A002897 -id:A002897 - cp's OEIS frontend

A181544 Triangle in which the g.f. for row n is [Sum_{k>=0} C(n+k-1,k)^3x^k](1-x)^(3n+1), read by rows of k=0..2n terms.

1, 1, 4, 1, 1, 20, 48, 20, 1, 1, 54, 405, 760, 405, 54, 1, 1, 112, 1828, 8464, 13840, 8464, 1828, 112, 1, 1, 200, 5925, 52800, 182700, 273504, 182700, 52800, 5925, 200, 1, 1, 324, 15606, 233300, 1424925, 4050864, 5703096, 4050864, 1424925, 233300, 15606, 324, 1, 1, 490, 35623, 818300, 7917371, 37215794, 91789005, 123519792, 91789005, 37215794, 7917371, 818300, 35623, 490, 1, 1, 704, 73200, 2430400, 34657700, 246781248, 955910032, 2116980800, 2751843600, 2116980800, 955910032, 246781248, 34657700, 2430400, 73200, 704, 1

Offset: 0

Paul D. Hanna, Oct 30 2010

			Triangle begins:
 1;
 1, 4, 1;
 1, 20, 48, 20, 1;
 1, 54, 405, 760, 405, 54, 1;
 1, 112, 1828, 8464, 13840, 8464, 1828, 112, 1;
 1, 200, 5925, 52800, 182700, 273504, 182700, 52800, 5925, 200, 1;
 1, 324, 15606, 233300, 1424925, 4050864, 5703096, 4050864, 1424925, 233300, 15606, 324, 1; ...
Row g.f.s begin:
 (1) = (1-x)*(1 + x + x^2 + x^3 + x^4 +...);
 (1 + 4*x + x^2) = (1-x)^4*(1 + 2^3*x + 3^3*x^2 + 4^3*x^3 +...);
 (1 + 20*x + 48*x^2 + 20*x^3 + x^4) = (1-x)^7*(1 + 3^3*x + 6^3*x^2 +...);
 (1 + 54*x + 405*x^2 + 760*x^3 + 405*x^4 + 54*x^5 + x^6) = (1-x)^10*(1 + 4^3*x + 10^3*x^2 + 20^3*x^3 + 35^3*x^4 +...); ...

Paul D. Hanna, Table of n, a(n) for n = 0..1088, as a flattened triangle of rows 0..32
Ilia Gaiur, Vladimir Rubtsov, and Duco van Straten, Product formulas for the Higher Bessel functions, arXiv:2405.03015 [math.AG], 2024. See p. 18.
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 14.

Cf. A181543, A181545, A006480, A002897, A000172.

Cf. A183204 (central terms), A183205.

t[n_, k_] := SeriesCoefficient[Sum[Binomial[n+j, j]^3*x^j, {j, 0, n+k}]*(1-x)^(3*n+1), {x,0, k}]; Table[t[n, k], {n, 0, 9}, {k, 0, 2*n}] // Flatten (* Jean-François Alcover, Feb 04 2014, after PARI *)

{T(n,k)=polcoeff(sum(j=0,n+k,binomial(n+j,j)^3*x^j)*(1-x)^(3*n+1),k)}
for(n=0,10,for(k=0,2*n,print1(T(n,k),", "));print(""))

Row sums equal A006480(n) = (3n)!/(n!)^3, which is de Bruijn's s(3,n).
From Yahia Kahloune, Jan 30 2014: (Start)
Using these coefficients we can obtain formulas for the sums
Sum_{i=1..n} C(e-1+i,e)^3. Let us define b(k,e,3) = sum_{i=0..k-e} (-1)^i*C(3*e+1,i)*C(k-i,e)^3, where k=e+i.
For example:
b(e,e,3) = 1;
b(e+1,e,3) = (e+1)^3-(3*e+1) = e^2*(e+3);
b(e+2,e,3) = C(e+2,2)^3 - (3*e+1)*(e+1)^3 + C(3*e+1,2);
b(e+3,e,3) = C(e+3,e)^3 - (3*e+1)*C(e+2,e)^3 + C(3*e+1,2)*C(e+1,e)^3 - C(3*e+1,3);
b(e+4,e,3) = C(e+4,e)^3 - (3*e+1)*C(e+3,e)^3  + C(3*e+1,2)*C(e+2,e) - C(3*e+1,3)*C(e+1,e)^3 + C(3*e+1,4).
Then we have the formula: Sum_{i=1..n} C(e-1+i,e)^3 = Sum_{i=0..2*e} b(e+i,e,3)*C(n+e+i,3*e+1).
Example: Sum_{i=1..7} C(2+i,3)^3 = C(10,10) + 54*C(11,10) + 405*C(12,10) + 760*C(13,10) + 405*C(14,10) + 54*C(15,10) + C(16,10) = 820260. (End)
Let E be the operator D*x*D*x*D, where D denotes the derivative operator d/dx. Then (1/(n)!^3)  * E^n(1/(1 - x)) = (row n generating polynomial)/(1 - x)^(3*n+1) = Sum_{k >= 0} binomial(n+k, k)^3*x^k. For example, when n = 2 we have (1/2!)^3*E^3(1/(1 - x)) = (1 + 20 x + 48 x^2 + 20 x^3 + x^4)/(1 - x)^7. - Sergii Voloshyn, Dec 03 2024

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

Original entry on oeis.org

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

A094639 Partial sums of squares of Catalan numbers (A000108).

Original entry on oeis.org

1, 2, 6, 31, 227, 1991, 19415, 203456, 2248356, 25887400, 307993016, 3763786812, 47032778956, 598933188956, 7751562502556, 101741582076581, 1351906409905481, 18159677984049581, 246298405721739581

Offset: 0

André F. Labossière, May 27 2004

Koshy and Salmassi give an elementary proof that the only prime Catalan numbers are A000108(2) = 2 and A000108(3) = 5. Franklin T. Adams-Watters showed that the only semiprime Catalan number is A000108(4) = 14. The subsequence of primes in the partial sum of squares of Catalan numbers begins: 2, 31, 227, 101741582076581. [Jonathan Vos Post, May 27 2010]

Conjecture: For any positive integer n, the polynomial P_n(x) = sum_{k = 0}^n(C_k)^2*x^k (with C_k = binomial(2k, k)/(k+1)) is irreducible over the field of rational numbers. [Zhi-Wei Sun, Mar 23 2013]

Michael De Vlieger, Table of n, a(n) for n = 0..838
Paul Barry, Jacobsthal Decompositions of Pascal's Triangle, Ternary Trees, and Alternating Sign Matrices, Journal of Integer Sequences, 19, 2016, #16.3.5.
Joel E. Cohen, Variance Functions of Asymptotically Exponentially Increasing Integer Sequences Go Beyond Taylor's Law, J. Int. Seq., Vol. 25 (2022), Article 22.9.3.

Cf. A000108, A094638, A014137, A001246, A033536, A000984, A006134, A082894, A002897, A079727.

Accumulate[CatalanNumber[Range[0,20]]^2] (* Harvey P. Dale, May 01 2011 *)

a(n) = Sum_{k=0..n} ((2k)!/(k!)^2/(k+1))^2. - Alexander Adamchuk, Feb 16 2008
Sum_{i=1..n} [c(i)]^2 = Sum_{i=1..n} [C(2*i-2, i-1)/i]^2 = (1/(n-1)!)^2 * [ n^C(2*n-4, 1) + {2*C(n-1, 2)}*n^(2*n-5) + {C(n-2, 0) + 4*C(n-2, 1) + 13*C(n-2, 2) + 22*C(n-2, 3) + 12*C(n-2, 4)}*n^C(2*n-6, 1) + {12*C(n-3, 1) + 152*C(n-3, 2) + 458*C(n-3, 3) + 640*C(n-3, 4) + 440*C(n-3, 5) + 120*C(n-3, 6)}*n^(2*n-7) + {40*C(n-4, 0) + 313*C(n-4, 1) + 2332*C(n-4, 2) + 9536*C(n-4, 3) + 21409*C(n-4, 4) + 28068*C(n-4, 5) + 21700*C(n-4, 6) + 9240*C(n-4, 7) + 1680*C(n-4, 8) + ... + C(n-3, 0)*((n-1)!)^2 ].
Recurrence: (n+1)^2*a(n) = (17*n^2 - 14*n + 5)*a(n-1) - 4*(2*n - 1)^2*a(n-2). - Vaclav Kotesovec, Jul 01 2016
a(n) ~ 2^(4*n+4) /(15*Pi*n^3). - Vaclav Kotesovec, Jul 01 2016

A288455 Chebyshev coefficients of density of states of BCC lattice.

Original entry on oeis.org

1, -48, 1728, -79872, 4058112, -216956928, 11977752576, -676117610496, 38792847949824, -2253773963526144, 132241430641901568, -7821943674035503104, 465750331610495975424, -27888626411947306254336, 1677958399935741979262976, -101375476324084742212288512, 6146869366762959307806867456

Offset: 0

Yen-Lee Loh, Jun 16 2017

sign

This is the sequence of integers z^n g_n for n=0,2,4,6,... where g_n are the coefficients in the Chebyshev polynomial expansion of the density of states of the body-centered cubic lattice (z=8), g(w) = 1 / (Pi*sqrt(1-w^2)) * Sum_{n>=0} (2-delta_n) g_n T_n(w). Here |w| <= 1 and delta is the Kronecker delta.
The Chebyshev coefficients, g_n, are related to the number of walks on the lattice that return to the origin, W_n, as g_n = Sum_{k=0..n} a_{nk} z^{-k} W_k, where z is the coordination number of the lattice and a_{nk} are the coefficients of Chebyshev polynomials such that T_n(x) = Sum_{k=0..n} a_{nk} x^k.
For the bcc lattice (z=8), the even Chebyshev coefficients can be expressed in closed form in terms of the hypergeometric function pFq, as z^{2N} g_{2N} = (1 + delta_N) * 2^(2N-1) Binomial(2N,N)^3 * 4F3 (-N, -N, -N, -N; 1-2N, 1/2-N, 1/2-N; 1).

Yen Lee Loh, A general method for calculating lattice Green functions on the branch cut, arXiv:1706.03083 [math-ph], 2017.

Related to numbers of walks returning to origin, W_n, on BCC lattice (A002897).

zng[n_] :=
  If[OddQ[n], 0,
  (1 + KroneckerDelta[m]) 2^(2 m - 1) *
  Binomial[2m,m]^3 HypergeometricPFQ[{-m,-m,-m,-m}, {1-2m, 1/2-m, 1/2-m}, 1] /. m -> n/2];
Table[zng[n], {n,0,50}]
Wchain[n_] := If[OddQ[n], 0, Binomial[n, n/2]];
Wbcc[n_] := Wchain[n]^3;
ank[n_, k_] := SeriesCoefficient[ChebyshevT[n, x], {x, 0, k}];
zng[n_] := Sum[ank[n, k]*8^(n-k)*Wbcc[k], {k, 0, n}];
Table[zng[n], {n,0,50}]

A186415 a(n) = binomial(2n,n)^3/(n+1).

Original entry on oeis.org

1, 4, 72, 2000, 68600, 2667168, 112698432, 5053029696, 236860767000, 11493303192800, 573327757086656, 29253930349198464, 1521079361361956032, 80361335659444000000, 4304087536829486400000, 233271979857187430688000, 12774642558686527109607000, 706008965215713532853436000, 39337406606398593529683000000

Offset: 0

Emanuele Munarini, Feb 21 2011

Seiichi Manyama, Table of n, a(n) for n = 0..557

Cf. A000108, A002894, A000888, A002897, A186414.

A186415 := proc(n) binomial(2*n,n)^3/(n+1) ; end proc: # R. J. Mathar, Feb 23 2011

Mathematica
```
Table[Binomial[2n,n]^3/(n+1),{n,0,40}]
```

makelist(binomial(2*n,n)^3/(n+1),n,0,40);

G.f.: 3F2(1/2,1/2,1/2;1,2;64x), where 3F2(.,.,.;.,.;.) is a generalized hypergeometric series.
a(n) = A000888(n)*A000984(n). - R. J. Mathar, Feb 23 2011
a(n) ~ 64^n/(Pi^(3/2)*n^(5/2)). - Ilya Gutkovskiy, Nov 01 2016

A186414 a(n) = binomial(2n,n)^3/(n+1)^2.

Original entry on oeis.org

1, 2, 24, 500, 13720, 444528, 16099776, 631628712, 26317863000, 1149330319280, 52120705189696, 2437827529099872, 117006104720150464, 5740095404246000000, 286939169121965760000, 14579498741074214418000

Offset: 0

Emanuele Munarini, Feb 21 2011

Seiichi Manyama, Table of n, a(n) for n = 0..559

Cf. A000108, A002894, A000888, A002897, A186415.

[Binomial(2*n,n)^3/(n+1)^2: n in [0..50]]; // Vincenzo Librandi, Mar 27 2011

Table[Binomial[2n, n]^3/(n + 1)^2, {n, 0, 20}]

makelist(binomial(2*n,n)^3/(n+1)^2,n,0,40);

G.f.: 3F2({1/2, 1/2, 1/2}, {2, 2}, 64x), where 3F2 is a hypergeometric function.

A127776 a(n) = ( (2^n / n!) * Product_{k=0..n-1} (4*k + 1) )^2.

Original entry on oeis.org

1, 4, 100, 3600, 152100, 7033104, 344622096, 17582760000, 924193822500, 49701090010000, 2721631688947600, 151241747739534400, 8507348310348810000, 483459012855561960000, 27715027900230072360000, 1600820011517288979513600, 93072675982122379574532900

Offset: 0

Michael Somos, Jan 14 2007

nonn

Cayley (1878) refers to Gauss, Werke, t. iii, p. 424 for a slightly different form of a square of a hypergeometric series being hypergeometric. - Michael Somos, Jun 25 2012

			G.f. = 1 + 4*x + 100*x^2 + 3600*x^3 + 152100*x^4 + 7033104*x^5 + ...

A. Cayley, An Identity, Messenger of Mathematics, 7 (1878), p. 69

Gheorghe Coserea, Table of n, a(n) for n = 0..301
Robert S. Maier, On Rationally Parametrized Modular Equations, arXiv:math/0611041 [math.NT], 2006.

Cf. A002897, A004981, A091401, A186284.

[1] cat [n le 1 select 4 else (4*(4*n-3)^2*Self(n-1)/ n^2): n in [1..30]]; // Vincenzo Librandi, Aug 26 2016

From Michael Somos, Jun 25 2012 (Start)
a[ n_] := If[ n < 0, 0, (Pochhammer[ 1/4, n] 8^n / n!)^2];
a[ n_] := SeriesCoefficient[ Hypergeometric2F1[ 1/4, 1/4, 1, 64 x], {x, 0, n}];
a[ n_] := SeriesCoefficient[ HypergeometricPFQ[ {1/2, 1/2, 1/2}, {1, 1}, 64 x]^(1/2), {x, 0, n}];
(End)
Join[{1}, RecurrenceTable[{a[1] == 4, a[n] == (4 (4 n - 3)^2 a[n-1] / n^2)}, a, {n, 20}]] (* Vincenzo Librandi, Aug 26 2016 *)

{a(n) = if( n<0, 0, prod( k=1, n, (8*k - 6) / k)^2)};

{a(n) = my(A); if( n<1, n==0, A = x * O(x^n); polcoeff( subst( 1 / agm(1, sqrt(1 - 16*x + A) ), x, serreverse( x*(1 - 16*x) + A )), n))};

seq(N) = {
  my(a = vector(N)); a[1] = 4;
  for (n=2, N, a[n] = 4*(4*n-3)^2*a[n-1]/n^2);
  concat(1,a);
};
seq(15) \\ Gheorghe Coserea, Aug 26 2016

Expansion of K(k) / (Pi/2) in powers of (k * k'/4)^2, where K(k) is the complete elliptic integral of first kind evaluated at modulus k.
Expansion of 1 / AGM( 1, (1 - 16*x)^(1/2) ) in powers of x * (1 - 16*x) where AGM() is the arithmetic-geometric mean.
G.f.: F(1/4, 1/4; 1; 64*x).
a(n) = A004981(n)^2. Convolution square is A002897.
a(n) ~ 64^n / (Gamma(1/4)^2 * n^(3/2)). - Vaclav Kotesovec, Sep 08 2015
From Gheorghe Coserea, Aug 26 2016: (Start)
n^2 * a(n) = 4*(4*n-3)^2 * a(n-1), with a(0) = 1.
0 = 16*x*(x+64)*y'' + 8*(3*x+128)*y' + y, where y(x) = A(x/-4096). (End)

A176285 a(1) = 1, and then 4(2n + 1)^2a(n+1) + n^2a(n) = (205n^2 + 160n + 32)binomial(2n-1, n)^3 (n = 1, 2, 3, ...).

Original entry on oeis.org

1, 11, 316, 12011, 522376, 24593348, 1219951188, 62798884331, 3323228619736, 179665076698136, 9880531254032176, 550994628527745476, 31084678988906064016, 1770908612898043660556, 101738260887234550287316

Offset: 1

Zhi-Wei Sun, Apr 14 2010

On Apr 04 2010, Zhi-Wei Sun introduced this sequence and conjectured that each term a(n) is a positive integer. He also guessed that a(n) is odd if and only if n is a power of two. It is easy to see that 8*n^2*binomial(2*n,n)^2*a(n) equals Sum_{k=0..n-1} (205*k^2 + 160*k + 32)*(-1)^(n-1-k)*binomial(2*k,k)^5. Sun also conjectured that for any prime p > 5 we have Sum_{k=0..p-1} (205*k^2 + 160*k + 32)(-1)^k*binomial(2*k,k)^5 == 32*p^2 + 64*p^3*Sum_{k=1..p-1} 1/k (mod p^7) and Sum_{k=0..(p-1)/2} (205*k^2 + 160*k + 32)(-1)^k*binomial(2*k,k)^5 == 32*p^2 + 896/3*p^5*B_{p-3} (mod p^6), where B_0, B_1, B_2, ... are Bernoulli numbers. It is also remarkable that Sum_{n>0} (-1)^n(205*n^2 - 160*n + 32)/(n^5*binomial(2*n,n)^5) = -2*zeta(3) as proved by T. Amdeberhan and D. Zeilberger via the WZ method.

			For n = 2 we have a(2) = 11 since 4*(2*1 + 1)^2*a(2) = -1^2*a(1) + (205*1^2 + 160*1 + 32)*binomial(2*1 - 1,1)^3 = 396.

T. Amdeberhan and D. Zeilberger, Hypergeometric series acceleration via the WZ method, Electron. J. Combin. 4(1997), no.2, #R3.
Kasper Andersen, Re: A somewhat surprising conjecture
J. Guillera, Hypergeometric identities for 10 extended Ramanujan-type series, arXiv:1104.0396 [math.NT], 2011; Ramanujan J. 15(2008), 219-234.
Zhi-Wei Sun, Open Conjectures on Congruences, preprint, arXiv:0911.5665 [math.NT], 2009-2011.
Zhi-Wei Sun, A somewhat surprising conjecture
Zhi-Wei Sun, Re: A somewhat surprising conjecture

Cf. A000984, A001700, A002897, A173774.

a := proc(n) option remember; if n = 1 then 1 elif n = 2 then 11 else
 ( (209715*n^7 - 1828715*n^6 + 6716166*n^5 - 13424850*n^4 + 15714735*n^3 - 10726375*n^2 + 3936520*n - 598220)*a(n-1) + 8*(n - 2)^2*(2*n - 3)^3*(205*n^2 - 250*n + 77)*a(n-2) )/( 4*(n - 1)^3*(2*n - 1)^2*(205*n^2 - 660*n + 532) ) end if; end:
seq(a(n), n = 1..20); # Peter Bala, Jul 12 2024

u[n_]:=u[n]=((205(n-1)^2+160(n-1)+32)Binomial[2n-3,n-1]^3-(n-1)^2*u[n-1])/(4(2n-1)^2); u[1]=1; Table[u[n],{n,1,50}]

a(n) = (Sum_{k=0..n-1} (205*k^2 + 160*k + 32)(-1)^(n-1-k)*binomial(2*k,k)^5)/(8*n^2*binomial(2*n,n)^2).
From Peter Bala, Jul 11 2024: (Start)
a(n+1) = Sum_{k = 0..n} binomial(n, k)^2 * binomial(n+k, k) * binomial(2*n+k+1, n).
Compare with the identity Sum_{k = 0..n} binomial(n, k)^2 * binomial(n+k, k) * binomial(2*n+k, n) = binomial(2*n, n)^3 = A002897(n).
a(n) = binomial(2*n-1, n-1)*hypergeom([-n+1, -n+1, n, 2*n], [1, 1, n+1], 1).
P-recursive: 4*(205*n^2 - 250*n + 77)*(2*n + 1)^2*n^3*a(n+1) = (209715*n^7 - 360710*n^6 + 147891*n^5 + 65280*n^4 - 57280*n^3 + 3680*n^2 + 5120*n - 1024)*a(n) + 8*(205*n^2 + 160*n + 32)*(n - 1)^2*(2*n - 1)^3*a(n-1) with a(1) = 1 and a(2) = 11.
Conjecture: the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and all positive integers n and r. (End)
a(n) ~ 2^(6*n-3) / (5*Pi^(3/2)*n^(3/2)). - Vaclav Kotesovec, Jul 17 2024
a(n) = Sum_{k = 0..n-1} (-1)^(n+k+1) * binomial(n-1, k) * binomial(2*n+k-1, k) * A108625(n-1, k). Cf. A002897. - Peter Bala, Oct 16 2024

A183068 Central terms of triangle A183065.

Original entry on oeis.org

1, 26, 3246, 606500, 137915470, 35218238076, 9702014515116, 2818627826459016, 851612982884556750, 265166341958122567820, 84556145346599067308596, 27489903606068331188121816, 9081510922185418532993154796

Offset: 0

Paul D. Hanna, Dec 22 2010

Peter Bala, A recurrence for A183068

Cf. A183065, A183066, A183067.

seq(simplify(binomial(2*n, n)*hypergeom([-n, -n, n+1, n+1/2], [1, 1, 1], 4)), n = 0..20);

{a(n)=polcoeff(polcoeff(sum(m=0,2*n,(4*m)!/m!^4*x^(2*m)*y^m/(1-x-x*y+x*O(x^(2*n)))^(4*m+1)),2*n,x),n,y)}

a(n) = A183065(2*n,n).
a(n) = [x^(2*n)*y^n] Sum_{m >= 0} (4*m)!/m!^4 * x^(2*m)*y^m/(1-x-x*y)^(4*m+1).
From Peter Bala, Jul 15 2024: (Start)
a(n) = binomial(2*n, n)*Sum_{k = 0..n} binomial(n, k)^2*binomial(2*n+2*k, 2*k)* binomial(2*k, k) = Sum_{k = 0..n} (2*n+2*k)!/(k!^4*(n-k)!^2). Cf. A002897(n) = Sum_{k = 0..n} (2*n+k)!/(k!^3*(n-k)!^2) and A005258(n) = n!*Sum_{k = 0..n} (n+k)!/(k!^3*(n-k)!^2).
a(n) = binomial(2*n, n)*hypergeom([-n, -n, n+1, n+1/2], [1, 1, 1], 4).
Conjecture: the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(2*r)) holds for all primes p and positive integers n and r (End)
a(n) ~ sqrt(3) * (2 + sqrt(6))^(4*n + 3/2) / (16*Pi^2*n^2). - Vaclav Kotesovec, Jul 16 2024

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A181544 Triangle in which the g.f. for row n is [Sum_{k>=0} C(n+k-1,k)^3x^k](1-x)^(3n+1), read by rows of k=0..2n terms.

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A176285 a(1) = 1, and then 4(2n + 1)^2a(n+1) + n^2a(n) = (205n^2 + 160n + 32)binomial(2n-1, n)^3 (n = 1, 2, 3, ...).