A054142 -id:A054142 - cp's OEIS frontend

A160657 a(n) is the period of a 2 X 4n rectangular oscillator in the 2 X 2 (B36/S125) Life-like cellular automaton.

2, 6, 14, 14, 62, 126, 30, 30, 1022, 126, 4094, 2046, 1022, 32766, 62, 62, 8190, 174762, 8190, 2046, 254, 8190, 16777214, 4194302, 510, 134217726, 2097150, 1022, 1073741822, 2147483646, 126, 126, 17179869182, 8388606, 68719476734, 1022, 2097150, 2147483646

Offset: 1

Nathaniel Johnston, May 22 2009

nonn

These oscillators work and have the same period in any rule from B3/S5 to B3678/S012567.
The Nathaniel Johnston rectangular oscillator link points to Sierpinski's gasket (Pascal's triangle mod 2) as a source for the chaotic terms of A003558. This is consistent with the comment of [Sep 21 2011, A003558] showing an alternative trigonometric connection to A054142, since the latter row terms are found as alternate ascending diagonals in Pascal's triangle. - Gary W. Adamson, Sep 21 2011
From Charlie Neder, Jan 11 2019: (Start)
a(n) = A268754(2n).
Proof: Decompose the phases of the oscillators into rectangles, as in the linked paper. Each of these rectangles has a corner on the exterior of the bounding diamond of the oscillator which determines the rectangle. As shown in the paper, these corners behave as Rule 90 on a width-n strip, which is exactly what A268754 emulates. Since the initial 2 X 4n block used in this sequence corresponds to the one-cell "seed" used in A268754, the resulting patterns must have the same period. (End)

			a(2) = 6 because a 2 X 8 box has period 6 in this rule.

Adam P. Goucher, Table of n, a(n) for n = 1..160
Nathaniel Johnston, Rectangular Oscillators in the 2*2 (B36/S125) Cellular Automaton, 2009.
Nathaniel Johnston, The B36/S125 "2×2" Life-Like Cellular Automaton, arXiv:1203.1644 [nlin.CG], 2012; also in Game of Life Cellular Automata, A. Adamatzky (ed.), Springer-UK, 2010, pages 99-114.
LifeWiki, 2x2

Cf. A003558, A054142, A268754, A298819.

g = Function[{sq, p}, Module[{l = Length[sq]},
Do[If[sq[[i]] == sq[[j]], Return[p^(j - 1) - p^(i - 1)]],
{j, 2, l}, {i, 1, j - 1}]]];
MPM = Algebra`MatrixPowerMod;
EventualPeriod = Function[{m, v, p},
Module[{n = Length[m], w, sq, k, primes},
sq = NestList[(MPM[#, p, p]) &, m, n];
w = Mod[Last[sq].v, p];
sq = Map[(Mod[#.w, p]) &, sq];
k = g[sq, p];
If[k == Null, k = p^n Apply[LCM, Table[p^r - 1, {r, 1, n}]]];
primes = Map[First, FactorInteger[k]];
primes = Select[primes, (# > 1) &];
While[Length[primes] > 0,
primes = Select[primes, (Mod[k, #] == 0) &];
primes = Select[primes, (Mod[MPM[m, k/#, p].w, p] == w) &];
k = k/Fold[Times, 1, primes];
]; k ]];
mat = Function[{n}, Table[Boole[Abs[i - j] == 1], {i, 1, n}, {j, 1, n}]];
vec = Function[{n}, Table[Boole[i == 1], {i, 1, n}]];
Table[EventualPeriod[mat[2 n], vec[2 n], 2], {n, 1, 100}]
(* Adam P. Goucher, Jan 13 2019 *)

a(n) divides 2^(A003558(n) + 1) - 2 for n >= 1. [Corrected by Charlie Neder, Jan 11 2019]

a(18) corrected by Charlie Neder, Jan 11 2019

A375450 Expansion of g.f. A(x) satisfying 0 = Sum_{k=0..n} (-1)^k * binomial(2n-k,k) ([x^k] A(x)^n) for n >= 1.

Original entry on oeis.org

1, 1, 2, 9, 75, 960, 17056, 398023, 11785624, 431999096, 19225521917, 1022238356603, 64053139874787, 4673871388801269, 393051019651091208, 37747728969729478069, 4106730533743416782815, 502513668471090178354603, 68716238916399477889072499, 10440633447359638146139853297

Offset: 0

Paul D. Hanna, Sep 11 2024

nonn

Note that 0 = Sum_{k=0..n} (-1)^k * binomial(n+k,2*k) * ([x^k] C(x)^n) for n >= 1 is satisfied by the Catalan function C(x) = 1 + x*C(x)^2 (A000108), where coefficient [x^k] C(x)^n = binomial(n+2*k-1,k)*n/(n+k).

			G.f.: A(x) = 1 + x + 2*x^2 + 9*x^3 + 75*x^4 + 960*x^5 + 17056*x^6 + 398023*x^7 + 11785624*x^8 + 431999096*x^9 + ...
RELATED TABLES.
The table of coefficients of x^k in A(x)^n begins:
  n=1: [1, 1,  2,   9,  75,  960,  17056, ...];
  n=2: [1, 2,  5,  22, 172, 2106,  36413, ...];
  n=3: [1, 3,  9,  40, 297, 3477,  58412, ...];
  n=4: [1, 4, 14,  64, 457, 5120,  83454, ...];
  n=5: [1, 5, 20,  95, 660, 7091, 112010, ...];
  n=6: [1, 6, 27, 134, 915, 9456, 144632, ...];
  ...
from which we may illustrate the defining property given by
0 = Sum_{k=0..n} (-1)^k * binomial(2*n-k,k) * ([x^k] A(x)^n).
Using the coefficients in the table above, we see that
  n=1: 0 = 1*1 - 1*1;
  n=2: 0 = 1*1 - 3*2 + 1*5;
  n=3: 0 = 1*1 - 5*3 + 6*9 - 1*40;
  n=4: 0 = 1*1 - 7*4 + 15*14 - 10*64 + 1*457;
  n=5: 0 = 1*1 - 9*5 + 28*20 - 35*95 + 15*660 - 1*7091;
  n=6: 0 = 1*1 - 11*6 + 45*27 - 84*134 + 70*915 - 21*9456 + 1*144632;
  ...
The triangle A054142(n,k) = binomial(2*n-k,k) begins:
  n=0: 1;
  n=1: 1,  1;
  n=2: 1,  3,  1;
  n=3: 1,  5,  6,  1;
  n=4: 1,  7, 15, 10,  1;
  n=5: 1,  9, 28, 35, 15,  1;
  n=6: 1, 11, 45, 84, 70, 21, 1;
  ...

Paul D. Hanna, Table of n, a(n) for n = 0..200

Cf. A054142, A375451.

{a(n) = my(A=[1]); for(i=1, n, A=concat(A, 0);
A[#A] = sum(k=0, #A-1, (-1)^(#A-k) * binomial(2*(#A-1)-k, 1*k) * polcoef(Ser(A)^(#A-1), k) )/(#A-1) ); A[n+1]}
for(n=0, 20, print1(a(n), ", "))

a(n) ~ c * 2^(2*n) * n^(2*n + 3/2) / (exp(2*n) * Pi^(2*n)), where c = 35.20725926431251936515... - Vaclav Kotesovec, Sep 11 2024

A027989 a(n) = self-convolution of row n of array T given by A027926.

Original entry on oeis.org

1, 3, 10, 33, 105, 324, 977, 2895, 8462, 24465, 70101, 199368, 563425, 1583643, 4430290, 12342849, 34262337, 94800780, 261545777, 719697255, 1975722326, 5412138033, 14796520365, 40380240528, 110016825025, 299285288499, 813011578522, 2205652007265, 5976479585817

Offset: 0

Clark Kimberling

a(n) is the number of all columns in stack polyominoes of perimeter 2n+4. - Emanuele Munarini, Apr 07 2011

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
Index entries for linear recurrences with constant coefficients, signature (6,-11,6,-1).

Cf. A027926, A054142, A172991, A188648.

Table[((5+4n)Fibonacci[1+2n]-(1+2n)Fibonacci[2n])/5,{n,0,28}] (* Emanuele Munarini, Apr 07 2011 *)

makelist(((5+4*n)*fib(1+2*n)-(1+2*n)*fib(2*n))/5,n,0,20); /* Emanuele Munarini, Apr 07 2011 */

Vec((1-3*x+3*x^2)/(1-3*x+x^2)^2+O(x^66)) /* Joerg Arndt, Apr 08 2011 */

a(n) = (2/5)*(n + 1)*F(2*n+3) + (1/5)*F(2*n+2) - (4/5)*(n + 1)*F(2*n), where F(n) = A000045(n). - Ralf Stephan, May 13 2004
From Emanuele Munarini, Apr 07 2011: (Start)
a(n) = ((4*n + 5)*F(2*n+1) - (2*n + 1)*F(2*n))/5, where F(n) = A000045(n).
a(n) = Sum_{k=0..n} binomial(2*n-k, k)*(k + 1).
G.f.: (1 - 3*x + 3*x^2)/(1 - 3*x + x^2)^2.
a(n) = 6*a(n-1) - 11*a(n-2) + 6*a(n-3) - a(n-4). (End)

A098435 Triangle of Salie numbers T(n,k) for negative n,k, n < k.

Original entry on oeis.org

1, -1, 1, 2, -3, 1, -8, 13, -6, 1, 56, -92, 45, -10, 1, -608, 1000, -493, 115, -15, 1, 9440, -15528, 7662, -1799, 245, -21, 1, -198272, 326144, -160944, 37817, -5180, 462, -28, 1, 5410688, -8900224, 4392080, -1032088, 141465, -12684, 798, -36, 1

Offset: 1

Ralf Stephan, Sep 08 2004

Inverse matrix of A054142. - Paul Barry, Jan 21 2005

Essentially the same as the triangle giving by [0,-1,-1,-4,-4,-9,-9,-16,-16,-25,...] DELTA[1,0,1,0,1,0,1,0,1,0,...] = 1; 0,1; 0,-1,1; 0,2,-3,1; 0,-8,13,-6,1; 0,56,-92,45,-10,1; ... where DELTA is the operator defined in A084938. - Philippe Deléham, Aug 30 2006

			   1;
  -1,   1;
   2,  -3,   1;
  -8,  13,  -6,   1;
  56, -92,  45, -10,  1;

D. Dumont and J. Zeng, Polynomes d'Euler et les fractions continues de Stieltjes-Rogers, Ramanujan J. 2 (1998) 3, 387-410.

T(-1, k) = (-1)^k*A005439(k-1). Row sums are zero.

rows = 9; A054142 = Table[ PadRight[ Table[ Binomial[2*n-k, k], {k, 0, n}], rows], {n, 0, rows-1}]; inv = Inverse[A054142]; Table[ Take[inv[[n]], n], {n, 1, rows}] // Flatten (* Jean-François Alcover, Oct 02 2013, after Paul Barry *)

See A065547 for formulas.

A109223 Number triangle related to the Fibonacci polynomials.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 3, 5, 1, 1, 1, 6, 5, 7, 1, 1, 1, 6, 15, 7, 9, 1, 1, 1, 10, 15, 28, 9, 11, 1, 1, 1, 10, 35, 28, 45, 11, 13, 1, 1, 1, 15, 35, 84, 45, 66, 13, 15, 1, 1, 1, 15, 70, 84, 165, 66, 91, 15, 17, 1, 1, 1, 21, 70, 210, 165, 286, 91, 120, 17, 19, 1, 1, 1, 21, 126, 210

Offset: 0

Paul Barry, Jun 22 2005

Riordan array (1/(1-x), x/(1-x^2)^2). Row-reversal of number triangle A109221. Diagonals form a repeated version of A054142. Row sums are A109222. Diagonal sums are A094967.

			Rows begin
  1;
  1,  1;
  1,  1,  1;
  1,  3,  1,  1;
  1,  3,  5,  1,  1;
  1,  6,  5,  7,  1,  1;
  1,  6, 15,  7,  9,  1,  1;

A109223 := proc(n,k): binomial(floor((n+k)/2)+k, 2*k) end: seq(seq(A109223(n,k), k=0..n), n=0..11); # Johannes W. Meijer, Aug 14 2011

T(n,k) = binomial(floor((n+k)/2)+k, 2*k)

T(n,k) = A065941(n+k,n-k). - Johannes W. Meijer, Aug 14 2011

A172991 Triangle of binomial sums read by rows: T(n,k) = sum(C(2n-2k-i,i) * C(2k-i,i), i=0..min(k,n-k)).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 4, 4, 1, 1, 6, 11, 6, 1, 1, 8, 22, 22, 8, 1, 1, 10, 37, 63, 37, 10, 1, 1, 12, 56, 136, 136, 56, 12, 1, 1, 14, 79, 249, 376, 249, 79, 14, 1, 1, 16, 106, 410, 849, 849, 410, 106, 16, 1, 1, 18, 137, 627, 1663, 2317, 1663, 627, 137, 18, 1, 1, 20, 172, 908, 2942, 5371, 5371, 2942, 908, 172, 20, 1, 1, 22, 211, 1261, 4826, 11017, 14545, 11017, 4826, 1261, 211, 22, 1

Offset: 0

Emanuele Munarini, Apr 07 2011

The matrix inverse starts
1;
-1,1;
1,-2,1;
-1,4,-4,1;
0,-8,13,-6,1;
7,12,-38,26,-8,1;
-35,-12,114,-101,43,-10,1; - R. J. Mathar, Mar 22 2013

			G.f. =
1 +
(y + 1)*x +
(y^2 + 2*y + 1)*x^2 +
(y^3 + 4*y^2 + 4*y + 1)*x^3 +
(y^4 + 6*y^3 + 11*y^2 + 6*y + 1)*x^4 + ...
Triangle begins:
1,
1,  1,
1,  2,  1,
1,  4,  4,   1,
1,  6, 11,   6,   1,
1,  8, 22,  22,   8,   1,
1, 10, 37,  63,  37,  10,  1,
1, 12, 56, 136, 136,  56, 12,  1,
1, 14, 79, 249, 376, 249, 79, 14,  1

Cf. A188648, A054142, A027989.

Flatten[Table[Sum[Binomial[2n-2k-i,i]Binomial[2k-i,i],{i,0,Min[k,n-k]}],{n,0,12},{k,0,n}]]

create_list(sum(binomial(2*n-2*k-i,i)*binomial(2*k-i,i),i,0,min(k,n-k)),n,0,10,k,0,n);

G.f.: (1 -x -x*y -2*x^2*y +x^3*y +x^3*y^2 +4*x^4*y^2 -x^6*y^3) / (1 -2*x +x^2 -2*x*y+2*x^3*y +x^2*y^2 +2*x^3*y^2 +3*x^4*y^2 -2*x^5*y^2 -2*x^5*y^3 -6*x^6*y^3 +x^8*y^4).

Central coefficients T(2n,n) = A188648.

A192398 a(n) = n^4 + 3n^3 - 3n.

Original entry on oeis.org

1, 34, 153, 436, 985, 1926, 3409, 5608, 8721, 12970, 18601, 25884, 35113, 46606, 60705, 77776, 98209, 122418, 150841, 183940, 222201, 266134, 316273, 373176, 437425, 509626, 590409, 680428, 780361, 890910, 1012801, 1146784, 1293633, 1454146, 1629145, 1819476

Offset: 1

Gary W. Adamson, Jun 30 2011

Related to the 9-gon (nonagon). Following Steinbach's strategy re: "Diagonal Product Formulas" and applied to the 9-gon (nonagon), we extract the constants (a, b, c, d) as e-vals of the 4 X 4 tridiagaonal matrix with (1's in the super and subdiagonals), (1,2,2,2), and the rest zeros. The charpoly of this matrix is row 4 of A054142, a Morgan-Voyce polynomial: x^4 - 7*x^3 + 15*x^2 + 10*x - 1 = 0. Following Steinbach's procedure, let the matrix = M; then find the first four rows of M^n * [1,0,0,0,...] getting (1; 1,1; 2,3,1; 5,9,5,1). Using the SIMULT operation, we equate each of these rows to successive powers of the constant c (largest e-val of matrix M), 3.5320888...as follows: SIMULT: [1,0,0,0] = 1; [1,1,0,0] = c; [2,3,1,0] = c^2; [5,9,5,1] = c^3. Solving, we obtain the four distinct diagonals of the 9-gon (nonagon) with edge = 1: (1, 2.5320888,..., 2.879385,..., and 1.879385,...).
The sequence is column 3 in the array of A162997.
Analogous sequences using the matrix M^k generator -M^2 generates A028387: (1, 5, 11, 19, 29, 41,...); M^3 generates A123972: (1, 13, 41, 91, 169,...).

			a(5) = 5^4 + 3*5^3 - 3*5 = (625 + 375 - 15) = 985.
a(4) = 436 = (1, 3, 3, 1) dot (1, 33, 86, 78) = (1 + 99 + 258 + 78) = 436.
a(7) = 3409 = lower right term in M^4, M = {{1,6}{1,7}}.
a(4) = 436 = (3 + a) * (3 + b) * (3 + c) * (3 + d), = (5.347296...) * (3.120614...) * (4) * (6.532088...) = 436.

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Peter Steinbach, Golden Fields: A Case for the Heptagon, Mathematics Magazine, Vol. 70, No. 1, Feb. 1997.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A028387, A054142, A123972, A162997.

[n^4 +3*n^3 -3*n: n in [1..45]]; // Vincenzo Librandi, Nov 25 2011

A192398:=n->n^4+3*n^3-3*n: seq(A192398(n), n=1..40); # Wesley Ivan Hurt, Sep 12 2014

LinearRecurrence[{5,-10,10,-5,1},{1,34,153,436,985},50] (* Vincenzo Librandi, Nov 25 2011 *)
Table[n^4+3n^3-3n,{n,40}] (* Harvey P. Dale, Feb 21 2023 *)

a(n) = n^4 +3*n^3 -3*n \\ Charles R Greathouse IV, Jun 30 2011

[n*(n^3+3*n^2-3) for n in range(1,51)] # G. C. Greubel, Jul 11 2023

G.f.: (1 +29*x -7*x^2 +x^3) / (1-x)^5. - R. J. Mathar, Jul 08 2011
a(n) = binomial transform of [1, 33, 86, 78, 24, 0, 0, 0,...].
a(n) = lower right term in the 2 X 2 matrix M^4, M = {{1,n-1}, {1,n}}.
a(n) = ((n-1) + a) * ((n-1) + b) * ((n-1) + c) * ((n-1) + d), where a, b, c, d, = {k=1,2,3,4} 4*cos^2 (2*Pi*k)/9.
E.g.f.: x*(1 + 16*x + 9*x^2 + x^3)*exp(x). - G. C. Greubel, Jul 11 2023

A121529 Triangle read by rows: T(n,k) is the number of nondecreasing Dyck paths of semilength n and having k double rises at an odd level (n >= 1, k >= 0).

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 10, 2, 1, 19, 14, 1, 33, 50, 5, 1, 55, 132, 45, 1, 90, 301, 205, 13, 1, 146, 631, 680, 139, 1, 236, 1255, 1892, 763, 34, 1, 381, 2409, 4717, 3019, 419, 1, 615, 4509, 10920, 9846, 2677, 89, 1, 993, 8283, 23974, 28292, 12241, 1241, 1, 1604, 14998

Offset: 1

Emeric Deutsch, Aug 05 2006

A nondecreasing Dyck path is a Dyck path for which the sequence of the altitudes of the valleys is nondecreasing.
Row n contains 1+floor(n/2) terms.
Row sums are the odd-indexed Fibonacci numbers (A001519).
T(2n,n) = Fibonacci(2n-1) (A001519).
Sum_{k>=0} k*T(n,k) = A121530(n).

			T(4,2)=2 because we have U/UDDU/UDD and U/UU/UDDDD, where U=(1,1) and D=(1,-1) (the double rises at an odd level are indicated by a /).
Triangle starts:
  1;
  1,  1;
  1,  4;
  1, 10,  2;
  1, 19, 14;
  1, 33, 50,  5;

E. Barcucci, A. Del Lungo, S. Fezzi and R. Pinzani, Nondecreasing Dyck paths and q-Fibonacci numbers, Discrete Math., 170, 1997, 211-217.

Cf. A001519, A121530, A121531, A054142.

G:=z*(1-t*z^2)*(1-z+t*z-z^2-t*z^2-t^2*z^3)/(1-z-t*z^2)/(1-z-z^2-3*t*z^2-t*z^3+t^2*z^4): Gser:=simplify(series(G,z=0,18)): for n from 1 to 15 do P[n]:=sort(coeff(Gser,z,n)) od: for n from 1 to 15 do seq(coeff(P[n],t,j),j=0..floor(n/2)) od; # yields sequence in triangular form

G.f.: G(t,z) = z*(1-tz^2)*(1 - z + tz - z^2 - tz^2 - t^2*z^3)/((1 - z - tz^2)*(1 - z - z^2 - 3tz^2 - tz^3 + t^2*z^4)).

A144252 Eigentriangle, row sums = A144251 shifted, right border = A144251.

Original entry on oeis.org

1, 1, 1, 1, 3, 2, 1, 5, 12, 6, 1, 7, 30, 60, 24, 1, 9, 56, 210, 360, 122, 1, 11, 90, 504, 1680, 2562, 758, 1, 13, 132, 990, 5040, 15372, 21224, 5606, 1, 15, 182, 1716, 11880, 36364, 159180, 201816, 47378, 1, 17, 240, 2730, 24024, 157014, 700392, 1849980, 2177010, 479532

Offset: 0

Gary W. Adamson, Sep 16 2008

Right border = A144251: (1, 1, 2, 6, 24, 122, 758,...) with row sums = the same sequence shifted. Sum of n-th row terms = rightmost term of next row.

			First few rows of the triangle =
  1;
  1, 1;
  1, 3, 2;
  1, 5, 12, 6;
  1, 7, 30, 60, 24;
  1, 9, 56, 210, 360, 122;
  1, 11, 90, 504, 1680, 2562, 758;
  1, 13, 132, 990, 5040, 15372, 21224, 5606;
  ...
The triangle is generated from A054142 and its own eigensequence, A144251.
Triangle A054142 =
  1;
  1, 1;
  1, 3, 1;
  1, 5, 6, 1;
  1, 7, 15, 10, 1;
  ...
The eigensequence of A054142 = A144251: (1, 1, 2, 6, 24, 122, 758, 5606,...);
Example: row 3 of A144252 = (1, 5, 12, 6) = termwise products of (1, 5, 6, 1) and (1, 1, 2, 6) = (1*1, 5*1, 6*2, 1*6).

Cf. A144251, A054142, A125273, A085478

A054142(n, k) = binomial(2*n-k, k);
V144251(nn) = my(v=vector(nn)); v[1] = 1; for (n=2, nn, v[n] = sum(k=0, n-1, A054142(n-2,k)*v[k+1]);); v;
row(n) = my(v=V144251(n+1)); vector(n+1, k, A054142(n,k-1) * v[k]); \\ Michel Marcus, Jan 18 2025

Eigentriangle by rows, T(n,k) = A054142(n,k) * A144251(k); were A144251 = the eigensequence of triangle A054142.

More terms from Michel Marcus, Jan 18 2025

A171822 Triangle T(n,k) = binomial(2n-k, k)binomial(n+k, 2*k), read by rows.

Original entry on oeis.org

1, 1, 1, 1, 9, 1, 1, 30, 30, 1, 1, 70, 225, 70, 1, 1, 135, 980, 980, 135, 1, 1, 231, 3150, 7056, 3150, 231, 1, 1, 364, 8316, 34650, 34650, 8316, 364, 1, 1, 540, 19110, 132132, 245025, 132132, 19110, 540, 1, 1, 765, 39600, 420420, 1288287, 1288287, 420420, 39600, 765, 1

Offset: 0

Roger L. Bagula, Dec 19 2009

			Triangle begins as:
  1;
  1,    1;
  1,    9,     1;
  1,   30,    30,       1;
  1,   70,   225,      70,       1;
  1,  135,   980,     980,     135,       1;
  1,  231,  3150,    7056,    3150,     231,       1;
  1,  364,  8316,   34650,   34650,    8316,     364,       1;
  1,  540, 19110,  132132,  245025,  132132,   19110,     540,     1;
  1,  765, 39600,  420420, 1288287, 1288287,  420420,   39600,   765,    1;
  1, 1045, 75735, 1166880, 5465460, 9018009, 5465460, 1166880, 75735, 1045, 1;

G. C. Greubel, Rows n = 0..100 of the triangle, flattened

Cf. A054142, A085478, A183160.

[Binomial(2*n-k, k)*Binomial(n+k, 2*k): k in [0..n], n in [0..10]]; // G. C. Greubel, Feb 22 2021

Table[Binomial[2*n-k, k]*Binomial[n+k, 2*k], {n,0,10}, {k,0,n}]//Flatten

flatten([[binomial(2*n-k, k)*binomial(n+k, 2*k) for k in (0..n)] for n in (0..10)]) # G. C. Greubel, Feb 22 2021

T(n, k) = binomial(2*n-k, k)*binomial(n+k, 2*k) = A054142(n, k)*A085478(n, k).

Sum_{k=0..n} T(n, k) = Hypergeometric 4F3([-n, -n, 1/2 -n, n+1], [1/2, 1, -2*n], 1) = A183160(n). - G. C. Greubel, Feb 22 2021

Edited by G. C. Greubel, Feb 22 2021

cp's OEIS Frontend

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