A002894 -id:A002894 - cp's OEIS frontend

A288454 Chebyshev coefficients of density of states of square lattice.

1, -8, 32, -512, 4608, -73728, 819200, -13107200, 160563200, -2569011200, 33294385152, -532710162432, 7161992183808, -114591874940928, 1580900152246272, -25294402435940352, 355702534255411200, -5691240548086579200, 81223136710964019200, -1299570187375424307200, 18765793505701126995968

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 square lattice (z=4), 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 square lattice (z=4), 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)^2 * 3F2 (-N, -N, -N; 1-2N, 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 square lattice (A002894).

zng[n_] := If[OddQ[n], 0, (1+KroneckerDelta[m]) 2^(2m-1) Binomial[2m, m]^2 HypergeometricPFQ[{-m, -m, -m}, {1-2m,1/2-m}, 1] /. m->n/2];
Table[zng[n], {n,0,50}]
Wchain[n_] := If[OddQ[n], 0, Binomial[n, n/2]];
Wsq[n_] := Wchain[n]^2;
ank[n_, k_] := SeriesCoefficient[ChebyshevT[n, x], {x, 0, k}];
zng[n_] := Sum[ank[n, k]*4^(n - k)*Wsq[k], {k, 0, n}];
Table[zng[n], {n,0,50}]

A364303 Square array read by ascending antidiagonals: T(n,k) = [x^k] (1 - x)^(2k) Legendre_P(n*k, (1 + x)/(1 - x)) for n, k >= 0.

Original entry on oeis.org

1, 1, -2, 1, 0, 6, 1, 4, -6, -20, 1, 10, 36, 0, 70, 1, 18, 300, 400, 90, -252, 1, 28, 1050, 11440, 4900, 0, 924, 1, 40, 2646, 77616, 485100, 63504, -1680, -3432, 1, 54, 5544, 316540, 6370650, 21841260, 853776, 0, 12870, 1, 70, 10296, 972400, 42031990, 554822268, 1022041020, 11778624, 34650, -48620

Offset: 0

Peter Bala, Jul 19 2023

The first row of the table is a signed version of the central binomial coefficients A000984. The central binomial coefficients satisfy the supercongruences u(n*p^r) == u(n*p^(r-1)) (mod p ^(3*r)) for all primes p >= 5 and all positive integers n and r (see Meštrović, equation 39). We conjecture that each row sequence of the table satisfies the same supercongruences.

			 Square array begins:
 n\k|  0   1      2        3           4             5
  - + - - - - - - - - - - - - - - - - - - - - - - - - -
  0 |  1  -2      6      -20          70          -252    ...  (-1)^k*A000984(k)
  1 |  1   0     -6        0          90             0    ...  A245086
  2 |  1   4     36      400        4900         63504    ...  A002894
  3 |  1  10    300    11440      485100      21841260    ...  A275652
  4 |  1  18   1050    77616     6370650     554822268    ...  A275653
  5 |  1  28   2646   316540    42031990    5921058528    ...  A275654
  6 |  1  40   5544   972400   189290920   39089615040    ...  A275655
  7 |  1  54  10296  2484000   665091000  188907932304    ...  A364304
  8 |  1  70  17550  5567380  1960044750  732012601320    ...  A364305

R. Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862--2012), arXiv:1111.3057 [math.NT], 2011.
Wikipedia, Dixon's identity

Cf. A000984 (row 0 unsigned), A245086 (row 1), A002894 (row 2), A275652 (row 3), A275653 (row 4), A275654 (row 5), A275655 (row 6), A364304 (row 7), A364305 (row 8).

Cf. A364113, A364298

T(n,k) := coeff(series( (1 - x)^(2*k) * LegendreP(n*k, (1 + x)/(1 - x)), x, 11), x, k):
# display as a square array
seq(print(seq(T(n, k), k = 0..10)), n = 0..10);
# display as a sequence
seq(seq(T(n-k, k), k = 0..n), n = 0..10);

T(n,k) = Sum_{i = 0..k} binomial(n*k, k-i)^2 * binomial((n-2)*k+i-1, i).
For n >= 2, T(n,k) = binomial((n-1)*k, k)^2 * hypergeom([a, b, b], [1 + a - b, 1 + a - b], 1), where a = (n - 3)*k and b = -k.
For n >= 3, T(n,k) = ((n - 1)*k)! * ((n + 1)*k/2)! * ((n - 3)*k/2)! / ( ((n - 1)*k/2)!^2 * k!^2 *  ((n - 3)*k)! ) by Dixon's 3F2 summation theorem, where fractional factorials are defined in terms of the gamma function.

A364509 Square array read by ascending antidiagonals: T(n,k) = (2k)!/k!^2 ( (2nk)! * ((n + 2)k)! )/( (nk)! * ((n + 1)*k)!^2 ) for n, k > = 0.

Original entry on oeis.org

1, 1, 4, 1, 6, 36, 1, 16, 90, 400, 1, 50, 784, 1680, 4900, 1, 168, 8910, 48400, 34650, 63504, 1, 588, 113256, 2011100, 3312400, 756756, 853776, 1, 2112, 1528436, 96993024, 503909070, 240374016, 17153136, 11778624, 1, 7722, 21395520, 5056527000, 92279796840, 133954543800, 18116083216

Offset: 0

Peter Bala, Jul 28 2023

Given two sequences of integers c = (c_1, c_2, ..., c_K) and d = (d_1, d_2, ..., d_L) where c_1 + ... + c_K = d_1 + ... + d_L  we can define the factorial ratio sequence u_k(c, d) = (c_1*k)!*(c_2*k)!* ... *(c_K*k)!/ ( (d_1*k)!*(d_2*k)!* ... *(d_L*k)! ) and ask whether it is integral for all k >= 0. The integer L - K is called the height of the sequence. Bober completed the classification of integral factorial ratio sequences of height 1. Soundararajan gives many examples of two-parameter families of integral factorial ratio sequences of height 2.
Each row sequence of the present table is an integral factorial ratio sequence of height 2.
It is known that both row 0, the squares of the central binomial numbers, and row 1, the de Bruijn numbers, satisfy the supercongruences u(n*p^r) == u(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and all positive integers n and r. We conjecture that all the row sequences of the table satisfy the same supercongruences [added Oct 11 2024: follows from Meštrović, Section 6, equation 39, since T(n, k) = binomial(2*k, k) * binomial(2*n*k, n*k) * binomial((n+2)*k, k)/binomial((n+1)*k, k)].

			 Square array begins:
 n\k|  0    1        2           3               4                  5
  - + - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
  0 |  1    4       36         400            4900              63504 ... A002894
  1 |  1    6       90        1680           34650             756756 ... A006480
  2 |  1   16      784       48400         3312400          240374016 ... A364510
  3 |  1   50     8910     2011100       503909070       133954543800 ... A364511
  4 |  1  168   113256    96993024     92279796840     93172920645168 ...
  5 |  1  588  1528436  5056527000  18592935952500  72567511917065088 ...

Winston de Greef, Table of n, a(n) for n = 0..3240 (80 antidiagonals)
J. W. Bober, Factorial ratios, hypergeometric series, and a family of step functions, arXiv:0709.1977 [math.NT], 2007; J. London Math. Soc. (2) 79 2009, 422-444.
Romeo Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2012), arXiv:1111.3057 [math.NT], (2011).
K. Soundararajan, Integral factorial ratios: irreducible examples with height larger than 1, Phil. Trans. Royal Soc., A378: 2018044, 2019.
Wikipedia, Dixon's identity

A002894 (row 0), A006480 (row 1), A364510 (row 3), A364511 (row 4).

Cf. A364303, A364506.

 # display as a square array
T(n,k) := (2*k)!/k!^2 * ( (2*n*k)! * ((n + 2)*k)! )/( (n*k)! * ((n + 1)*k)!^2 ):
seq( print(seq(T(n,k), k = 0..10)), n = 0..10):
# display as a sequence
seq( seq(T(n-k,k), k = 0..n), n = 0..10);

T(n,k) = (2*k)!/k!^2 * ( (2*n*k)! * ((n + 2)*k)! )/( (n*k)! * ((n + 1)*k)!^2 ) \\ Winston de Greef, Oct 05 2023

T(n,k) = Sum_{i = -k..k} (-1)^i * binomial(2*k, k+i)^2 * binomial(2*n*k, n*k+i) (shows that the table entries are integers).
For n >= 1, T(n,k) = (-1)^k * binomial(2*n*k, (n+1)*k)^2 * hypergeom([-2*k, -2*k, -(n+1)*k], [1, 1 + (n-1)*k], 1) = (2*k)!/k!^2 * ( (2*n*k)! * ((n + 2)*k)! )/( (n*k)! * ((n + 1)*k)!^2 ) by Dixon's 3F2 summation theorem.
T(n,k) = (-1)^(n*k) * [x^((n+1)*k)] ( (1 - x)^(2*(n+1)*k) * Legendre_P(2*k, (1 + x)/(1 - x)) ). - Peter Bala, Aug 14 2023

A010370 a(n) = binomial(2n, n)^2 / (1-2n).

Original entry on oeis.org

1, -4, -12, -80, -700, -7056, -77616, -906048, -11042460, -139053200, -1796567344, -23696871744, -317933029232, -4326899214400, -59605244280000, -829705000377600, -11654762427179100, -165021757273414800, -2353088020380174000, -33764531705178120000

Offset: 0

Joe Keane (jgk(AT)jgk.org)

Expansion of hypergeometric function F(-1/2, 1/2; 1; 16*x).

Expansion of E(m)/(Pi/2) in powers of m/16 = (k/4)^2, where E(m) is the complete elliptic integral of the second kind evaluated at m. - Michael Somos, Mar 04 2003

			G.f. = 1 - 4*x - 12*x^2 - 80*x^3 - 700*x^4 - 7056*x^5 - 77616*x^6 - ...

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 591.
J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 8.

Robert Israel, Table of n, a(n) for n = 0..835
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].

Cf. A000891, A000894, A002420, A002894

seq(binomial(2*n,n)^2/(1-2*n), n=0..30); # Robert Israel, Jul 10 2017

CoefficientList[Series[EllipticE[16x]2/Pi, {x, 0, 20}], x]
Table[Binomial[2n,n]^2/(1-2n),{n,0,30}] (* Harvey P. Dale, Mar 07 2013 *)

{a(n) = binomial(2*n, n)^2 / (1 - 2*n)}; /* Michael Somos, Dec 13 2002 */

a(n) ~ -1/2*Pi^-1*n^-2*2^(4*n). [corrected by Vaclav Kotesovec, Oct 04 2019]
a(n) = -4 * A000891(n-1), n>0. - Michael Somos, Dec 13 2002
G.f.: F(-1/2, 1/2; 1; 16x) = E(16*x) / (Pi/2). a(n) = binomial(2*n, n)^2 / (1 - 2*n). - Michael Somos, Mar 04 2003
E.g.f.: Sum_{n>=0} a(n) * (x/2)^(2n)/(2n)! = I0^2*(1-2*x^2) +2*x*I0*I1 +2*x^2*I1^2 where I0=BesselI(0, x), I1=BesselI(1, x). - Michael Somos, Jun 22 2005
n^2*a(n) -4*(2*n-1)*(2*n-3)*a(n-1)=0. - R. J. Mathar, Feb 15 2013
0 = a(n)*(+1048576*a(n+2) + 2695168*a(n+3) - 989568*a(n+4) + 65340*a(n+5)) + a(n+1)*(-8192*a(n+2) - 99840*a(n+3) + 52652*a(n+4) - 4236*a(n+5)) + a(n+2)*(-128*a(n+2) + 280*a(n+3) - 484*a(n+4) + 57*a(n+5)) for all n in Z. - Michael Somos, Jan 21 2017
a(n) = A002894(n) - 8 * A000894(n-1). - Michael Somos, Jul 10 2017

Additional comments from Michael Somos, Dec 13 2002

A110257 Numerators in the coefficients that form the odd-indexed partial quotients of the continued fraction representation of the inverse tangent of 1/x.

Original entry on oeis.org

1, 5, 81, 325, 20825, 83349, 1334025, 5337189, 1366504425, 5466528925, 87470372561, 349899121845, 22394407746529, 89580335298125, 1433319858545625, 5733391194015525, 5871086572691471625

Offset: 1

Paul D. Hanna, Jul 18 2005

Limit a(n)/A110258(n) = limit A110255(2*n-1)/A110256(2*n-1) = 4/Pi.

			arctan(1/x) = 1/x - 1/(3*x^3) + 1/(5*x^5) - 1/(7*x^7) +-...
= [0; x, 3*x, 5/4*x, 28/9*x, 81/64*x, 704/225*x, 325/256*x, 768/245*x, 20825/16384*x, 311296/99225*x, 83349/65536*x, 1507328/480249*x, 1334025/1048576*x, 3145728/1002001*x,...]
= 1/(x + 1/(3*x + 1/(5/4*x + 1/(28/9*x + 1/(81/64*x +...))))).
The coefficients of x in the even-indexed partial quotients converge to Pi: {3, 28/9, 704/225, 768/245, 311296/99225, ...}.
The coefficients of x in the odd-indexed partial quotients converge to 4/Pi: {1, 5/4, 81/64, 325/256, 20825/16384, ...}.

Paul D. Hanna, Table of n, a(n) for n = 1..200

Cf. A110258 (denominators), A110255/A110256 (continued fraction), A110259/A110260.

a := n -> (4*n+1)*binomial(2*n,n)^2/4^(add(i,i=convert(n,base,2)));
seq(a(n), n=0..16);  # Peter Luschny, Mar 23 2014

a[n_] := (4n+1) Binomial[2n, n]^2 / 4^DigitCount[n, 2, 1];
Array[a, 16] (* Jean-François Alcover, Jun 13 2019, from Maple *)

{a(n)=numerator(subst((contfrac( sum(k=0,2*n+1,(-1)^k/x^(2*k+1)/(2*k+1)),2*n+2))[2*n],x,1))}

a(n) = A110255(2*n-1).

a(n) = (4*n+1)*A002894(n)/4^A000120(n). - Peter Luschny, Mar 23 2014

A135389 Number of walks of length 2*n+2 from origin to (1,1) in a square lattice.

Original entry on oeis.org

2, 24, 300, 3920, 52920, 731808, 10306296, 147232800, 2127513960, 31031617760, 456164781072, 6749962774464, 100445874620000, 1502052155856000, 22557604697766000, 340044833169460800, 5143178101688094600

Offset: 0

Stefan Hollos (stefan(AT)exstrom.com), Dec 11 2007

a(n) is the number of walks of length 2n+2 in an infinite square lattice that begin at the origin and end at (1,1) using steps (1,0), (-1,0), (0,1), (0,-1).

			G.f. = 2 + 24*x + 300*x^2 + 3920*x^3 + 731808*x^4 + 10306296*x^5 + ... - _Michael Somos_, Oct 17 2019

S. Hollos and R. Hollos, Lattice Paths and Walks.

Cf. A002894, A060150, A337900-A337902.

series( 2*hypergeom([3/2, 3/2],[3],16*x), x=0, 20);  # Mark van Hoeij, Apr 06 2013

Table[Binomial[2n + 2, n] Binomial[2n + 2, n + 1], {n, 0, 19}] (* Alonso del Arte, Apr 06 2013 *)

a(n) = binomial(2n+2,n) * binomial(2n+2,n+1) = A001791(n+1)*A000984(n+1).
G.f.: 2*2F1(3/2,3/2; 3; 16*x). - Mark van Hoeij, Apr 06 2013
D-finite with recurrence n*(n+2)*a(n) -4*(2*n+1)^2*a(n-1)=0. - R. J. Mathar, Jul 14 2013
E.g.f.: Sum_{n>0} a(n-1) * x^(2*n)/(2*n)! = BesselI(1, 2*x)^2. - Michael Somos, Oct 17 2019

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

A275654 a(n) = (5n)!/((3n)!n!^2) ((3n/2)!(7n/2)!)/(5n/2)!^2.

Original entry on oeis.org

1, 28, 2646, 316540, 42031990, 5921058528, 866486466720, 130220534668224, 19958454291525750, 3105489721784166640, 489023391870111994896, 77758775451291032116200, 12464212878673327376454304, 2011515147856766922731424000

Offset: 0

Peter Bala, Aug 04 2016

Right-hand side of the binomial sum identity Sum_{k = 0..n} (-1)^(n+k)*binomial(5*n + k,5*n - k)*binomial(2*k,k)*binomial(2*n - k,n) = (5*n)!/((3*n)!*n!^2) * ((3*n/2)!*(7*n/2)!)/(5*n/2)!^2.
We also have Sum_{k = 0..5*n} (-1)^k*binomial(5*n + k,5*n - k)* binomial(2*k,k) *binomial(2*n - k,n) = (5*n)!/((3*n)!*n!^2) * ((3*n/2)!*(7*n/2)!)/(5*n/2)!^2.
Compare with Sum_{k = 0..n} (-1)^(n+k)*binomial(2*n + k,2*n - k)*binomial(2*k,k)*binomial(2*n - k,n) = binomial(2*n,n)^2 = A002894(n). See also A275652, A275653 and A275655.

Cf. A275652, A275653, A275655, A276098, A276100, A276101, A276102, A352651, A352652.

seq(simplify(factorial(3*n/2)*factorial(5*n)*factorial(7*n/2)/(factorial(n)^2*factorial(3*n)*factorial(5*n/2)^2)), n = 0 .. 20);

Table[(5 n)!/((3 n)! n!^2) ((3 n/2)! (7 n/2)!)/(5 n/2)!^2, {n, 0, 13}] (* Michael De Vlieger, Aug 07 2016 *)

a(n) = sum(k = 0, n, binomial(4*n-k-1,n-k)*binomial(5*n,k)^2); \\ Michel Marcus, Apr 21 2022

from math import factorial
from sympy import factorial2
def A275654(n): return int(factorial(5*n)*factorial2(3*n)*factorial2(7*n)//factorial(3*n)//factorial(n)**2//factorial2(5*n)**2) # Chai Wah Wu, Aug 08 2023

a(n) = (3*n/2)!*(5*n)!*(7*n/2)!/(n!^2*(3*n)!*(5*n/2)!^2).
Recurrence: 3*a(n)*n^2*(n - 1)^2*(3*n - 1)*(3*n - 5)*(5*n - 2)*(5*n - 4)*(5*n - 6)*(5*n - 8) = 7*(5*n - 1)*(5*n - 3)*(5*n - 7)*(5*n - 9)*(7*n - 2)*(7*n - 4)*(7*n - 6)*(7*n - 8)*(7*n - 10)*(7*n - 12)*a(n-2).
a(n) = [x^n] G(x)^(7*n) where G(x) = 1 + 4*x + 85*x^2+ 4220*x^3 + 283285*x^4 + 22308156*x^5 + 1939419083*x^6 + ... appears to have integer coefficients.
exp( Sum_{n >= 1} a(n)*x^n/n ) = F(x)^7, where F(x) = 1 + 4*x + 197*x^2 + 15840*x^3 + 1580819*x^4 + 178220584*x^5 + 21729476664*x^6 + ... appears to have integer coefficients.
a(n) ~ 7^(7*n/2+1/2)/(2*sqrt(5)*Pi*3^(3*n/2)*n). - Ilya Gutkovskiy, Aug 07 2016
From Peter Bala, Mar 23 2022: (Start)
a(n) = Sum_{k = 0..n} binomial(4*n-k-1,n-k)*binomial(5*n,k)^2.
For n >= 1, a(n) = (7/5)*binomial(m*n,2*n)*binomial(m*n/2,2*n)* binomial(2*n,n)^2/binomial(m*n/2,n)^2 at m = -3. Se A352651 for the case m = 1.
a(n) = [x^n] (1 - x)^(2*n) * P(5*n,(1 + x)/(1 - x)), where P(n,x) denotes the n-th Legendre polynomial. Cf. A275652.
a(p) == a(1) (mod p^3) for primes p >= 5.
Conjecture: The supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and all positive integers n and k. (End)

A067804 Triangle read by rows: T(n,k) is the number of walks (each step +-1) of length 2n which have a cumulative value of 0 last at step 2k.

Original entry on oeis.org

1, 2, 2, 6, 4, 6, 20, 12, 12, 20, 70, 40, 36, 40, 70, 252, 140, 120, 120, 140, 252, 924, 504, 420, 400, 420, 504, 924, 3432, 1848, 1512, 1400, 1400, 1512, 1848, 3432, 12870, 6864, 5544, 5040, 4900, 5040, 5544, 6864, 12870, 48620, 25740, 20592, 18480

Offset: 0

Henry Bottomley, Feb 07 2002

Since the triangle is symmetric, the probability that a one-dimensional random walk returns to the origin at all in the steps m through to 2m is 1/2 (for m odd).

Diagonal sums are A106183. - Paul Barry, Apr 24 2005

			Triangle begins:
    1;
    2,   2;
    6,   4,   6;
   20,  12,  12,  20;
   70,  40,  36,  40,  70;
  252, 140, 120, 120, 140, 252;
  ...
For a walk of length 4 (=2*2), 6 are only ever 0 at step 0, 4 are zero at step 2 but not step 4 and 6 are 0 at step 4.
For n=3,k=2, T(3,2)=12 since there are 12 monotonic paths from (0,0) to (2,2) and then on to (3,3). Using E for eastward steps and N for northward steps, the 12 paths are given by EENNNE, ENENNE, ENNENE, NNEENE, NENENE, NEENNE, EENNEN, ENENEN, ENNEEN, NNEEEN, NENEEN, NEENEN. - _Dennis P. Walsh_, Mar 23 2012

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 79, ex. 3f.

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
B. C. Carlson, Power series for inverse Jacobian elliptic functions, Math. Comp., 77 (2008), 1615-1621, see p. 1617, equation (2.20).
C. M. Grinstead and J. L. Snell, Introduction to Probability p. 482.
R. P. Kelisky, Inverse elliptic functions and Legendre polynomials, Amer. Math. Monthly 66 (1959), pp. 480-483. MR0103993 (21 #2755).
Michael Z. Spivey, A Combinatorial Proof for the Alternating Convolution of the Central Binomial Coefficients, The American Mathematical Monthly 121.6 (2014): 537-540. [Suggested by _Roger L. Bagula_, Jun 21 2014]

Columns include A000984, A028329. Central diagonal is A002894.

Cf. A002426, A120406.

/* As triangle */ [[Binomial(2*k, k)*Binomial(2*n-2*k, n-k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Jul 19 2015

Table[Table[Binomial[2k,k]Binomial[2(n-k),n-k],{k,0,n}],{n,0,10}]//Grid  (* Geoffrey Critzer, Jun 30 2013 *)
T[ n_, k_] := SeriesCoefficient[ D[ InverseJacobiSN[2 x, m] / 2, x], {x, 0, 2 n}, {m, 0, k}]; (* Michael Somos, May 06 2017 *)

T(n, k) = binomial(2*k, k) * binomial(2*n-2*k, n-k) /* Michael Somos, Aug 20 2005 */

T(n, k) = C(2k, k)*C(2n-2k, n-k) = C(2n, n)*C(n, k)^2/C(2n, 2k) = A000984(k)*A000984(n-k) = A000984(n)*A008459(n, k)/A007318(2n, 2k).
Row sums are 4^n = A000302(n).
G.f.: A(x,y) = 1/sqrt((1-4*x)*(1-4*x*y)). - Vladeta Jovovic, Dec 12 2003
Sum{k>=0} T(n, k)*(-3)^k = (-4)^n * A002426(n). Sum_{k>=0} T(n, k)/(2*k+1) = 2^(4*n)/((2*n+1)*C(2*n, n)). - Philippe Deléham, Dec 31 2003
O.g.f.: A(x,y) = 1 + x*d/dx(log(B(x,y))), where B(x,y) is the o.g.f. of A120406. - Peter Bala, Jul 17 2015

A069466 Triangle T(n, k) of numbers of square lattice walks that start and end at origin after 2*n steps and contain exactly k steps to the east, possibly touching origin at intermediate stages.

Original entry on oeis.org

1, 2, 2, 6, 24, 6, 20, 180, 180, 20, 70, 1120, 2520, 1120, 70, 252, 6300, 25200, 25200, 6300, 252, 924, 33264, 207900, 369600, 207900, 33264, 924, 3432, 168168, 1513512, 4204200, 4204200, 1513512, 168168, 3432, 12870, 823680, 10090080, 40360320, 63063000, 40360320, 10090080, 823680, 12870

Offset: 0

Martin Wohlgemuth, Mar 24 2002

A Pólya plane walk takes steps (N,E,S,W) along cardinal directions in the plane, visiting only points of Z^2 (cf. Links). T(n,k) is the number of walks departing from and returning to the origin, with exactly 2*k steps along the NS axis and 2*(n-k) steps along the EW direction. Equivalently, triangle T(n,k) is the number of distinct permutations of a 2*n-letter word with letters (N,E,S,W) in multiplicity (k,n-k,k,n-k). Moving only along either NS or EW directions, T(n,0) = T(n,n) = A000894(n). Row sums appear as Equation 4 in the original Pólya article, Sum_{k=0..n} T(n,k) = A002894(n). This identity is proven routinely using Zeilberger's algorithm. - Bradley Klee, Aug 12 2018

			Triangle begins:
    1,
    2,    2,
    6,   24,     6,
   20,  180,   180,    20,
   70, 1120,  2520,  1120,   70,
  252, 6300, 25200, 25200, 6300, 252
  ...
T(4,2) = 2520 because there are 2520 distinct lattice walks of length 2*4=8 starting and ending at the origin and containing exactly 2 steps to the east.
For T(2,k), the lattice-path words are:
T(2,0):{EEWW, WEEW, WWEE, EWWE, WEWE, EWEW}
T(2,1):{NESW, NEWS, NSEW, NSWE, NWES, NWSE, ENSW, ENWS, ESNW, ESWN, EWNS, EWSN, SNEW, SNWE, SENW, SEWN, SWNE, SWEN, WNES, WNSE, WENS, WESN, WSNE, WSEN}
T(2,2):{NNSS, SNNS, SSNN, NSSN, SNSN, NSNS}

Muniru A Asiru, Table of n, a(n) for n = 0..1325 (Rows n=0..50, flattened)
A. Bostan, Calcul Formel pour la Combinatoire des Marches, Habilitation à Diriger des Recherches, Université Paris 13, 2017, page 11.
G. Pólya, Über eine Aufgabe der Wahrscheinlichkeitsrechnung betreffend die Irrfahrt im Straßennetz, Mathematische Annalen, 84 (1921), 149-160.

T(2*n, n) = A008977(n).
Cf. A002894, A000984.
Cf. A007318 (Pascal, m=1), this sequence (m=2), A320824 (m=3).

T:=Flat(List([0..8],n->List([0..n],k->Binomial(2*n,n)*(Binomial(n,k))^2))); # Muniru A Asiru, Oct 21 2018

T:=(n,k)->binomial(2*n,n)*(binomial(n,k))^2: seq(seq(T(n,k),k=0..n),n=0..8); # Muniru A Asiru, Oct 21 2018

T[k_, r_] := Binomial[2k, k]*Binomial[k, r]^2; Table[T[k, r], {k, 0, 8}, {r, 0, k}] // Flatten (* Jean-François Alcover, Nov 21 2012, from explicit formula *)

Recurrences: T(1, 0) = T(1, 1)=2; T(k, r) = 2*k*(2*k-1)/(k-r)^2 * T(k-1, r); T(k, r) = (k+1-r)^2/r^2 * T(k, r-1).
T(n, k) = binomial(2*n, n) * binomial(n, k)^2.
Sum_{k=0..n} T(n, k) = A002894(n).
From Bradley Klee, Aug 12 2018: (Start)
T(n,k) = (2*n)!/((n-k)!*k!)^2.
T(n,k) = C(2*n,2*k)*C(2*(n-k),n-k)*C(2*k,k).
Sum_{k=0..n} T(n,k) = Sum_{k=0..n} C(2*n,2*k)*C(2*(n-k),n-k)*C(2*k,k) = C(2*n,n)^2.
Sum_{k=0..n} T(n,k) = Sum_{k=0..n} (2n)!/(k!(n-k)!)^2 = C(2*n,n)^2.
(End)

cp's OEIS Frontend

A288454 Chebyshev coefficients of density of states of square lattice.

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A364303 Square array read by ascending antidiagonals: T(n,k) = [x^k] (1 - x)^(2*k) * Legendre_P(n*k, (1 + x)/(1 - x)) for n, k >= 0.

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A364509 Square array read by ascending antidiagonals: T(n,k) = (2*k)!/k!^2 * ( (2*n*k)! * ((n + 2)*k)! )/( (n*k)! * ((n + 1)*k)!^2 ) for n, k > = 0.

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A010370 a(n) = binomial(2*n, n)^2 / (1-2*n).

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A110257 Numerators in the coefficients that form the odd-indexed partial quotients of the continued fraction representation of the inverse tangent of 1/x.

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A135389 Number of walks of length 2*n+2 from origin to (1,1) in a square lattice.

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

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A364303 Square array read by ascending antidiagonals: T(n,k) = [x^k] (1 - x)^(2k) Legendre_P(n*k, (1 + x)/(1 - x)) for n, k >= 0.

A364509 Square array read by ascending antidiagonals: T(n,k) = (2k)!/k!^2 ( (2nk)! * ((n + 2)k)! )/( (nk)! * ((n + 1)*k)!^2 ) for n, k > = 0.

A010370 a(n) = binomial(2n, n)^2 / (1-2n).

A275654 a(n) = (5n)!/((3n)!n!^2) ((3n/2)!(7n/2)!)/(5n/2)!^2.