A000897 -id:A000897 - cp's OEIS frontend

A186420 a(n) = binomial(2n,n)^4.

1, 16, 1296, 160000, 24010000, 4032758016, 728933458176, 138735983333376, 27435582641610000, 5588044012339360000, 1165183173971324375296, 247639903129149250277376, 53472066459540320483696896, 11701285507234585729600000000, 2589980371199606611713600000000

Offset: 0

Emanuele Munarini, Feb 21 2011

			G.f.: 4F3({1/2,1/2,1/2,1/2},{1,1,1},256x) where 4F3 is a hypergeometric series.

Seiichi Manyama, Table of n, a(n) for n = 0..417

Cf. binomial(2n,n)^k: A000984 (k=1), A002894 (k=2), A002897 (k=3), this sequence (k=4).

Cf. A000108, A000888, A186414, A186415, A186416, A186418, A186419, A000897, A006480, A008977, A188662.

Table[Binomial[2n,n]^4,{n,0,20}]
Table[Coefficient[Series[HypergeometricPFQ[{1/2, 1/2, 1/2, 1/2}, {1, 1, 1}, 256 x], {x, 0, n}], x, n], {n, 0, 14}] (* Michael De Vlieger, Jul 13 2016 *)

Maxima
```
makelist(binomial(2*n,n)^4,n,0,40);
```

a(n) = A000984(n)^4 = A002894(n)^2.
a(n) = binomial(2*n,n)^4 = ( [x^n](1 + x)^(2*n) )^4 = [x^n](F(x)^(16*n)), where F(x) = 1 + x + 25*x^2 + 1798*x^3 + 183442*x^4 + 22623769*x^5 + 3142959012*x^6 + ... appears to have integer coefficients. For similar results see A000897, A002894, A002897, A006480, A008977 and A188662. - Peter Bala, Jul 14 2016
a(n) ~ 256^n/(Pi*n)^2. - Ilya Gutkovskiy, Jul 13 2016

A001451 a(n) = (5n)!/((3n)!n!n!).

Original entry on oeis.org

1, 20, 1260, 100100, 8817900, 823727520, 79919739900, 7962100660800, 808906548235500, 83426304143982800, 8707404737345073760, 917663774856743842200, 97491279924241456098300, 10427604345391237790688000, 1121786259855036145008408000

Offset: 0

N. J. A. Sloane

From Peter Bala, Jul 15 2016: (Start)
This sequence occurs as the right-hand side of the binomial sum identity Sum_{k = 0..n} (-1)^(k)*binomial(n,k)*binomial(2*n + k,n)*binomial(3*n - k,n) = (-1)^m*a(m) for n = 2*m. The sum vanishes for n odd. Cf. A273628 and A273629.
Note the similar results:
Sum_{k = 0..n} (-1)^k*binomial(n,k)* binomial(2*n + k,n)*binomial(3*n + k,n) = (-1)^n*(3*n)!/n!^3 = (-1)^n*A006480(n);
Sum_{k = 0..n} (-1)^k*binomial(n,k)*binomial(2*n - k,n)*binomial(3*n - k,n) = binomial(2*n,n)^2 = A002894(n);
Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(2*n + k,n)*binomial(3*n + k,n) = Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(2*n - k,n)*binomial(3*n - k,n) = binomial(2*n,n) = A000984(n);
Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(2*n + k,n)*binomial(3*n - k,n) = Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(2*n - k,n)*binomial(3*n + k,n) = (-1)^n*binomial(2*n,n) = (-1)^n*A000984(n). (End)
Choose three noncollinear step vectors to satisfy the zero sum, 3*v_1 + v_2 + v_3 = 0. Then a(n) is the number of loop plane walks of length 5*n which depart from and return to the origin. Equivalently, a(n) counts distinct permutations of a (5*n)-digit integer with digits 1,2,3 of multiplicity 3*n,n,n respectively. - Bradley Klee, Aug 12 2018

			G.f. = 1 + 20*x + 1260*x^2 + 100100*x^3 + 8817900*x^4 + 823727520*x^5 + ... - _Michael Somos_, Aug 12 2018

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A000897, A002894, A002897, A006480, A008977, A186420, A188662, A273628, A273629.

List([0..15],n->Factorial(5*n)/(Factorial(3*n)*Fact0rial(n)*Factorial(n))); # Muniru A Asiru, Aug 12 2018

[Factorial(5*n)/(Factorial(3*n)*Factorial(n)*Factorial(n)): n in [0..30]]; // Vincenzo Librandi, May 22 2011

Maple
```
f := n->(5*n)!/((3*n)!*n!*n!);
```

Table[(5*n)!/((3*n)!*n!*n!), {n, 0, 20}] (* Vincenzo Librandi, Sep 04 2012 *)

a(n) = binomial(4*n,n)*binomial(5*n,n) = ( [x^n](1 + x)^(4*n) ) * ( [x^n](1 + x)^(5*n) ) = [x^n](F(x)^(20*n)), where F(x) = 1 + x + 12*x^2 + 390*x^3 + 16984*x^4 + 867042*x^5 + 48848541*x^6 + ... appears to have integer coefficients. For similar results see A000897, A002894, A002897, A006480, A008977, A186420 and A188662. - Peter Bala, Jul 14 2016
a(n) ~ 3^(-3*n-1/2)*5^(5*n+1/2)/(2*Pi*n). - Ilya Gutkovskiy, Jul 13 2016
G.f.: G(x) = 4F3(1/5,2/5,3/5,4/5;1/3,2/3,1;(5^5/3^3)*x). Let G^(n)(x) = d^n/dx^n G(x), and c = {120, 15000*x-6, 45000*x^2-114*x, 25000*x^3-135*x^2, 3125*x^4-27*x^3}, then Sum_{n=0..4} c_n*G^(n)(x) = 0. - Bradley Klee, Aug 12 2018
From Peter Bala, Mar 20 2022: (Start)
Right-hand side of the following identities valid for n >= 1:
Sum_{k = 0..3*n} 2*n*(2*n+k-1)!/(k!*n!^2) = (5*n)!/((3*n)!*n!^2);
Sum_{k = 0..n} 4*n*(4*n+k-1)!/(k!*n!*(3*n)!) = (5*n)!/((3*n)!*n!^2). Cf. A000897 and A113424. (End)
a(n) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n, k) * A108625(4*n, k) (verified using the MulZeil procedure in Doron Zeilberger's MultiZeilberger package). - Peter Bala, Oct 12 2024

A184423 a(n) = (2n)!(3*n)!/n!^5.

Original entry on oeis.org

1, 12, 540, 33600, 2425500, 190702512, 15849497664, 1369618398720, 121821136479900, 11079206239530000, 1025579963180813040, 96310511463483233280, 9152842704012278107200, 878622906816654279840000

Offset: 0

Paul D. Hanna, Jan 13 2011

Denoted by h_3[n] by T. Piezas III. He also gives formulas for 1/Pi such as 1/Pi = 2 * Sum_{n>=0} a(n) * (-1)^n * (51*n + 7) / (12^3)^(n + 1/2). - Michael Somos, May 31 2012
Diagonal of the rational function R(x,y,z,w) = 1/(1-(w*y+w*z+x+y+z)). - Gheorghe Coserea, Jul 15 2016
From Peter Bala, Jun 28 2023: (Start)
The supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) holds for primes p >= 5 and positive integers n and r. This follows from Meštrović equation 39, since a(n) = binomial(3*n,n) * binomial(2*n,n)^2.
Inductively define a family of sequences {a(i,n) : n >= 0}, i >= 1, by setting a(1,n) = a(n) and, for i >= 2, a(i,n) = [x^n] ( exp(Sum_{k >= 1} a(i-1,k)*x^k/k) )^n. We conjecture that the sequences {a(i,n) : n >= 0}, i >= 2, also satisfy the supercongruences u(n*p^r) == u(n*p^(r-1)) (mod p^(3*r)) for primes p >= 5 and positive integers n and r. Cf. A362730 and A362732. (End)

			G.f.: A(x) = 1 + 12*x + 540*x^2 + 33600*x^3 + 2425500*x^4 +...
G.f. of A184424 equals A(x)^(1/2):
A(x)^(1/2) = 1 + 6*x + 252*x^2 + 15288*x^3 + 1089270*x^4 + 84963060*x^5 +...+ [(3^n/n!^2)*Product_{k=1..n} (6*k-4)*(6*k-5)]*x^n +...

Gheorghe Coserea, Table of n, a(n) for n = 0..200
Timothy Huber, Daniel Schultz, and Dongxi Ye, Ramanujan-Sato series for 1/pi, Acta Arith. (2023) Vol. 207, 121-160. See p. 11.
Romeo Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862--2012), arXiv:1111.3057 [math.NT], 2011.
Titus Piezas III, 0013: Article 3 (Pi Formulas and the Monster Group).

Cf. A184424, A275047.

Related to diagonal of rational functions: A268545-A268555.

Table[((2n)!(3n)!)/(n!)^5,{n,0,20}] (* Harvey P. Dale, Dec 18 2018 *)

PARI
```
{a(n)=(3*n)!*(2*n)!/n!^5}
```

{a(n)=polcoeff(sum(m=0,n,3^m*prod(k=1,m,(6*k-4)*(6*k-5))/m!^2*x^m+x*O(x^n))^2,n)}

Self-convolution of A184424:
a(n) = Sum_{k=0..n} A184424(k)*A184424(n-k) where A184424(n) = (3^n/n!^2)*Product_{k=1..n} (6*k-4)*(6*k-5).
a(n) = 6 * (2*n - 1) * (3*n - 1) * (3*n - 2) / n^3 * a(n-1) if n>0. - Michael Somos, May 31 2012
0 = (x^2-108*x^3)*y''' + (3*x-486*x^2)*y''+ (1-348*x)*y' - 12*y, where y is g.f. - Gheorghe Coserea, Jul 15 2016
a(n) ~ 3^(1/2)/(2*Pi^(3/2)) * n^(-3/2) * 108^n. - Ilya Gutkovskiy, Jul 15 2016
a(n) = C(2*n,n)^2 * C(3*n,n) = ( [x^n](1 + x)^(2*n) )^2 * ( [x^n](1 + x)^(3*n) ) = [x^n]( F(x)^(12*n) ), where [x^n] is the coefficient extraction operator and where F(x) = 1 + x + 11*x^2 + 350*x^3  + 15293*x^4  + 794433*x^5  + 45958617*x^6 + ... appears to have integral coefficients. Cf. A000897 and A001451. - Peter Bala, Dec 30 2019

A361657 Constant term in the expansion of (1 + x^2 + y^2 + 1/(x*y))^n.

Original entry on oeis.org

1, 1, 1, 1, 13, 61, 181, 421, 1261, 5293, 21421, 73261, 232321, 789361, 2954953, 11127481, 39961741, 139908301, 499315501, 1835933293, 6792310153, 24827506873, 90058277233, 328509505633, 1210097040769, 4473191880961, 16495696956961, 60721903812961

Offset: 0

Seiichi Manyama, Mar 19 2023

nonn

Cf. A002426, A361658.

Cf. A000897, A344560, A361637.

Table[n!*Sum[1/(k!^2*(2*k)!*(n - 4*k)!), {k, 0, n/4}], {n, 0, 30}] (* Vaclav Kotesovec, Mar 20 2023 *)

a(n) = n!*sum(k=0, n\4, 1/(k!^2*(2*k)!*(n-4*k)!));

a(n) = n! * Sum_{k=0..floor(n/4)} 1/(k!^2 * (2*k)! * (n-4*k)!) = Sum_{k=0..floor(n/4)} binomial(n,4*k) * A000897(k).
From Vaclav Kotesovec, Mar 20 2023: (Start)
Recurrence: (n-2)*n^2*a(n) = (4*n^3 - 12*n^2 + 10*n - 3)*a(n-1) - (n-1)*(6*n^2 - 18*n + 13)*a(n-2) + 4*(n-2)^2*(n-1)*a(n-3) + 63*(n-3)*(n-2)*(n-1)*a(n-4).
a(n) ~ (1 + 2*sqrt(2))^(n+1) / (4*Pi*n). (End)

A178819 Pascal's prism (3-dimensional array) read by folded antidiagonal cross-sections: (h+i; h, i-j, j), h >= 0, i >= 0, 0 <= j <= i.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 2, 1, 2, 1, 1, 3, 3, 1, 3, 6, 3, 3, 3, 1, 1, 4, 4, 6, 12, 6, 4, 12, 12, 4, 1, 4, 6, 4, 1, 1, 5, 10, 10, 5, 1, 5, 20, 30, 20, 5, 10, 30, 30, 10, 10, 20, 10, 5, 5, 1, 1, 6, 6, 15, 30, 15, 20, 60, 60, 20, 15, 60, 90, 60, 15, 6, 30, 60, 60, 30, 6, 1, 6, 15, 20, 15, 6, 1

Offset: 0

Harlan J. Brothers, Jun 16 2010

P_h = level h of Pascal's prism where P_1 = Pascal's triangle (A007318) and P_2 = denominators of Leibniz harmonic triangle (A003506). A sequence of length k through P is defined by P for n = {1, 2, 3, ..., k}.

			Prism begins (levels 1-4):
1
1 1
1 2 1
1 3 3 1
1
2 2
3 6 3
4 12 12 4
1
3 3
6 12 6
10 30 30 10
1
4 4
10 20 10
20 60 60 20

H. J. Brothers, Pascal's prism, The Mathematical Gazette, 96 (July 2012), 213-220.
H. J. Brothers, Pascal's Prism: Supplementary Material

Level 1 = A007318.
Level 2 = A003506.
Level 3 = A094305.
Level 4 = A178820.
Level 5 = A178821.
Level 6 = A178822.
Sums of shallow diagonals for each level correspond to rows of square A037027.
Contains A109649 and A046816.
P = A000984.
P = A006480.
P = A000897.
P<3n-2, 3n-2, n> = A113424.

end = 5; Column/@Table[Multinomial[h, i-j, j], {h, 0, end}, {i, 0, end}, {j, 0, i}]

a_(h, i, j) = (h+i-2; h-1, i-j, j-1), h >= 1, i >= 1, 1 <= j <= i.
Recurrence:
For P_h, element a is given by: a_(1, 1) = 1; a_(i, j) = ((i+h-2)/(i-1)) (a_(i-1, j) + a_(i-1, j-1)).

Keyword tabf by Michel Marcus, Oct 22 2017

A295870 a(n) = binomial(3n,n)*CQC(n), where CQC(n) = A005721(n) = A005190(2n) is a central quadrinomial coefficient.

Original entry on oeis.org

1, 12, 660, 48720, 4005540, 349260912, 31626298704, 2940502593600, 278788387440420, 26831860080682800, 2613367831568654160, 257012469788428710720, 25479526081439438845200, 2543092744417831625342400, 255292245777771431285140800, 25755871314484468746363582720

Offset: 0

Bradley Klee, Feb 23 2018

nonn

Compare with EllipticK A002894 and the Ramanujan period-energy functions A113424, A006480, A000897. The series expansion "T(x) = 2*Pi*Sum_{n>=0} a_n*x^n" determines the real period T of elliptic curves in the family "x=p^2+q^2-4*(q^2-p^2)*q, 0 < x < 1/108". This sequence serves as a counterexample to the naive idea that elliptic integrals will always evaluate to a hypergeometric function such as 2F1(a,b;c;x).
A300058 is the complex period-energy function, after scaling energy and time dimensions such that all a(n) are integers and a(0)=1. The Picard-Fuchs equation is "(12-288*x+9216*x^2)*T(x) + (-1+232*x-8160*x^2+82944*x^3)*T'(x) + (-x+164*x^2-6432*x^3+41472*x^4)*T''(x)".
Although the sequence is not generated by a hypergeometric function, it can be formulated in terms of Hypergeometric numbers, specifically the binomial coefficients. Then Zeilberger's algorithm outputs a second order recurrence with polynomial coefficients.
The contour plot is nice to look at, with reflection symmetry, three critical points, and two separatrices dividing the phase plane into eight distinct regions.
Hyperbolic Critical points are located at (q,p) locations (1/6,0) and (-1/4,sqrt(5)/4) and (-1/4,-sqrt(5)/4). Is it possible to use chord-and-tangent addition rules to produce an exponentially-convergent Diophantine approximation to sqrt(5) that moves along the upper separatrix x=1/8?
Does there exist a period-preserving transformation that takes any one of the curves with 0 < x < 1/108 into a particular Weierstrass curve from the L-function and Modular Forms Database?

D. Husemöller, Elliptic Curves, 2nd ed., New York: Springer, 2004.
J. H. Silverman, The Arithmetic of Elliptic Curves, 2nd ed., New York: Springer, 2009.

J. Cremona, Elliptic Curves over Q, LMFDB 2017.
B. Klee, The Virtues of X_{n+1} = (4+3*X_{n})/(3+2*X_{n}), seqfans mailing list, 2017.
B. Klee, Geometric G.F. for Ramanujan Periods, seqfans mailing list, 2017.
Brad Klee, Deriving Hypergeometric Picard-Fuchs Equations, Wolfram Demonstrations Project (2018).
Bradley Klee, Phase Plane Geometry.
M. Kontsevich and D. Zagier, Periods, Institut des Hautes Etudes Scientifiques 2001 IHES/M/01/22.
P. Paule and M. Schorn, FastZeil: the Paule/Schorn implementation of Gosper's and Zeilberger's algorithm, RISC 2017; Local copy, pdf file only, no active links
D. Zeilberger, The Method of Creative Telescoping, Journal of Symbolic Computation, 11.3 (1991), 195-204.

Cf. A000897, A002894, A006480, A113424.
Factors: A005190, A005809, A005721.
Complex Period: A300058.

b[NN_]:=Total/@Table[((-1)^k)*Binomial[3*n,n]*Binomial[2*n,k]*Binomial[5*n-4*k-1,3*n-4*k],{n,0,NN},{k,0,Floor[3*n/4]}];
c1=8*(-30+201*n-319*n^2+145*n^3);c2=-8640*(n-5/3)*(n-4/3)*(n-1/5);c3=10*(n-6/5)*n^2;a[0]=1;a[1]=12;a[n0_]:=ReplaceAll[(c1/c3)*a[n0-1]+(c2/c3)*a[n0-2],{n->n0}];
({#,SameQ[a/@Range[0, 15],#]}&@b[15])[[1]]

a(n) = A005809(n)*A005721(n).
a(n) = Sum_{k=0..floor(3n/4)} ((-1)^k)*binomial(3*n,n)*binomial(2 *n, k)*binomial(5*n - 4*k - 1, 3*n - 4*k).
c1 = 8 *(-30 + 201*n - 319*n^2 + 145*n^3); c2 = -8640*(n - 5/3)*(n - 4/3)*(n - 1/5); c3 = 10*(n - 6/5)*n^2; a(0)=1; a(1)=12; a(n) = (c1/c3)*a(n-1) + (c2/c3)*a(n-2).

A300058 a(n) = binomial(3n,n)/(2Pi)Integral_{x=0..2Pi} (12cos^2(x)sin(x) + 20sin^3(x))^(2n) dx.

Original entry on oeis.org

1, 492, 707220, 1298204880, 2654173160100, 5765723073622512, 13021894087331233104, 30217387890886676251200, 71532102917478013611243300, 171944976047709681477985038000, 418347201888204996027087975427920

Offset: 0

Bradley Klee, Feb 23 2018

nonn

Compare with A295870. The series expansion "T(x)=2*Pi*sqrt(3/5)*Sum_{n>=0} a_n*(x/25)^n" determines the period T of anharmonic oscillation along a contour of the Hamiltonian energy surface "x=2H=(5/3)*p^2+q^2+4*(p^2+q^2)*q,0

The period-energy function T(x) satisfies the Picard-Fuchs equation "(2460+28512*x+2239488*x^2)*T(x)-(125-24840*x-1423008*x^2-20155392*x^3)*T'(x)+(-125*x+1620*x^2+1189728*x^3+10077696 x^4)*T''(x)", also the P.F.Eq. of A295870 under transformation x->x'=1/108-x.

A300057 has a similar definition to A005721, with a couple of extra integers appearing in the integrand. This makes a nice analogy between real and complex periods A295870, A300058. Second-order recurrences with polynomial coefficients define both sequences.

Bradley Klee, Phase Plane Geometry.
Brad Klee, Deriving Hypergeometric Picard-Fuchs Equations, Wolfram Demonstrations Project (2018).
M. Kontsevich and D. Zagier, Periods, Institut des Hautes Etudes Scientifiques 2001 IHES/M/01/22.

Cf. A002894, A113424, A006480, A000897. Factors: A005809, A300057. Real Period: A295870.

a := n -> 36^n*(3*n)!/n!^3*hypergeom([-2*n, n+1/2], [n+1], -2/3):
seq(simplify(a(n)), n=0..10); # Peter Luschny, Apr 19 2018

c1=12*(-230+2259*n-3933*n^2+1863*n^3);c2=5248800*(n-5/3)*(n-4/3)*(n-1/9);c3=9*n^2*(n-10/9);a[0]=1;a[1]=492;a[n0_]:=ReplaceAll[(c1/c3)*a[n0-1]+(c2/c3)*a[n0-2],n->n0];
b[NN_]:=Expand[Total[Flatten[#]]&/@Table[Binomial[3*n,n]*Binomial[2*n,k2]*Binomial[2*n,k1]*Binomial[2*n,3*n-k1-k2]*((4+Sqrt[15])^(2*n-k1))*((4-Sqrt[15])^(2*n-k2)),{n,0,NN},{k1,0,2*n},{k2,0,2*n}]]; ({#,SameQ[#,a/@Range[0,10]]}&@b[10])[[1]]
Table[Binomial[3*n, n] * SeriesCoefficient[(1 + 9*z + 9*z^2 + z^3)^(2*n), {z, 0, 3*n}], {n, 0, 15}] (* Vaclav Kotesovec, Apr 18 2018 *)

a(n) = A005809(n)*A300057(n).
a(n) = Sum_{k1=0..2n} Sum_{k2=0..2n} binomial(3*n,n)*binomial(2*n,k2)*binomial(2*n,k1)*binomial(2*n,3*n-k1-k2)*((4+sqrt(15))^(2*n-k1))*((4-sqrt(15))^(2*n-k2))
a(0) = 1; a(1) = 492; a(n):=(c1/c3)*a(n-1)+(c2/c3)*a(n-2); with
  c1 = 12*(-230+2259*n-3933*n^2+1863*n^3);
  c2 = 5248800*(n-5/3)*(n-4/3)*(n-1/9);
  c3 = 9*n^2*(n-10/9);
a(n) ~ 2^(2*n - 1) * 3^(3*n - 1/2) * 5^(2*n + 1/2) / (Pi*n). - Vaclav Kotesovec, Apr 18 2018

A308836 The nome q=exp(T_C/T_R)=Sum_{n>=0} a(n)(x/64)^n follows from the series solutions of 3T-d/dx(16(1-x)x*dT/dx)=0.

Original entry on oeis.org

0, 1, 40, 1876, 95072, 5045474, 276107408, 15444602248, 878268335296, 50588345910799, 2944021398570264, 172780225616034252, 10211876493716693664, 607169816926036666486, 36286222314596227018672, 2178246170438379512947864, 131270483744089714062036032

Offset: 0

Bradley Klee, Jun 27 2019

nonn

Also appears in Ramanujan's theory of elliptic functions, signature 4 (cf. A000897). Almkvist et al. give a real and complex Ansatz for the second-order, ordinary differential equation: T_R = 1 + x*{Z[[x]]}, T_C = T_R*log(x) + x*{Z[[x]]}.

B.C. Berndt, "Ramanujan's Notebooks Part II", Springer, 2012, pages 80-82.

G. Almkvist et al., Generalizations of Clausen's Formula and Algebraic Transformations of Calabi-Yau Differential Equations, Proceedings of the Edinburgh Mathematical Society, 54 (2011), p. 275.

Cf. A005797, A308835, A308837.

G[nMax_] := Dot[RecurrenceTable[ {Dot[{(4*n - 7)^2 (4*n - 5)^2 (8*n - 3), -16 (n - 1) (105 - 562*n + 1056*n^2 - 864*n^3 + 256*n^4), 256 (n - 1) n^3 (8*n - 11)},  a[n - #] & /@ Reverse[Range[0, 2]]] == 0, a[0] == 0, a[1] == 5/8}, a, {n, 0, nMax}], x^Range[0, nMax]];
qSer[nMax_] := Expand[Times[x, Normal[ Series[Exp[ Divide[G[nMax], Hypergeometric2F1[1/4, 3/4, 1, x]]], {x, 0, nMax}]]]];
CoefficientList[(1/k)*qSer[20] /. {x -> k*x}, x] /. {k -> 64}

A356546 Triangle read by rows. T(n, k) = RisingFactorial(n + 1, n) / (k! * (n - k)!).

Original entry on oeis.org

1, 2, 2, 6, 12, 6, 20, 60, 60, 20, 70, 280, 420, 280, 70, 252, 1260, 2520, 2520, 1260, 252, 924, 5544, 13860, 18480, 13860, 5544, 924, 3432, 24024, 72072, 120120, 120120, 72072, 24024, 3432, 12870, 102960, 360360, 720720, 900900, 720720, 360360, 102960, 12870

Offset: 0

Peter Luschny, Aug 12 2022

The counterpart using the falling factorial is Leibniz's Harmonic Triangle A003506.

			Triangle T(n, k) begins:
[0]     1;
[1]     2,      2;
[2]     6,     12,      6;
[3]    20,     60,     60,     20;
[4]    70,    280,    420,    280,     70;
[5]   252,   1260,   2520,   2520,   1260,    252;
[6]   924,   5544,  13860,  18480,  13860,   5544,    924;
[7]  3432,  24024,  72072, 120120, 120120,  72072,  24024,   3432;
[8] 12870, 102960, 360360, 720720, 900900, 720720, 360360, 102960, 12870;

cf. A000984, A059304 (row sums, see also A343842), A265609 (rising factorial).

Cf. A003506, A173018 (Eulerian numbers), A000108, A000897 (central terms).

A356546 := (n, k) -> pochhammer(n+1, n)/(k!*(n-k)!):
for n from 0 to 8 do seq(A356546(n, k), k=0..n) od;

T[ n_, k_] := Binomial[2*n, n] * Binomial[n, k]; (* Michael Somos, Aug 18 2022 *)

{T(n, k) = binomial(2*n, n) * binomial(n, k)}; /* Michael Somos, Aug 18 2022 */

def A356546(n, k):
    return rising_factorial(n+1,n) // (factorial(k) * factorial(n-k))
for n in range(9): print([A356546(n, k) for k in range(n+1)])

Bernoulli(n) / Catalan(n) = Sum_{k=0..n} (-1)^k*A173018(n, k) / T(n, k), (with Bernoulli(1) = 1/2).

G.f.: 1/sqrt(1 - 4*x*(y + 1)). - Vladimir Kruchinin, Feb 15 2023

A295864 a(n) = hypergeom([-n, -n], [1], 1) * n! / (floor(n/2)!)^2.

Original entry on oeis.org

1, 2, 12, 120, 420, 7560, 18480, 480480, 900900, 30630600, 46558512, 1955457504, 2498640144, 124932007200, 137680171200, 7985449929600, 7735904619300, 510569704873800, 441233078286000, 32651247793164000, 25467973278667920, 2088373808850769440, 1484298740174927040

Offset: 0

Peter Luschny, Feb 13 2018

nonn

Cf. A000897, A000984, A056040.

a := n -> binomial(2*n, n)*n!/iquo(n, 2)!^2: seq(a(n), n=0..22);

a[n_] := Multinomial[Quotient[n,2], Quotient[n,2], Mod[n,2]] Multinomial[n,n];
Table[a[n], {n, 0, 22}]

def A295864():
    r, c, n = 1, 1, 0
    while True:
        yield r * c
        n += 1
        c = c*(4*n-2)//n
        r = (r*4)//n if n % 2 == 0 else r*n
a = A295864(); [next(a) for i in range(23)]

a(2*n) = A000897(n).
a(n) = A000984(n) * A056040(n).
a(n) = (2*n)!/(n!*floor(n/2)!^2).
a(n) = (2^(2*n)*Gamma(n+1/2))/(sqrt(Pi)*Gamma(floor(n/2)+1)^2).
a(n) = multinomial([n/2], [n/2], n mod 2)*multinomial(n, n).
a(n) = 4^(n+floor(n/2))*hypergeom([-n,1/2],[1],1)*hypergeom([-floor(n/2),(-1)^n/2],[1],1).
a(n) = c(n)*8^n*Pochhammer(1/4, [n/2])*Pochhammer(3/4, [n/2])/[n/2]!^2 where c(n) = 1 if n is even else c(n) = (2*n-1)/4.
a(n) ~ (8^n/(sqrt(2)*Pi*n))*c(n) where c(n) = 2 - 3/(4*n) if n is even else c(n) = n + 1/8.

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A186420 a(n) = binomial(2n,n)^4.

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A001451 a(n) = (5*n)!/((3*n)!*n!*n!).

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A184423 a(n) = (2*n)!*(3*n)!/n!^5.

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A361657 Constant term in the expansion of (1 + x^2 + y^2 + 1/(x*y))^n.

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A178819 Pascal's prism (3-dimensional array) read by folded antidiagonal cross-sections: (h+i; h, i-j, j), h >= 0, i >= 0, 0 <= j <= i.

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A295870 a(n) = binomial(3n,n)*CQC(n), where CQC(n) = A005721(n) = A005190(2n) is a central quadrinomial coefficient.

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A300058 a(n) = binomial(3*n,n)/(2*Pi)*Integral_{x=0..2*Pi} (12*cos^2(x)*sin(x) + 20*sin^3(x))^(2*n) dx.

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A001451 a(n) = (5n)!/((3n)!n!n!).

A184423 a(n) = (2n)!(3*n)!/n!^5.

A300058 a(n) = binomial(3n,n)/(2Pi)Integral_{x=0..2Pi} (12cos^2(x)sin(x) + 20sin^3(x))^(2n) dx.

A308836 The nome q=exp(T_C/T_R)=Sum_{n>=0} a(n)(x/64)^n follows from the series solutions of 3T-d/dx(16(1-x)x*dT/dx)=0.