A091527 -id:A091527 - cp's OEIS frontend

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

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

A244039 a(n) = 2^(2n-1) ( binomial(3n/2,n) + binomial((3n-1)/2,n) ).

Original entry on oeis.org

1, 5, 39, 338, 3075, 28770, 274134, 2645844, 25781283, 253068530, 2498678754, 24788450076, 246889978062, 2467197059124, 24725226928140, 248396412496488, 2500825206700323, 25225687837101330, 254877697946626410, 2579123090162503500, 26133512970919973850, 265126176290618366460

Offset: 0

N. J. A. Sloane, Jun 28 2014

nonn

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..200
I. M. Gessel, A short proof of the Deutsch-Sagan congruence for connected non crossing graphs, arXiv preprint arXiv:1403.7656 [math.CO], 2014. See f_3(n).

Cf. A045741, A091527, A244038, A244469.

[Round(2^(2*n-1)*( Gamma(3*n/2+1)/Gamma(n/2+1) + Gamma((3*n+1)/2)/Gamma((n+1)/2) )/Factorial(n)): n in [0..25]]; // G. C. Greubel, Aug 20 2019

a := n -> 2^(2*n-1)*(binomial(3*n/2,n) + binomial((3*n-1)/2,n));
seq(a(n), n=0..25);

Table[2^(2n-1)*(Binomial[3n/2, n] + Binomial[(3n-1)/2, n]), {n, 0, 25}] (* Vincenzo Librandi, Jun 29 2014 *)

a(n) = 2^(2*n-1)*(binomial(3*n/2, n) + binomial((3*n-1)/2, n));
vector(25, n, n--; a(n)) \\ G. C. Greubel, Aug 20 2019

[2^(2*n-1)*(binomial(3*n/2, n) + binomial((3*n-1)/2, n)) for n in (0..25)] # G. C. Greubel, Aug 20 2019

From Peter Bala, Mar 04 2022: (Start)
a(n) = [x^n] ( (1 + 2*x)^3/(1 + x) )^n. Cf. A091527.
a(n) = Sum_{k = 0..n} (-1)^k * 2^(n-k) * binomial(3*n,n-k) * binomial(n+k-1,k).
n*(n-1)*(6*n-11)*a(n) = - 18*(n-1)*a(n-1) + 12*(3*n-4)*(3*n-5)*(6*n-5)*a(n-2) with a(0) = 1 and a(1) = 5.
The o.g.f. A(x) = 1 + 5*x + 39*x^2 + 338*x^3 + ... is the diagonal of the bivariate rational function 1/(1 - t*(1 + 2*x)^3/(1 + x)) and hence is an algebraic function over the field of rational functions Q(x) by Stanley 1999, Theorem 6.33, p. 197.
Calculation gives (1 - 108*x^2)*A(x)^3 - (1 + 9*x)*A(x) = x.
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. (End)
a(n) = 2^n*binomial(3*n, n)*hypergeom([-n, n], [2*n + 1], 1/2). - Peter Luschny, Mar 07 2022
From Seiichi Manyama, Aug 08 2025: (Start)
a(n) = Sum_{k=0..n} binomial(3*n,k) * binomial(2*n-k,n-k).
a(n) = [x^n] (1+x)^(3*n)/(1-x)^(n+1).
a(n) = [x^n] 1/((1-x)^n * (1-2*x)^(n+1)).
a(n) = Sum_{k=0..n} 2^k * binomial(n+k,k) * binomial(2*n-k-1,n-k). (End)

A370097 a(n) = Sum_{k=0..n} binomial(3n,k) binomial(3*n-k-1,n-k).

Original entry on oeis.org

1, 5, 49, 545, 6401, 77505, 956929, 11976193, 151388161, 1928363009, 24712450049, 318255628289, 4115300220929, 53396370030593, 694845537386497, 9064787191660545, 118516719269445633, 1552528215946035201, 20372392543502991361, 267736366910401413121

Offset: 0

Seiichi Manyama, Feb 10 2024

nonn

Cf. A091527, A370098.
Cf. A119259, A370101.
Cf. A365842.

Table[Sum[2^k*(-1)^(n-k)*Binomial[3*n, k], {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Jul 31 2025 *)

a(n) = sum(k=0, n, binomial(3*n, k)*binomial(3*n-k-1, n-k));

a(n) = [x^n] ( (1+x)^3/(1-x)^2 )^n.
The g.f. exp( Sum_{k>=1} a(k) * x^k/k ) has integer coefficients and equals (1/x) * Series_Reversion( x*(1-x)^2/(1+x)^3 ). See A365842.
a(n) = Sum_{k=0..n} 2^k * (-1)^(n-k) * binomial(3*n,k). - Seiichi Manyama, Jul 31 2025
a(n) ~ 3^(3*n + 1/2) / (5 * sqrt(Pi*n) * 2^(n-1)). - Vaclav Kotesovec, Jul 31 2025
a(n) = Sum_{k=0..n} 2^k * binomial(2*n+k-1,k). - Seiichi Manyama, Aug 01 2025
a(n) = [x^n] 1/((1-x) * (1-2*x)^(2*n)). - Seiichi Manyama, Aug 09 2025

A330843 Square array T(n,k) = [x^n] ((1+x)^(k+1) / (1-x)^(k-1))^n, n>=0, k>=0, read by descending antidiagonals.

Original entry on oeis.org

1, 1, 0, 1, 2, -2, 1, 4, 6, 0, 1, 6, 30, 20, 6, 1, 8, 70, 256, 70, 0, 1, 10, 126, 924, 2310, 252, -20, 1, 12, 198, 2240, 12870, 21504, 924, 0, 1, 14, 286, 4420, 41990, 184756, 204204, 3432, 70, 1, 16, 390, 7680, 104006, 811008, 2704156, 1966080, 12870, 0

Offset: 0

Seiichi Manyama, Feb 07 2020

			Square array begins:
    1,   1,     1,      1,      1,       1, ...
    0,   2,     4,      6,      8,      10, ...
   -2,   6,    30,     70,    126,     198, ...
    0,  20,   256,    924,   2240,    4420, ...
    6,  70,  2310,  12870,  41990,  104006, ...
    0, 252, 21504, 184756, 811008, 2521260, ...

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

Columns k=1..7 give A000984, A091527, A001448, A262732, A211419, A262733, A211421.

Main diagonal is A332231.

T[n_, k_] := Sum[Binomial[(k + 1)*n, j] * Binomial[k*n - j - 1, n - j], {j, 0, n}]; Table[T[k, n - k], {n, 0, 9}, {k, 0, n}] // Flatten (* Amiram Eldar, May 05 2021 *)

T(n,k) = Sum_{j=0..n} binomial((k+1)*n,j) * binomial(k*n-j-1,n-j).

T(n,k) = 1/n! * ((k+1)*n)!/Gamma(1 + (k+1)*n/2) * Gamma(1 + (k-1)*n/2)/((k-1)*n)!.

A370098 a(n) = Sum_{k=0..n} binomial(3n,k) binomial(4*n-k-1,n-k).

Original entry on oeis.org

1, 6, 72, 978, 14016, 207006, 3116952, 47568618, 733189632, 11387193846, 177923724072, 2793666465090, 44042615547456, 696708049377294, 11053262513080440, 175800225426741978, 2802193910116429824, 44752001810800994022, 715924864099841086728

Offset: 0

Seiichi Manyama, Feb 10 2024

nonn

Cf. A091527, A370097.
Cf. A370099, A370102.
Cf. A365843.

a(n) = sum(k=0, n, binomial(3*n, k)*binomial(4*n-k-1, n-k));

a(n) = [x^n] ( (1+x)^3/(1-x)^3 )^n.
The g.f. exp( Sum_{k>=1} a(k) * x^k/k ) has integer coefficients and equals (1/x) * Series_Reversion( x*(1-x)^3/(1+x)^3 ). See A365843.
From Seiichi Manyama, Aug 09 2025: (Start)
a(n) = [x^n] (1-x)^(n-1)/(1-2*x)^(3*n).
a(n) = Sum_{k=0..n} 2^k * binomial(3*n,k) * binomial(n-1,n-k).
a(n) = Sum_{k=0..n} 2^k * (-1)^(n-k) * binomial(3*n+k-1,k) * binomial(n-1,n-k). (End)

A244469 a(0) = 0, thereafter, a(n) = 2^(2n-1)( binomial((3n-1)/2,n) - binomial(3n/2, n)/3 ).

Original entry on oeis.org

0, 1, 7, 58, 515, 4746, 44758, 428772, 4154403, 40599130, 399429602, 3950996556, 39255152846, 391466112324, 3916110379020, 39281346256008, 394942611929379, 3978982062756090, 40160256911157610, 405995113593507900, 4110284071450416090, 41666530928566504620, 422876855107176561780

Offset: 0

N. J. A. Sloane, Jun 28 2014

nonn

Vincenzo Librandi, Table of n, a(n) for n = 0..200
I. M. Gessel, A short proof of the Deutsch-Sagan congruence for connected non crossing graphs, arXiv preprint arXiv:1403.7656 [math.CO], 2014. See f_4(n).

Cf. A000108, A007297, A064062, A091527, A244039.

[n eq 0 select 0 else Round(2^(2*n-1)*(Gamma((3*n+1)/2)/Gamma((n+1)/2) - Gamma((3*n+2)/2)/(3*Gamma((n+2)/2)))/Factorial(n)): n in [0..30]]; // G. C. Greubel, Apr 17 2019

f4:=n->-(2^(2*n-1)/3)*binomial(3*n/2,n) + 2^(2*n-1)*binomial((3*n-1)/2,n);
[seq(f4(n),n=1..40)]; # then prepend f4(0)=0.

Join[{0}, Table[-(2^(2 n - 1)/3) Binomial[3 n/2, n] + 2^(2 n - 1) Binomial[(3 n - 1)/2, n], {n, 1, 30}]] (* Vincenzo Librandi, Jun 29 2014 *)

{a(n) = if(n==0,0, 2^(2*n-1)*(binomial((3*n-1)/2, n) - binomial(3*n/2, n)/3) )}; \\ G. C. Greubel, Apr 17 2019

def a(n):
   if n==0: return 0
   else: return 2^(2*n-1)*(binomial((3*n-1)/2, n) - binomial(3*n/2, n)/3)
[a(n) for n in (0..30)] # G. C. Greubel, Apr 17 2019

G.f.: g'(x)/g(x)-1, g(x)=(2*sqrt(9*x+1)*sin(arcsin((54*x^2+27*x+2)/(2*(9*x+1)^(3/2)))/3))/3-1/3. - Vladimir Kruchinin, Apr 14 2019
From Peter Bala, Mar 05 2022: (Start)
a(n) = (1/n)*Sum_{k = 0..n} k*2^(n-k)*binomial(n+k-1,k)*binomial(2*n-k-1,n-k) for n >= 1.
a(n) = [x^n] G(x)^n = [x^n] 1/(1 - x*C(2*x))^n, where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the g.f. of the Catalan numbers A000108 and G(x) is the g.f. of A064062.
n*(n-1)*(6*n-7)*a(n) = - 18*(n-1)*a(n-1) + 12*(3*n-5)*(6*n-1)*(3*n-4)*a(n-2) with a(1) = 1 a(2) = 7.
exp(Sum_{n >= 1} a(n)*x^n/n) = 1 + x + 4*x^2 + 23*x^3 + 156*x^4 + ... is the g.f. of A007297.
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. (End)

A294454 a(n) = n! * [x^n] exp(2nx)BesselI(0,2x)^n.

Original entry on oeis.org

1, 2, 20, 324, 7336, 213500, 7593744, 319195800, 15481238224, 850968357228, 52279073479120, 3549850939488392, 263999303861731200, 21340730504572110008, 1863120652816098506432, 174706136370865217610000, 17512175948995988236164000, 1868638289932305589084614220, 211478046685658614366937497296

Offset: 0

Ilya Gutkovskiy, Nov 23 2017

nonn

The n-th term of the n-fold exponential convolution of A000984 with themselves.

N. J. A. Sloane, Transforms

Cf. A000984, A081085, A091527, A293491.

Table[n! SeriesCoefficient[Exp[2 n x] BesselI[0, 2 x]^n, {x, 0, n}], {n, 0, 18}]

a(n) ~ c * d^n * n! / sqrt(n), where d = 6.46710510392662827829435747085578126903789467159876086... and c = 0.36028050364743885143298970162021762094091934461095... - Vaclav Kotesovec, May 04 2024

A331969 T(n, k) = [x^(n-k)] 1/(((1 - 2x)^k)(1 - x)^(k + 1)). Triangle read by rows, for 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 11, 7, 1, 1, 26, 30, 10, 1, 1, 57, 102, 58, 13, 1, 1, 120, 303, 256, 95, 16, 1, 1, 247, 825, 955, 515, 141, 19, 1, 1, 502, 2116, 3178, 2310, 906, 196, 22, 1, 1, 1013, 5200, 9740, 9078, 4746, 1456, 260, 25, 1

Offset: 0

Peter Luschny, Feb 03 2020

The triangle is the matrix inverse of the Riordan square (see A321620) generated by (1 + x - sqrt(1 - 6*x + x^2))/(4*x) (see A172094), where we take the absolute value of the terms.

T(n,k) is the number of evil-avoiding (2413, 3214, 4132, and 4213 avoiding) permutations of length (n+2) that start with 1 and whose inverse has k descents. - Donghyun Kim, Aug 16 2021

			Triangle starts:
[0] [1]
[1] [1,    1]
[2] [1,    4,    1]
[3] [1,   11,    7,    1]
[4] [1,   26,   30,   10,    1]
[5] [1,   57,  102,   58,   13,    1]
[6] [1,  120,  303,  256,   95,   16,    1]
[7] [1,  247,  825,  955,  515,  141,   19,   1]
[8] [1,  502, 2116, 3178, 2310,  906,  196,  22,  1]
[9] [1, 1013, 5200, 9740, 9078, 4746, 1456, 260, 25, 1]
...
Seen as a square array (the triangle is formed by descending antidiagonals):
1,  1,   1,    1,    1,     1,      1,      1,       1, ... [A000012]
1,  4,  11,   26,   57,   120,    247,    502,    1013, ... [A000295]
1,  7,  30,  102,  303,   825,   2116,   5200,   12381, ... [A045889]
1, 10,  58,  256,  955,  3178,   9740,  28064,   77093, ... [A055583]
1, 13,  95,  515, 2310,  9078,  32354, 106970,  333295, ...
1, 16, 141,  906, 4746, 21504,  87374, 326084, 1136799, ...
1, 19, 196, 1456, 8722, 44758, 204204, 849180, 3275931, ...

Donghyun Kim and Lauren Williams, Schubert polynomials and the inhomogeneous TASEP on a ring, arXiv:2102.00560 [math.CO], 2021.

Row sums A006012, alternating row sums A118434 with different signs, central column A091527.
T(n, 1) = A000295(n+1) for n >= 1, T(n, 2) = A045889(n-2) for n >= 2, T(n, 3) = A055583(n-3) for n >= 3.
Cf. A172094 (inverse up to sign).

gf := k -> 1/(((1-2*x)^k)*(1-x)^(k+1)): ser := k -> series(gf(k), x, 32):
# Prints the triangle:
seq(lprint(seq(coeff(ser(k), x, n-k), k=0..n)), n=0..6);
# Prints the square array:
seq(lprint(seq(coeff(ser(k), x, n), n=0..8)), k=0..6);

(* The function RiordanSquare is defined in A321620; returns the triangle as a lower triangular matrix. *)
M := RiordanSquare[(1 + x - Sqrt[1 - 6 x + x^2])/(4 x), 9];
Abs[#] & /@ Inverse[PadRight[M]]

A386937 a(n) = Sum_{k=0..n} binomial(3n+1,k) binomial(2*n-k-1,n-k).

Original entry on oeis.org

1, 5, 38, 325, 2934, 27314, 259356, 2496813, 24281510, 237978598, 2346750900, 23257207714, 231438363324, 2311082461380, 23146003391352, 232402586792061, 2338665721556742, 23579860411878110, 238157209512898500, 2409099858256570710, 24403155769842168660

Offset: 0

Seiichi Manyama, Aug 10 2025

nonn

Cf. A091527, A386833.

a(n) = sum(k=0, n, binomial(3*n+1, k)*binomial(2*n-k-1, n-k));

a(n) = [x^n] (1+x)^(3*n+1)/(1-x)^n.
a(n) = [x^n] 1/((1-x)^(n+2) * (1-2*x)^n).
a(n) = Sum_{k=0..n} 2^k * (-1)^(n-k) * binomial(3*n+1,k) * binomial(2*n-k+1,n-k).
a(n) = Sum_{k=0..n} 2^k * binomial(n+k-1,k) * binomial(2*n-k+1,n-k).
D-finite with recurrence n*(n+1)*a(n) +42*n*(n-2)*a(n-1) +12*(-33*n^2+120*n-95)*a(n-2) +72*(-63*n^2+189*n-110)*a(n-3) +3456*(3*n-8)*(3*n-10)*a(n-4)=0. - R. J. Mathar, Aug 19 2025

cp's OEIS Frontend

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

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A244039 a(n) = 2^(2*n-1) * ( binomial(3*n/2,n) + binomial((3*n-1)/2,n) ).

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A370097 a(n) = Sum_{k=0..n} binomial(3*n,k) * binomial(3*n-k-1,n-k).

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A330843 Square array T(n,k) = [x^n] ((1+x)^(k+1) / (1-x)^(k-1))^n, n>=0, k>=0, read by descending antidiagonals.

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A370098 a(n) = Sum_{k=0..n} binomial(3*n,k) * binomial(4*n-k-1,n-k).

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PARI

Formula

A244469 a(0) = 0, thereafter, a(n) = 2^(2*n-1)*( binomial((3*n-1)/2,n) - binomial(3*n/2, n)/3 ).

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Magma

Maple

Mathematica

PARI

Sage

Formula

A294454 a(n) = n! * [x^n] exp(2*n*x)*BesselI(0,2*x)^n.

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Mathematica

Formula

A331969 T(n, k) = [x^(n-k)] 1/(((1 - 2*x)^k)*(1 - x)^(k + 1)). Triangle read by rows, for 0 <= k <= n.

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A244039 a(n) = 2^(2n-1) ( binomial(3n/2,n) + binomial((3n-1)/2,n) ).

A370097 a(n) = Sum_{k=0..n} binomial(3n,k) binomial(3*n-k-1,n-k).

A370098 a(n) = Sum_{k=0..n} binomial(3n,k) binomial(4*n-k-1,n-k).

A244469 a(0) = 0, thereafter, a(n) = 2^(2n-1)( binomial((3n-1)/2,n) - binomial(3n/2, n)/3 ).

A294454 a(n) = n! * [x^n] exp(2nx)BesselI(0,2x)^n.

A331969 T(n, k) = [x^(n-k)] 1/(((1 - 2x)^k)(1 - x)^(k + 1)). Triangle read by rows, for 0 <= k <= n.

A386937 a(n) = Sum_{k=0..n} binomial(3n+1,k) binomial(2*n-k-1,n-k).