A059304 -id:A059304 - cp's OEIS frontend

A098658 a(n) = 3^n(2n)!/(n!)^2.

1, 6, 54, 540, 5670, 61236, 673596, 7505784, 84440070, 956987460, 10909657044, 124965162504, 1437099368796, 16581915793800, 191876454185400, 2225766868550640, 25874539846901190, 301362287628613860

Offset: 0

Paul Barry, Sep 20 2004

Number of lattice paths from (0,0) to (n,n) using steps (0,1) and three kinds of steps (1,0). - Joerg Arndt, Jul 01 2011
Sixth binomial transform of 1/sqrt(1-36*x^2).
Diagonal of the rational function 1 / (1 - 3*x - y). - Ilya Gutkovskiy, Apr 24 2025

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Hacène Belbachir and Abdelghani Mehdaoui, Recurrence relation associated with the sums of square binomial coefficients, Quaestiones Mathematicae (2021) Vol. 44, Issue 5, 615-624.
Hacène Belbachir, Abdelghani Mehdaoui, and László Szalay, Diagonal Sums in the Pascal Pyramid, II: Applications, J. Int. Seq., Vol. 22 (2019), Article 19.3.5.
Tony D. Noe, On the Divisibility of Generalized Central Trinomial Coefficients, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.7.

Cf. A000984, A059304, A101600, A322246.

[3^n*Factorial(2*n)/Factorial(n)^2: n in [0..20]]; // Vincenzo Librandi, Jul 05 2011

Table[3^n (2n)!/(n!)^2,{n,0,20}] (* Harvey P. Dale, Dec 14 2011 *)

/* same as in A092566 but use */
steps=[[1,0], [1,0], [1,0], [0,1]]; /* note the triple [1,0] */
/* Joerg Arndt, Jun 30 2011 */

G.f.: 1/sqrt((1-6*x)^2-36*x^2) = 1/sqrt(1-12*x).
E.g.f.: exp(6*x)*BesselI(0, 6x).
a(n) = [t^n](1+6*t+9*t^2)^n.
a(n) = 3^n*A000984(n). - R. J. Mathar, Oct 10 2012
G.f.: Q(0), where Q(k) = 1 + 12*x*(4*k+1)/( 4*k+2 - 12*x*(4*k+2)*(4*k+3)/(12*x*(4*k+3) + 4*(k+1)/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Sep 14 2013
n*a(n) +6*(-2*n+1)*a(n-1)=0. - R. J. Mathar, Nov 27 2014
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 6*x + 45*x^2 + 378*x^3 + ... is the o.g.f. for A101600. - Peter Bala, Jul 16 2015
From Amiram Eldar, Jul 21 2020: (Start)
Sum_{n>=0} 1/a(n) = 12/11 + 12*sqrt(11)*arcsin(1/sqrt(12))/121.
Sum_{n>=0} (-1)^n/a(n) = 12/13 - 12*sqrt(13)*arcsinh(1/sqrt(12))/169. (End)
From Peter Bala, Oct 12 2024: (Start)
a(n) = Integral_{x = 0..12} x^n * w(x) dx, where w(x) = 1/( Pi*sqrt(x*(12 - x)) ) is positive on the interval (0, 12). The weight function w(x) is singular at x = 0 and at x = 12 and is the solution of the Hausdorff moment problem.
Binomial transform of A322246.
The Gauss congruences hold: a(n*p^r) == a(n*p^(r-1)) (mod p^r) for all primes p and positive integers n and r. (End)

A154690 Triangle read by rows: T(n, k) = (2^(n-k) + 2^k)*binomial(n,k), 0 <= k <= n.

Original entry on oeis.org

2, 3, 3, 5, 8, 5, 9, 18, 18, 9, 17, 40, 48, 40, 17, 33, 90, 120, 120, 90, 33, 65, 204, 300, 320, 300, 204, 65, 129, 462, 756, 840, 840, 756, 462, 129, 257, 1040, 1904, 2240, 2240, 2240, 1904, 1040, 257, 513, 2322, 4752, 6048, 6048, 6048, 6048, 4752, 2322, 513

Offset: 0

Roger L. Bagula and Gary W. Adamson, Jan 14 2009

From G. C. Greubel, Jan 18 2025: (Start)
A more general triangle of coefficients may be defined by T(n, k, p, q) = (p^(n-k)*q^k + p^k*q^(n-k))*A007318(n, k). When (p, q) = (2, 1) this sequence is obtained.
Some related triangles are:
  (p, q) = (1, 1) : 2*A007318(n,k).
  (p, q) = (2, 2) : 2*A038208(n,k).
  (p, q) = (3, 2) : A154692(n,k).
  (p, q) = (3, 3) : 2*A038221(n,k). (End)

			Triangle begins as:
     2;
     3,    3;
     5,    8,     5;
     9,   18,    18,     9;
    17,   40,    48,    40,    17;
    33,   90,   120,   120,    90,    33;
    65,  204,   300,   320,   300,   204,    65;
   129,  462,   756,   840,   840,   756,   462,   129;
   257, 1040,  1904,  2240,  2240,  2240,  1904,  1040,   257;
   513, 2322,  4752,  6048,  6048,  6048,  6048,  4752,  2322,  513;
  1025, 5140, 11700, 16320, 16800, 16128, 16800, 16320, 11700, 5140, 1025;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened
A. Lakhtakia, R. Messier, V. K. Varadan, and V. V. Varadan, Use of combinatorial algebra for diffusion on fractals, Physical Review A, volume 34, Number 3 (1986) p. 2502, Fig. 3.

Cf. A000129, A001045, A025192, A038208, A038221, A069720, A107920, A154692.
Cf. A215149.
Sums include: A008776 (row), A010673 (alternating sign row).
Columns k: A000051 (k=0).
Main diagonal: A059304.

A154690:= func< n,k | (2^(n-k)+2^k)*Binomial(n,k) >;
[A154690(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jan 18 2025

A154690 := proc(n,m) binomial(n,m)*(2^(n-m)+2^m) ; end proc: # R. J. Mathar, Jan 13 2011

T[n_, m_]:= (2^(n-m) + 2^m)*Binomial[n,m];
Table[T[n,m], {n,0,12}, {m,0,n}]//Flatten

from sage.all import *
def A154690(n,k): return (pow(2,n-k)+pow(2,k))*binomial(n,k)
print(flatten([[A154690(n,k) for k in range(n+1)] for n in range(13)])) # G. C. Greubel, Jan 18 2025

T(n, k) = (2^(n-k) + 2^k)*A007318(n, k).
Sum_{k=0..n} T(n, k) = A008776(n) = A025192(n+1).
From G. C. Greubel, Jan 18 2025: (Start)
T(n, n-k) = T(n, k) (symmetry).
T(n, 1) = n + A215149(n), n >= 1.
T(2*n-1, n) = 3*A069720(n).
Sum_{k=0..n} (-1)^k*T(n, k) = A010673(n+1).
Sum_{k=0..floor(n/2)} T(n-k, k) = A000129(n+1) + A001045(n+1).
Sum_{k=0..floor(n/2)} (-1)^k*T(n-k, k) = n+1 + A107920(n+1). (End)

A303872 Triangle read by rows: T(0,0) = 1; T(n,k) = -T(n-1,k) + 2 T(n-1,k-1) for k = 0,1,...,n; T(n,k)=0 for n or k < 0.

Original entry on oeis.org

1, -1, 2, 1, -4, 4, -1, 6, -12, 8, 1, -8, 24, -32, 16, -1, 10, -40, 80, -80, 32, 1, -12, 60, -160, 240, -192, 64, -1, 14, -84, 280, -560, 672, -448, 128, 1, -16, 112, -448, 1120, -1792, 1792, -1024, 256, -1, 18, -144, 672, -2016, 4032, -5376, 4608, -2304, 512

Offset: 0

Shara Lalo, May 25 2018

Row n gives coefficients in expansion of (-1+2x)^n. Row sums=1.

In the center-justified triangle, the numbers in skew diagonals pointing top-Left give the triangle in A133156 (coefficients of Chebyshev polynomials of the second kind), and the numbers in skew diagonals pointing top-right give the triangle in A305098. The coefficients in the expansion of 1/(1-x) are given by the sequence generated by the row sums. The generating function of the central terms is 1/sqrt(1+8x), signed version of A059304.

			Triangle begins:
   1;
  -1,   2;
   1,  -4,   4;
  -1,   6, -12,    8;
   1,  -8,  24,  -32,   16;
  -1,  10, -40,   80,  -80,    32;
   1, -12,  60, -160,  240,  -192,   64;
  -1,  14, -84,  280, -560,   672, -448,   128;
   1, -16, 112, -448, 1120, -1792, 1792, -1024, 256;

Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 389-391.

Shara Lalo, Skew diagonals in center-justified triangle
Paweł Lorek and Piotr Markowski, Absorption time and absorption probabilities for a family of multidimensional gambler models, arXiv:1812.00690 [math.PR], 2018.

Row sums give A000012.
Signed version of A013609 ((1+2*x)^n).
Cf. A033999 (column 0).

T[0, 0] = 1; T[n_, k_] := If[n < 0 || k < 0, 0, - T[n - 1, k] + 2 T[n - 1, k - 1]]; Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten
For[i = 0, i < 4, i++, Print[CoefficientList[Expand[(-1 +2 x)^i], x]]]

T(n, k) = if ((n<0) || (k<0), 0, if ((n==0) && (k==0), 1, -T(n-1, k) + 2*T(n-1, k-1)));
tabl(nn) = for (n=0, nn, for (k=0, n, print1(T(n,k), ", ")); print); \\ Michel Marcus, May 26 2018

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

T(n,k) = (-1)^(n+k)*2^k*binomial(n,k). - Stefano Spezia, Aug 08 2025

A069722 Number of rooted unicursal planar maps with n edges and exactly one vertex of valency 1 (unicursal means that exactly two vertices are of odd valency; there is an Eulerian path).

Original entry on oeis.org

0, 4, 24, 160, 1120, 8064, 59136, 439296, 3294720, 24893440, 189190144, 1444724736, 11076222976, 85201715200, 657270374400, 5082890895360, 39392404439040, 305870434467840, 2378992268083200, 18531097667174400, 144542561803960320, 1128808577897594880

Offset: 1

Valery A. Liskovets, Apr 07 2002

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Valery A. Liskovets and Timothy R. S. Walsh, Enumeration of Eulerian and unicursal planar maps, Discr. Math., 282 (2004), 209-221.

Cf. A059304, A069720, A069721, A069723, A089156.

[0] cat[2^(n-1)*Binomial(2*n-2, n-1): n in [2..20]]; // Vincenzo Librandi, Nov 17 2011

Z:=(1-sqrt(1-z))*8^n/sqrt(1-z): Zser:=series(Z, z=0, 32): seq(coeff(Zser, z, n), n=0..19); # Zerinvary Lajos, Jan 01 2007

Join[{0},Table[2^(n-1) Binomial[2n-2,n-1],{n,2,20}]] (* Harvey P. Dale, Nov 16 2011 *)

a(n) = 2^(n-1)*binomial(2n-2, n-1), n>1.
a(n) = 2*A069723(n), n>1.
G.f. for a(n)^2: 1/AGM(1, (1-64*x)^(1/2)). - Benoit Cloitre, Jan 01 2004
a(n) = A059304(n-1), n>1. [R. J. Mathar, Sep 29 2008]
a(n) ~ 2^(3*n-3)/sqrt(Pi*n). - Vaclav Kotesovec, Sep 28 2019
E.g.f.: x * (exp(4*x) * (BesselI(0,4*x) - BesselI(1,4*x)) - 1). - Ilya Gutkovskiy, Nov 03 2021
From Amiram Eldar, Jan 16 2024: (Start)
Sum_{n>=2} 1/a(n) = 1/7 + 8*arcsin(1/(2*sqrt(2)))/(7*sqrt(7)).
Sum_{n>=2} (-1)^n/a(n) = 1/9 + 4*log(2)/27. (End)

A119309 a(n) = binomial(2n,n) 6^n.

Original entry on oeis.org

1, 12, 216, 4320, 90720, 1959552, 43110144, 960740352, 21616657920, 489977579520, 11171488813056, 255928652808192, 5886359014588416, 135839054182809600, 3143703825373593600, 72933928748667371520

Offset: 0

Reinhard Zumkeller, May 14 2006

nonn

Number of lattice paths from (0,0) to (n,n) using three kinds of steps (1,0) and two kinds of steps (0,1). - Joerg Arndt, Jul 01 2011

Central terms of the triangles in A013620 and A038220.

			a(3) = binomial(2*3,3) * (6^3) = 20 * 216 = 4320. - _Indranil Ghosh_, Mar 03 2017

Indranil Ghosh, Table of n, a(n) for n = 0..400
Hacène Belbachir and Abdelghani Mehdaoui, Recurrence relation associated with the sums of square binomial coefficients, Quaestiones Mathematicae (2021) Vol. 44, Issue 5, 615-624.

Cf. A000984, A013620, A038220, A059304, A098658.

Table[Binomial[2n,n]*(6^n), {n, 0, 15}] (* Indranil Ghosh, Mar 03 2017 *)

/* same as in A092566 but use */
steps=[[1,0], [1,0], [1,0], [0,1], [0,1]]; /* note repeated entries */
/* Joerg Arndt, Jun 30 2011 */

a(n)=binomial(2*n,n)*6^n \\ Charles R Greathouse IV, Mar 03 2017

import math
f=math.factorial
def C(n,r): return f(n)//f(r)//f(n-r)
def A119309(n): return C(2*n,n)*(6**n) # Indranil Ghosh, Mar 03 2017

a(n) = 6^n * A000984(n).
G.f.: 1/sqrt(1-24*x). - Zerinvary Lajos, Dec 20 2008 [Corrected by Joerg Arndt, Jul 01 2011]
D-finite with recurrence: n*a(n) +12*(-2*n+1)*a(n-1)=0. - R. J. Mathar, Jan 20 2020
a(n) = 2^n*A098658(n) = 3^n*A059304(n). - R. J. Mathar, Jan 20 2020
From Amiram Eldar, Jul 21 2020: (Start)
Sum_{n>=0} 1/a(n) = 24/23 + 24*sqrt(23)*arcsin(1/sqrt(24))/529.
Sum_{n>=0} (-1)^n/a(n) = 24/25 - 24*arcsinh(1/sqrt(24))/125. (End)
E.g.f.: exp(12*x) * BesselI(0,12*x). - Ilya Gutkovskiy, Sep 14 2021

A248168 Expansion of g.f. 1/sqrt((1-3x)(1-11*x)).

Original entry on oeis.org

1, 7, 57, 511, 4849, 47607, 477609, 4862319, 50026977, 518839783, 5414767897, 56795795679, 598213529809, 6322787125207, 67026654455433, 712352213507151, 7587639773475777, 80977812878889927, 865716569022673401, 9269461606674304959, 99387936492243451569, 1066975862517563301303

Offset: 0

Paul D. Hanna, Oct 03 2014

nonn

			G.f.: A(x) = 1 + 7*x + 57*x^2 + 511*x^3 + 4849*x^4 + 47607*x^5 +...
where A(x)^2 = 1/((1-3*x)*(1-11*x)):
A(x)^2 = 1 + 14*x + 163*x^2 + 1820*x^3 + 20101*x^4 + 221354*x^5 +...

Seiichi Manyama, Table of n, a(n) for n = 0..961
Hacène Belbachir, Abdelghani Mehdaoui, and László Szalay, Diagonal Sums in the Pascal Pyramid, II: Applications, J. Int. Seq., Vol. 22 (2019), Article 19.3.5.
Paveł Szabłowski, Beta distributions whose moment sequences are related to integer sequences listed in the OEIS, Contrib. Disc. Math. (2024) Vol. 19, No. 4, 85-109. See p. 97.

Cf. A248167, A084771.

[n le 2 select 7^(n-1) else (7*(2*n-3)*Self(n-1) - 33*(n-2)*Self(n-2))/(n-1) : n in [1..40]]; // G. C. Greubel, May 31 2025

CoefficientList[Series[1/Sqrt[(1-3*x)*(1-11*x)], {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 03 2014 *)

{a(n)=polcoeff( 1 / sqrt((1-3*x)*(1-11*x) +x*O(x^n)), n) }
for(n=0, 25, print1(a(n), ", "))

{a(n)=polcoeff( (1 + 7*x + 4*x^2 +x*O(x^n))^n, n) }
for(n=0, 25, print1(a(n), ", "))

{a(n)=sum(k=0,n, 3^(n-k)*2^k*binomial(n,k)*binomial(2*k,k))}
for(n=0, 25, print1(a(n), ", "))

@CachedFunction
def A248168(n):
     if (n<2): return 7^n
     else: return (7*(2*n-1)*A248168(n-1) - 33*(n-1)*A248168(n-2))//n
print([A248168(n) for n in range(41)]) # G. C. Greubel, May 31 2025

a(n) equals the central coefficient in (1 + 7*x + 4*x^2)^n, n>=0.
a(n) = Sum_{k=0..n} 3^(n-k) * 2^k * C(n,k) * C(2*k,k).
a(n) = Sum_{k=0..n} 11^(n-k) * (-2)^k * C(n,k) * C(2*k,k). - Paul D. Hanna, Apr 20 2019
a(n)^2 = A248167(n), which gives the coefficients in 1 / AGM(1-3*11*x, sqrt((1-3^2*x)*(1-11^2*x))).
Equals the binomial transform of 2^n*A026375(n).
Equals the second binomial transform of A084771.
Equals the third binomial transform of A059304(n) = 2^n*(2*n)!/(n!)^2.
a(n) ~ 11^(n+1/2)/(2*sqrt(2*Pi*n)). - Vaclav Kotesovec, Oct 03 2014
D-finite with recurrence: n*a(n) +7*(-2*n+1)*a(n-1) +33*(n-1)*a(n-2)=0. [Belbachir]
a(n) = (1/4)^n * Sum_{k=0..n} 3^k * 11^(n-k) * binomial(2*k,k) * binomial(2*(n-k),n-k). - Seiichi Manyama, Aug 18 2025

A307910 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of 1/sqrt(1 - 2kx + k(k-4)x^2).

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 3, 0, 1, 3, 8, 7, 0, 1, 4, 15, 32, 19, 0, 1, 5, 24, 81, 136, 51, 0, 1, 6, 35, 160, 459, 592, 141, 0, 1, 7, 48, 275, 1120, 2673, 2624, 393, 0, 1, 8, 63, 432, 2275, 8064, 15849, 11776, 1107, 0, 1, 9, 80, 637, 4104, 19375, 59136, 95175, 53344, 3139, 0

Offset: 0

Seiichi Manyama, May 05 2019

			Square array begins:
   1,   1,     1,     1,      1,       1,       1, ...
   0,   1,     2,     3,      4,       5,       6, ...
   0,   3,     8,    15,     24,      35,      48, ...
   0,   7,    32,    81,    160,     275,     432, ...
   0,  19,   136,   459,   1120,    2275,    4104, ...
   0,  51,   592,  2673,   8064,   19375,   40176, ...
   0, 141,  2624, 15849,  59136,  168125,  400896, ...
   0, 393, 11776, 95175, 439296, 1478125, 4053888, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened

Columns k=0..4 give A000007, A002426, A006139, A122868, A059304.
Main diagonal gives A092366.
Cf. A107267, A292627, A307819, A307847, A307855, A307883.

A[n_, k_] := k^n Hypergeometric2F1[(1-n)/2, -n/2, 1, 4/k]; A[0, ] = 1; A[, 0] = 0; Table[A[n-k, k], {n, 0, 10}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, May 07 2019 *)

A(n,k) is the coefficient of x^n in the expansion of (1 + k*x + k*x^2)^n.
A(n,k) = Sum_{j=0..floor(n/2)} k^(n-j) * binomial(n,j) * binomial(n-j,j) = Sum_{j=0..floor(n/2)} k^(n-j) * binomial(n,2*j) * binomial(2*j,j).
n * A(n,k) = k * (2*n-1) * A(n-1,k) - k * (k-4) * (n-1) * A(n-2,k).

A098660 E.g.f. BesselI(0,2sqrt(2)x) + BesselI(1,2sqrt(2)x)/sqrt(2).

Original entry on oeis.org

1, 1, 4, 6, 24, 40, 160, 280, 1120, 2016, 8064, 14784, 59136, 109824, 439296, 823680, 3294720, 6223360, 24893440, 47297536, 189190144, 361181184, 1444724736, 2769055744, 11076222976, 21300428800, 85201715200, 164317593600

Offset: 0

Paul Barry, Sep 20 2004

Third binomial transform (shifted right) is A047781. Hankel transform is A166232(n+1).

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A059304, A069720 (bisections).

m:=30; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!((1+4*x-Sqrt(1-8*x^2))/(4*x*Sqrt(1-8*x^2)))); // G. C. Greubel, Aug 17 2018

nmax = 30; CoefficientList[Series[BesselI[0, 2*Sqrt[2]*x] + BesselI[1, 2*Sqrt[2]*x]/Sqrt[2], {x, 0, nmax}], x] * Range[0, nmax]! (* Vaclav Kotesovec, Nov 13 2017 *)

x='x+O('x^30); Vec((1+4*x-sqrt(1-8*x^2))/(4*x*sqrt(1-8*x^2))) \\ G. C. Greubel, Aug 17 2018

G.f.: 1/sqrt(1-8*x^2)+(1-sqrt(1-8*x^2))/(4*x*sqrt(1-8*x^2)) = (1+4*x-sqrt(1-8*x^2))/(4*x*sqrt(1-8*x^2)).
a(n) = binomial(n, floor(n/2))2^floor(n/2).
a(n+1) = (1/Pi)*int(x^n*(x+4)/sqrt(8-x^2),x,-2*sqrt(2),2*sqrt(2)) if n is odd [corrected by Vaclav Kotesovec, Nov 13 2017].
Conjecture: (n+1)*a(n) +(n-1)*a(n-1) -n*a(n-2) +(2-n)*a(n-3) = 0. - R. J. Mathar, Nov 15 2011

A135838 Triangle read by rows: T(n,k) = 2^floor(n/2)*binomial(n-1,k-1).

Original entry on oeis.org

1, 2, 2, 2, 4, 2, 4, 12, 12, 4, 4, 16, 24, 16, 4, 8, 40, 80, 80, 40, 8, 8, 48, 120, 160, 120, 48, 8, 16, 112, 336, 560, 560, 336, 112, 16, 16, 128, 448, 896, 1120, 896, 448, 128, 16, 32, 288, 1152, 2688, 4032, 4032, 2688, 1152, 288, 32

Offset: 1

Gary W. Adamson, Dec 01 2007

			First few rows of the triangle are:
  1;
  2,  2;
  2,  4,  2;
  4, 12, 12,  4;
  4, 16, 24, 16,  4;
  8, 40, 80, 80, 40, 8;
  ...

Gheorghe Coserea, Rows n = 1..100, flattened

Cf. A016116, A059304, A069720, A093968, A094015 (row sums), A135837.

A135838 := proc(n,k)
    2^floor(n/2)*binomial(n-1,k-1) ;
end proc:
seq(seq( A135838(n,k),k=1..n),n=1..10) ; # R. J. Mathar, Aug 15 2022

T[n_, k_]:= 2^Floor[n/2]*Binomial[n-1, k-1];
Table[T[n, k], {n,12}, {k,n}] //Flatten (* G. C. Greubel, Feb 07 2022 *)

A(n,k) = 2^(n\2)*binomial(n-1,k-1);
concat(vector(10, n, vector(n, k, A(n,k))))  \\ Gheorghe Coserea, May 18 2016

flatten([[2^(n//2)*binomial(n-1, k-1) for k in (1..n)] for n in (1..12)]) # G. C. Greubel, Feb 07 2022

M * Pascal's triangle as infinite lower triangular matrices, where M = a triangle with (1, 2, 2, 4, 4, 8, 8, 16, 16, ...) in the main diagonal and the rest zeros.
Sum_{k=1..n} T(n, k) = A094015(n-1).
From G. C. Greubel, Feb 07 2022: (Start)
T(n, n-k) = T(n, k).
T(n, 1) = A016116(n).
T(n, 2) = 2*A093968(n-1).
T(2*n-1, n) = A059304(n-1).
T(2*n, n) = 2*A069720(n). (End)

A158687 Riordan array (1/(1-x),x(1+x)^2/(1-x)).

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 8, 7, 1, 1, 12, 24, 10, 1, 1, 16, 56, 49, 13, 1, 1, 20, 104, 160, 83, 16, 1, 1, 24, 168, 400, 351, 126, 19, 1, 1, 28, 248, 832, 1120, 656, 178, 22, 1, 1, 32, 344, 1520, 2912, 2561, 1102, 239, 25, 1

Offset: 0

Paul Barry, Mar 24 2009

Row sums are A077936. Diagonal sums are A129847. Central terms are A059304.

Inverse of alternating signed version is A100326.

			Number triangle begins
1,
1, 1,
1, 4, 1,
1, 8, 7, 1,
1, 12, 24, 10, 1,
1, 16, 56, 49, 13, 1,
1, 20, 104, 160, 83, 16, 1

Cf. A059304, A077936, A100326, A129847.

Number triangle T(n,k) = Sum_{j=0..n-k} C(n-j,k)*C(2k,j).
T(n,k) = T(n-1,k) + T(n-1,k-1) + 2*T(n-2,k-1) + T(n-3,k-1), T(0,0) = T(1,0) = T(1,1) = 1, T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Nov 11 2013
G.f.: 1/(1-y-x*(1+y)^2). - Vladimir Kruchinin, Apr 21 2015

cp's OEIS Frontend

A098658 a(n) = 3^n*(2*n)!/(n!)^2.

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A154690 Triangle read by rows: T(n, k) = (2^(n-k) + 2^k)*binomial(n,k), 0 <= k <= n.

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A303872 Triangle read by rows: T(0,0) = 1; T(n,k) = -T(n-1,k) + 2 T(n-1,k-1) for k = 0,1,...,n; T(n,k)=0 for n or k < 0.

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A069722 Number of rooted unicursal planar maps with n edges and exactly one vertex of valency 1 (unicursal means that exactly two vertices are of odd valency; there is an Eulerian path).

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A119309 a(n) = binomial(2*n,n) * 6^n.

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A248168 Expansion of g.f. 1/sqrt((1-3*x)*(1-11*x)).

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A098658 a(n) = 3^n(2n)!/(n!)^2.

A119309 a(n) = binomial(2n,n) 6^n.

A248168 Expansion of g.f. 1/sqrt((1-3x)(1-11*x)).

A307910 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of 1/sqrt(1 - 2kx + k(k-4)x^2).

A098660 E.g.f. BesselI(0,2sqrt(2)x) + BesselI(1,2sqrt(2)x)/sqrt(2).