Function for the Diagonal T_{2n,n} in Triangles, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.6.

Cf. A002003, A006318, A027307, A103882, A103884, A123164, A144097, A156894, A266213, A370102.

A103885:= func< n | n eq 0 select 1 else (&+[ Binomial(n, k)*Binomial(2*n+k-1, n-1): k in [0..n]]) >;
[A103885(n): n in [0..40]]; // G. C. Greubel, Oct 27 2024

a := n -> `if`(n=0, 1, 2*n*hypergeom([1 - 2*n, 1 - n], [2], 2)):
seq(simplify(a(n)), n=0..17); # Peter Luschny, Dec 30 2019
# Alternative (after Peter Bala ):
gf := n -> ( (1 + x)/(1 - x) )^n: ser := n -> series(gf(n), x, 40):
seq(coeff(ser(n), x, 2*n), n=0..17); # Peter Luschny, Mar 20 2020

Prepend[Table[Sum[2^i Binomial[n, i] Binomial[2n-1, i-1], {i, 1, 2n}], {n,1,20}], 1] (* Vaclav Kotesovec, Jul 01 2015 *)

a(n) = if (n==0, 1, sum(i=0, n, 2^i * binomial(n, i) * binomial(2*n-1, i-1))); \\ Michel Marcus, Mar 21 2020

def A103885(n): return 1 if n==0 else sum(binomial(n, k)*binomial(2*n+k-1, n-1) for k in range(n+1))
[A103885(n) for n in range(41)] # G. C. Greubel, Oct 27 2024

a(n) = Sum_{i=0..n} 2^i * binomial(n,i) * binomial(2*n-1,i-1). [Original definition, with summation range {i=1..n}.]
a(n) = A103884(n, n).
G.f.: A(x) = x*B(x)'/B(x), where B(x) is g.f. of A027307. - Vladimir Kruchinin, Jun 30 2015
From Vaclav Kotesovec, Jul 01 2015: (Start)
Recurrence: n*(2*n-1)*(5*n^2 - 15*n + 11)*a(n) = 2*(55*n^4 - 220*n^3 + 296*n^2 - 152*n + 24)*a(n-1) + (n-2)*(2*n-3)*(5*n^2 - 5*n + 1)*a(n-2).
a(n) ~ ((11 + 5*sqrt(5))/2)^n / (2 * 5^(1/4) * sqrt(Pi*n)). (End)
a(n) = [x^n] (1/(1 - x - x/(1 - x - x/(1 - x - x/(1 - x - x/(1 - ...))))))^n, a continued fraction. - Ilya Gutkovskiy, Sep 29 2017
a(n) = 2*n*hypergeom([1 - 2*n, 1 - n], [2], 2) for n >= 1. - Peter Luschny, Dec 30 2019
From Peter Bala, Mar 01 2020: (Start)
a(n) = Sum_{k = 0..n} C(n, k)*C(2*n+k-1, n-1), with a(0) = 1.
a(n) = Sum_{k = 0..n} C(2*n, 2*k)*C(2*n-k-1, n-1), with a(0) = 1.
a(n) = (1/2)*Sum_{k = 0..n} C(2*n, n-k)*C(2*n+k-1, k). Cf. A156894.
a(n) = [x^n] S(x)^n, where S(x) = (1 - x - sqrt(1 - 6*x + x^2))/(2*x) is the o.g.f. of the sequence of large Schröder numbers A006318.
a(n) = (1/2) * [x^(n)] ( (1 + x)/(1 - x) )^(2*n). Cf. A002003(n) = [x^n] ( (1 + x)/(1 - x) )^n.
Conjecture: a(n) = - [x^n] G(x)^(-n), where G(x) = 1 + 2*x + 14*x^2 + 134*x^3 + 1482*x^4 + ... is the o.g.f. of A144097.
a(p) == 2 ( mod p^3 ) for prime p >= 5. (End)
From Peter Bala, Sep 22 2021: (Start)
a(n) = Sum_{k = 0..n} 4^k*binomial(n+k-1,n)*binomial(n,k)^2 / binomial(2*k,k).
Equivalently, a(n) = [x^n] T(n,(1+x)/(1-x)), where T(n,x) is the n-th Chebyshev polynomial of the first kind. Cf. A103882. (End)
For n>0, a(n) = (1/3) * [x^n] (1/S(-x))^(3*n), where S(x) = (1 - x - sqrt(1 - 6*x + x^2))/(2*x) is the o.g.f. of the sequence of large Schröder numbers A006318. Cf. A370102. - Peter Bala, Jul 29 2024

a(0) = 1 added and new name by Peter Bala, Mar 01 2020

A003169 Number of 2-line arrays; or number of P-graphs with 2n edges.

Original entry on oeis.org

1, 3, 14, 79, 494, 3294, 22952, 165127, 1217270, 9146746, 69799476, 539464358, 4214095612, 33218794236, 263908187100, 2110912146295, 16985386737830, 137394914285538, 1116622717709012, 9113225693455362, 74659999210200292

Offset: 1

N. J. A. Sloane

First column of triangle A100326. - Paul D. Hanna, Nov 16 2004

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 1..200
M. Bicknell and V. E. Hoggatt, Jr., Sequences of matrix inverses from Pascal, Catalan and related convolution arrays, Fib. Quart., 14 (1976), 224-232.
L. Carlitz, Enumeration of two-line arrays, Fib. Quart., Vol. 11 Number 2 (1973), 113-130.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 416
R. C. Read, On the enumeration of a class of plane multigraphs, Aequat. Math. 31 (1986) no 1, 47-63.
Jun Yan, Lattice paths enumerations weighted by ascent lengths, arXiv:2501.01152 [math.CO], 2025. See p. 8.
Anssi Yli-Jyrä and Carlos Gómez-Rodríguez, Generic Axiomatization of Families of Noncrossing Graphs in Dependency Parsing, arXiv:1706.03357 [cs.CL], 2017.

Cf. A003168, A100324, A100326, A156894.

a003169 = flip a100326 0  -- Reinhard Zumkeller, Nov 21 2015

a[0]:=0:a[1]:=1:a[2]:=3:for n from 3 to 30 do a[n]:=((324*n^2-708*n+360)*a[n-1] -(371*n^2-1831*n+2250)*a[n-2]+(20*n^2-130*n+210)*a[n-3])/(16*n*(2*n-1)) od:seq(a[n],n=1..25); # Emeric Deutsch, Jan 31 2005

lim = 21; t[0, 0] = 1; t[n_, 0] := t[n, 0] = Sum[(k + 1)*t[n - 1, k], {k, 0, n - 1}]; t[n_, k_] := t[n, k] = Sum[t[j + 1, 0]*t[n - j - 1, k - 1], {j, 0, n - k}]; Table[ t[n, 0], {n, lim}] (* Jean-François Alcover, Sep 20 2011, after Paul D. Hanna's comment *)

{a(n)=if(n==0,0,if(n==1,1,if(n==2,3,( (324*n^2-708*n+360)*a(n-1) -(371*n^2-1831*n+2250)*a(n-2)+(20*n^2-130*n+210)*a(n-3))/(16*n*(2*n-1)) )))} \\ Paul D. Hanna, Nov 16 2004

{a(n)=local(A=x+x*O(x^n));if(n==1,1, for(i=1,n,A=x*(1+A)/(1-A)^2); polcoeff(A,n))}

seq(n)=Vec(serreverse(x*(1 - x)^2/(1 + x) + O(x*x^n))) \\ Andrew Howroyd, Mar 07 2023

For formula see Read reference.
D-finite with recurrence a(n) = ( (324*n^2-708*n+360)*a(n-1) - (371*n^2-1831*n+2250)*a(n-2) + (20*n^2-130*n+210)*a(n-3) )/(16*n*(2*n-1)) for n>2, with a(0)=0, a(1)=1, a(2)=3. - Paul D. Hanna, Nov 16 2004
G.f. satisfies: A(x) = x*(1+A(x))/(1-A(x))^2 where A(0)=0. G.f. satisfies: (1+A(x))/(1-A(x)) = 2*G003168(x)-1, where G003168 is the g.f. of A003168. - Paul D. Hanna, Nov 16 2004
a(n) = (1/n)*Sum_{i=0..n-1} binomial(n,i)*binomial(3*n-i-2,n-i-1). - Vladeta Jovovic, Sep 13 2006
Appears to be (1/n)*Jacobi_P(n-1,1,n-1,3). If so then a(n) = (1/(2*n-1))*Sum_{k = 0..n-1} binomial(n-1,k)*binomial(2*n+k-1,k+1) = (1/n)*Sum_{k = 0..n} binomial(n,k)*binomial(2*n-2,n+k-1)*2^k. [Added Jun 11 2023: these are correct, and can be proved using the WZ algorithm.] - Peter Bala, Aug 01 2012
a(n) ~ sqrt(33/sqrt(17)-7) * ((71+17*sqrt(17))/16)^n / (4*sqrt(2*Pi)*n^(3/2)). - Vaclav Kotesovec, Aug 09 2013
The o.g.f. A(x) = 1 + 3*x + 14*x^2 + ... taken with offset 0, satisfies 1 + x*A'(x)/A(x) = 1 + 3*x + 19*x^2 + 138*x^3 + ..., the o.g.f. for A156894. - Peter Bala, Oct 05 2015
From Peter Bala, Jun 11 2023: (Start)
a(n) = (1/n)*Sum_{k = 0..n-1} binomial(n,k+1)*binomial(2*n+k-1,k) (Carlitz, equation 3.19).
4*n*(17*n - 29)*(2*n - 1)*a(n) = (1207*n^3 - 4473*n^2 + 5258*n - 1920)*a(n-1) - 2*(2*n - 5)*(17*n - 12)*(n - 2)*a(n-2) with a(1) = 1 and a(2) = 3. (End)

More terms from Emeric Deutsch, Jan 31 2005

A119259 Central terms of the triangle in A119258.

Original entry on oeis.org

1, 3, 17, 111, 769, 5503, 40193, 297727, 2228225, 16807935, 127574017, 973168639, 7454392321, 57298911231, 441739706369, 3414246490111, 26447737520129, 205272288591871, 1595964714385409, 12427568655368191, 96905907580960769, 756583504975757311, 5913649000782757889

Offset: 0

Reinhard Zumkeller, May 11 2006

The Gauss congruences a(n*p^k) == a(n^p^(k-1)) (mod p^k) hold for prime p and positive integers n and k. - Peter Bala, Jan 06 2022

R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
J. Abate and W. Whitt, Brownian Motion and the Generalized Catalan Numbers, J. Int. Seq. 14 (2011) # 11.2.6, eq (42).

Cf. A005408, A056220, A000225, A000337, A055580, A027608, A119258, A064062, A178792, A014300, A098665.

a119259 n = a119258 (2 * n) n  -- Reinhard Zumkeller, Aug 06 2014

Table[Binomial[2k - 1, k] Hypergeometric2F1[-2k, -k, 1 - 2k, -1], {k, 0, 10}] (* Vladimir Reshetnikov, Feb 16 2011 *)

from itertools import count, islice
def A119259_gen(): # generator of terms
    yield from (1,3)
    a, c = 2, 1
    for n in count(1):
        yield (a<>1
A119259_list = list(islice(A119259_gen(),20)) # Chai Wah Wu, Apr 26 2023

a(n) = A119258(2*n,n).
a(n) = Sum_{k=0..n} C(2*n,k)*C(2*n-k-1,n-k). - Paul Barry, Sep 28 2007
a(n) = Sum_{k=0..n} C(n+k-1,k)*2^k. - Paul Barry, Sep 28 2007
2*a(n) = A064062(n)+A178792(n). - Joseph Abate, Jul 21 2010
G.f.: (4*x^2+3*sqrt(1-8*x)*x-5*x)/(sqrt(1-8*x)*(2*x^2+x-1)-8*x^2-7*x+1). - Vladimir Kruchinin, Aug 19 2013
a(n) = (-1)^n - 2^(n+1)*binomial(2*n,n-1)*hyper2F1([1,2*n+1],[n+2],2). - Peter Luschny, Jul 25 2014
a(n) = (-1)^n + 2^(n+1)*A014300(n). - Peter Luschny, Jul 25 2014
a(n) = [x^n] ( (1 + x)^2/(1 - x) )^n. Exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 3*x + 13*x^2 + 67*x^3 + ... is essentially the o.g.f. for A064062. - Peter Bala, Oct 01 2015
The o.g.f. is the diagonal of the bivariate rational function 1/(1 - t*(1 + x)^2/(1 - x)) and hence is algebraic by Stanley 1999, Theorem 6.33, p.197. - Peter Bala, Aug 21 2016
n*(3*n-4)*a(n) +(-21*n^2+40*n-12)*a(n-1) -4*(3*n-1)*(2*n-3)*a(n-2)=0. - R. J. Mathar, Aug 09 2017
From Peter Bala, Mar 23 2020: (Start)
a(p) == 3 ( mod p^3 ) for prime p >= 5. Cf. A002003, A103885 and A156894.
More generally, we conjecture that a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. (End)
G.f.: (8*x)/(sqrt(1-8*x)*(1+4*x)-1+8*x). - Fabian Pereyra, Jul 20 2024
a(n) = 2^(n+1)*binomial(2*n,n) - A178792(n). - Akiva Weinberger, Dec 06 2024
a(n) = Sum_{k=0..n} 2^k * (-1)^(n-k) * binomial(2*n,k). - Seiichi Manyama, Jul 31 2025

A114496 a(n) = Sum of binomial(n,k)*binomial(2n+k,k) over all k.

Original entry on oeis.org

1, 4, 26, 190, 1462, 11584, 93536, 765314, 6323270, 52638760, 440815036, 3709445084, 31340292076, 265683004240, 2258793820988, 19251776923210, 164440378882630, 1407266585304760, 12063701803046300, 103571977632247076

Offset: 0

Eric Rowland, Dec 01 2005

Modification of A001850 inspired by the Apéry numbers A005259.
From Paul Barry, Feb 17 2009: (Start)
Central coefficient of (1 + 4x + 5x^2 + 2x^3)^n. The coefficients are the 4th row of A029635.
The third row of A029635 corresponds to the central Delannoy numbers A001850. (End)

G. C. Greubel, Table of n, a(n) for n = 0..1000
P. Barry, A Note on a Family of Generalized Pascal Matrices Defined by Riordan Arrays, Journal of Integer Sequences, 16 (2013), #13.5.4.

Cf. A114497, A114498, A003168, A156894.

Cf. A156886. - Paul Barry, Feb 17 2009

Table[Sum[Binomial[n, k]*Binomial[2n+k, k], {k, 0, n}], {n,0,25}]

a(n) = sum(k=0, n, 2^(n-k)*binomial(2*n,k)*binomial(n,k));
vector(50, n, a(n-1)) \\ Altug Alkan, Oct 05 2015

a(n) = Sum_{k=0..n} (binomial(n,k)*binomial(2n+k,k)).
Recurrence: 20*n*(2*n - 1)*a(n) = (371*n^2 - 411*n + 120)*a(n-1) -2*(81*n^2 - 299*n + 278)*a(n-2) + 4*(n-2)*(2*n-5)*a(n-3). - Vaclav Kotesovec, Oct 19 2012
a(n) ~ sqrt(1734 + 442*sqrt(17))*((71 + 17*sqrt(17))/16)^n/(68*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 19 2012
From Peter Bala, Oct 05 2015: (Start)
a(n) = Sum_{i = 0..n} 2^(n-i)*binomial(2*n,i)*binomial(n,i).
4*n*(2*n - 1)*(17*n - 23)*a(n) = (1207*n^3 - 2840*n^2 + 1897*n - 360)*a(n-1) - 2*(n - 1)*(17*n - 6)*(2*n - 3)*a(n-2) with a(0) = 1 and a(1) = 4.
1 + x*exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + x + 4*x^2 + 21*x^3 + 126*x^4 + ... is the o.g.f. for A003168. (End)

A352373 a(n) = [x^n] ( 1/((1 - x)^2*(1 - x^2)) )^n for n >= 1.

Original entry on oeis.org

2, 12, 74, 484, 3252, 22260, 154352, 1080612, 7621526, 54071512, 385454940, 2758690636, 19810063392, 142662737376, 1029931873824, 7451492628260, 54013574117106, 392188079586468, 2851934621212598, 20766924805302984, 151403389181347160, 1105047483656041080

Offset: 1

Peter Bala, Mar 14 2022

Suppose n identical objects are distributed in 3*n labeled baskets, 2*n colored white and n colored black. White baskets can contain any number of objects (or be empty), while black baskets must contain an even number of objects (or be empty). a(n) is the number of distinct possible distributions.
Number of nonnegative integer solutions to n = x_1 + x_2 + ... + x_(2*n) + 2*y_1 + 2*y_2 + ... + 2*y_n.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and positive integers n and k.
Calculation suggests that, in fact, stronger congruences may hold.
Conjecture: the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and positive integers n and k.
More generally, let r and s be integers and define a sequence (a(r,s;n))n>=1 by a(r,s;n) = [x^n] ( (1 + x)^r * (1 - x)^s )^n.
Conjecture: for each r and s the above supercongruences hold for the sequence (a(r,s;n))n>=1.
The present sequence is the case r = -1 and s = -3. Other cases include A000984 (r = 2, s = 0), A001700 with offset 1 (r = 0, s = -1), A002003 (r = 1, s = -1), A091527 (r = 3, s = -1), A119259 (r = 2, s = -1), A156894 (r = 1, s = -2), A165817 (r = 0, s = -2), A234839 (r = 1, s = 2), A348410 (r = -1, s = -2) and A351857 (r = -2, s = -4).

			n = 2: 12 distributions of 2 identical objects in 4 white and 2 black baskets
             White         Black
   1)   (0) (0) (0) (0)   [2] [0]
   2)   (0) (0) (0) (0)   [0] [2]
   3)   (2) (0) (0) (0)   [0] [0]
   4)   (0) (2) (0) (0)   [0] [0]
   5)   (0) (0) (2) (0)   [0] [0]
   6)   (0) (0) (0) (2)   [0] [0]
   7)   (1) (1) (0) (0)   [0] [0]
   8)   (1) (0) (1) (0)   [0] [0]
   9)   (1) (0) (0) (1)   [0] [0]
  10)   (0) (1) (1) (0)   [0] [0]
  11)   (0) (1) (0) (1)   [0] [0]
  12)   (0) (0) (1) (1)   [0] [0]
Examples of supercongruences:
a(7) - a(1) = 154352 - 2 = 2*(3^2)*(5^2)*(7^3) == 0 (mod 7^3);
a(2*11) - a(2) = 1105047483656041080 - 12 = (2^2)*3*(11^3)*13*101*103*2441* 209581 == 0 (mod 11^3).

R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.

Paolo Xausa, Table of n, a(n) for n = 1..1000

Cf. A000984, A001448, A001700, A002003, A091527, A119259, A156894, A165817, A211419, A211421, A234839, A262733, A276098, A348410, A351856, A351857.

seq(add( binomial(3*n-2*k-1,n-2*k)*binomial(n+k-1,k), k = 0..floor(n/2)), n = 1..25);

nterms=25;Table[Sum[Binomial[3n-2k-1,n-2k]Binomial[n+k-1,k],{k,0,Floor[n/2]}],{n,nterms}] (* Paolo Xausa, Apr 10 2022 *)

a(n) = Sum_{k = 0..floor(n/2)} binomial(3*n-2*k-1,n-2*k)*binomial(n+k-1,k).
a(n) = Sum_{k = 0..n} (-1)^k*binomial(4*n-k-1,n-k)*binomial(n+k-1,k).
a(n) = binomial(4*n-1,n)*hypergeom([n, -n], [1-4*n], -1).
48*n*(n-1)*(3*n-1)*(3*n-2)*(93*n^3-434*n^2+668*n-339)*a(n) = 12*(n-1)*(21762*n^6-134199*n^5+323805*n^4-386685*n^3+237728*n^2-70336*n+7680)*a(n-1) + 5*(5*n-9)*(5*n-8)*(5*n-7)*(5*n-6)*(93*n^3-155*n^2+79*n-12)*a(n-2) with a(1) = 2 and a(2) = 12.
The o.g.f. A(x) = 2*x + 12*x^2 + 74*x^3 + ... is the diagonal of the bivariate rational function x*t/(1 - t/((1 - x)^2*(1 - x^2))) and hence is an algebraic function over Q(x) by Stanley 1999, Theorem 6.33, p. 197.
A(x) = x*d/dx(log(F(x))), where F(x) = (1/x)*Series_Reversion( x*(1 - x)^2*(1 - x^2) ).
a(n) ~ sqrt(4 + sqrt(6)) * (13/4 + 31*sqrt(6)/18)^n / (2*sqrt(5*Pi*n)). - Vaclav Kotesovec, Mar 15 2022

A208904 Triangle of coefficients of polynomials v(n,x) jointly generated with A208660; see the Formula section.

Original entry on oeis.org

1, 3, 1, 5, 6, 1, 7, 19, 9, 1, 9, 44, 42, 12, 1, 11, 85, 138, 74, 15, 1, 13, 146, 363, 316, 115, 18, 1, 15, 231, 819, 1059, 605, 165, 21, 1, 17, 344, 1652, 2984, 2470, 1032, 224, 24, 1, 19, 489, 3060, 7380, 8378, 4974, 1624, 292, 27, 1, 21, 670, 5301, 16488

Offset: 1

Clark Kimberling, Mar 03 2012

For a discussion and guide to related arrays, see A208510.
Riordan array ((1+x)/(1-x)^2, x(1+x)/(1-x)^2) (follows from Kruchinin formula). - Ralf Stephan, Jan 02 2014
From Peter Bala, Jul 21 2014: (Start)
Let M denote the lower unit triangular array A099375 and for k = 0,1,2,... define M(k) to be the lower unit triangular block array
/I_k 0\
\ 0  M/
having the k x k identity matrix I_k as the upper left block; in particular, M(0) = M. Then the present triangle equals the infinite matrix product M(0)*M(1)*M(2)*... (which is clearly well-defined). See the Example section. (End)

			First five rows:
1
3...1
5...6....1
7...19...9....1
9...44...42...12...1
First five polynomials v(n,x):
1
3 + x
5 + 6x + x^2
7 + 19x + 9x^2 + x^3
9 + 44x + 42x^2 + 12x^3 + x^4
From _Peter Bala_, Jul 21 2014: (Start)
With the arrays M(k) as defined in the Comments section, the infinite product M(0*)M(1)*M(2)*... begins
/1        \/1        \/1        \      /1            \
|3 1      ||0 1      ||0 1      |      |3  1         |
|5 3 1    ||0 3 1    ||0 0 1    |... = |5  6  1      |
|7 5 3 1  ||0 5 3 1  ||0 0 3 1  |      |7 19  9  1   |
|9 7 5 3 1||0 7 5 3 1||0 0 5 3 1|      |9 44 42 12 1 |
|...      ||...      ||...      |      |...
(End)

Cf. A208660, A208510. A099375.

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := u[n - 1, x] + 2 x*v[n - 1, x];
v[n_, x_] := u[n - 1, x] + (x + 1)*v[n - 1, x] + 1;
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]    (* A208660 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]    (* A208904 *)

u(n,x)=u(n-1,x)+2x*v(n-1,x),
v(n,x)=u(n-1,x)+(x+1)*v(n-1,x)+1,
where u(1,x)=1, v(1,x)=1.
From Vladimir Kruchinin, Mar 11 2013: (Start)
T(n,k) = sum(i=0..n, binomial(i+k-1,2*k-1)*binomial(k,n-i))
((x+x^2)/(1-x)^2)^k = sum(n>=k, T(n,k)*x^n).
T(n,2)=A005900(n).
T(2*n-1,n) / n = A003169(n).
T(2*n,n) = A156894(n), n>1.
sum(k=1..n, T(n,k)) = A003946(n).
sum(k=1..n, T(n,k)*(-1)^(n+k)) = A078050(n).
n*sum(k=1..n, T(n,k)/k) = A058481(n). (End)
Recurrence: T(n+1,k+1) = sum {i = 0..n-k} (2*i + 1)*T(n-i,k). - Peter Bala, Jul 21 2014

cp's OEIS Frontend

A103885 a(n) = [x^(2*n)] ((1 + x)/(1 - x))^n.

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A003169 Number of 2-line arrays; or number of P-graphs with 2n edges.

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A119259 Central terms of the triangle in A119258.

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A114496 a(n) = Sum of binomial(n,k)*binomial(2n+k,k) over all k.

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A352373 a(n) = [x^n] ( 1/((1 - x)^2*(1 - x^2)) )^n for n >= 1.

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A208904 Triangle of coefficients of polynomials v(n,x) jointly generated with A208660; see the Formula section.

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