A000515 -id:A000515 - cp's OEIS frontend

A002894 a(n) = binomial(2n, n)^2.

1, 4, 36, 400, 4900, 63504, 853776, 11778624, 165636900, 2363904400, 34134779536, 497634306624, 7312459672336, 108172480360000, 1609341595560000, 24061445010950400, 361297635242552100, 5445717990022688400, 82358080713306090000, 1249287673091590440000

Offset: 0

N. J. A. Sloane

a(n) is the number of monotonic paths (only moving N and E) in the lattice [0..2n] X [0..2n] that contain the points (0,0), (n,n) and (2n,2n). - Joe Keane (jgk(AT)jgk.org), Jun 06 2002
This is the Taylor expansion of a special point on a curve described by Beauville. - Matthijs Coster, Apr 28 2004
Expansion of K(k) / (Pi/2) in powers of m/16 = (k/4)^2, where K(k) is the complete elliptic integral of the first kind evaluated at k. - Michael Somos, Mar 04 2003
Square lattice walks that start and end at origin after 2n steps. - Gareth McCaughan and Michael Somos, Jun 12 2004
If A is a random matrix in USp(4) (4 X 4 complex matrices that are unitary and symplectic) then a(n)=E[(tr(A^k))^{2n}] for any k > 4. - Andrew V. Sutherland, Apr 01 2008
From R. H. Hardin, Feb 03 2016 and R. J. Mathar, Feb 18 2016: (Start)
Also, number of 2 X (2n) arrays of permutations of 2n copies of 0 or 1 with row sums equal.
For example, some solutions for n=3:
  0 1 0 1 0 1   0 1 0 1 1 0   0 0 1 0 1 1   1 1 1 0 0 0
  1 0 0 0 1 1   1 1 0 1 0 0   0 0 0 1 1 1   0 0 1 1 0 1
There is a simple combinatorial argument to show that this is a(n): We have 2n copies of 0's and 1's and need equal row sums. Therefore there must be n 1's in each of the two rows. Otherwise there are no constraints, so there are C(2n,n) ways of placing the 1's in the first row and independently C(2n,n) ways of placing the 1's in the second. The product is clearly C(2n,n)^2. (End)
Also the even part of the bisection of A241530. One half of the odd part is given in A000894. - Wolfdieter Lang, Sep 06 2016
From Peter Bala, Jan 26 2018: (Start)
Let S = {[1,0,0], [0,1,0], [1,0,1], [0,1,1]} be a set of four column vectors. Then a(n) equals the number of 3 X k arrays whose columns belong to the set S and whose row sums are all equal to n (apply Eger, Theorem 3). An example is given below. Equivalently, a(n) equals the number of lattice paths from (0,0,0) to (n,n,n) using steps (1,0,0), (0,1,0), (1,0,1) and (0,1,1).
The o.g.f. for the sequence equals the diagonal of the rational function 1/(1 - (x + y + x*z + y*z)).
Row sums of A069466. (End)
Also, the constant term in the expansion of (x + 1/x + y + 1/y)^(2n). - Christopher J. Smyth, Sep 26 2018
Number of ways to place 2n^2 nonattacking pawns on a 2n x 2n board. - Tricia Muldoon Brown, Dec 12 2018
For n>0, a(n) is the number of Littlewood polynomials of degree 4n-1 that have a closed Lill path. A polynomial p(x) has a closed Lill path if and only if p(x) is divisible by x^(2)+1. - Raul Prisacariu, Aug 28 2024

			G.f. = 1 + 4*x + 36*x^2 + 400*x^3 + 4900*x^4 + 63504*x^5 + 853776*x^6 + ... - _Michael Somos_, Aug 06 2014
From _Peter Bala_, Jan 26 2018: (Start)
a(2) = 36: The thirty six 3 x k arrays with columns belonging to the set of column vectors S = {[1,0,0], [0,1,0], [1,0,1], [0,1,1]} and having all row sums equal to 2 are the 6 distinct arrays obtained by permuting the columns of
  /1 1 0 0\
  |0 0 1 1|,
  \0 0 1 1/
the 6 distinct arrays obtained by permuting the columns of
  /0 0 1 1\
  |1 1 0 0|
  \0 0 1 1/
and the 24 arrays obtained by permuting the columns of
  /1 0 1 0\
  |0 1 0 1|. (End)
  \0 0 1 1/

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 591,828.
J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 8.
Matthijs Coster, Over 6 families van krommen [On 6 families of curves], Master's Thesis (unpublished), Aug 26 1983.
Leonard Lipshitz and A. van der Poorten. "Rational functions, diagonals, automata and arithmetic." In Number Theory, Richard A. Mollin, ed., Walter de Gruyter, Berlin (1990): 339-358.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..100
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].
R. Bacher, Meander algebras, Institut Fourier, 1999.
E. Barcucci, A. Frosini and S. Rinaldi, On directed-convex polyominoes in a rectangle, Discr. Math., 298 (2005). 62-78.
Arnaud Beauville, Les familles stables de courbes elliptiques sur P^1 admettant quatre fibres singulières, Comptes Rendus, Académie Sciences Paris, no. 294, May 24 1982, 657-660. MR0664643 (83h:14008)
Alin Bostan, Armin Straub, and Sergey Yurkevich, On the representability of sequences as constant terms, arXiv:2212.10116 [math.NT], 2022.
Tricia Muldoon Brown, The Problem of Pawns, The Electronic Journal of Combinatorics (2019) Vol. 26, Issue 3, #P3.21. Also arXiv:1811.09606, [math.CO], 2018.
John Maxwell Campbell, New series involving harmonic numbers and squared central binomial coefficients, Rocky Mountain J. Math., 49 (2019), 2513-2544.
C. Domb, On the theory of cooperative phenomena in crystals, Advances in Phys., 9 (1960), 149-361.
Steffen Eger, On the Number of Many-to-Many Alignments of N Sequences, arXiv:1511.00622 [math.CO], 2015.
Murray Elder, Cogrowth, 2011.
M. Elder, A. Rechnitzer, E. J. Janse van Rensburg, and T. Wong, The cogrowth series for BS(N,N) is D-finite, arXiv:1309.4184 [math.GR], 2013.
Philippe Flajolet and Robert Sedgewick, Analytic Combinatorics, 2009; see page 90.
Florian Fürnsinn and Sergey Yurkevich, Algebraicity of hypergeometric functions with arbitrary parameters, arXiv:2308.12855 [math.CA], 2023.
Davidson Noby Joseph and Igor Boettcher, Walking on Archimedean Lattices: Insights from Bloch Band Theory, arXiv:2507.12662 [cond-mat.stat-mech], 2025. See p. 18.
Kiran S. Kedlaya and Andrew V. Sutherland, Hyperelliptic curves, L-polynomials and random matrices, arXiv:0803.4462 [math.NT], 2010.
Markus Kuba and Alois Panholzer, Lattice paths and the diagonal of the cube, arXiv:2411.03930 [math.CO], 2024. See p. 14.
L. Lipshitz and A. J. van der Poorten, Rational functions, diagonals, automata and arithmetic
Raul Prisacariu, Littlewood Polynomials of Degree n with Closed Lill Paths
Eric M. Rains, High powers of random elements of compact Lie groups, Probability Theory and Related Fields 107 (1997), 219-241.
Grzegorz Siudem and Agata Fronczak, Bell polynomials in the series expansions of the Ising model, arXiv:2007.16132 [math-ph], 2020.
Eric Weisstein's World of Mathematics, Lattice Path.
D. Zagier, Integral solutions of Apéry-like recurrence equations. See line G in sporadic solutions table of page 5.

Row sums of A069466.
Row 2 of A268367 (even terms).
Equals 4*A060150.
Cf. A000984, A000515, A010370, A054474 (INVERTi transform), A172390, A000897, A002897, A006480, A008977, A186420, A188662, A000894, A241530, A002898 (walks hex lattice).

[Binomial(2*n, n)^2: n in [0..20]]; // Vincenzo Librandi, Aug 07 2014

Maple
```
A002894 := n-> binomial(2*n,n)^2.
```

CoefficientList[Series[Hypergeometric2F1[1/2, 1/2, 1, 16x], {x, 0, 20}], x]
Table[Binomial[2n,n]^2,{n,0,20}] (* Harvey P. Dale, Jul 06 2011 *)
a[ n_] := SeriesCoefficient[ EllipticK[16 x] / (Pi/2), {x, 0, n}]; (* Michael Somos, Aug 06 2014 *)
a[n_] := 16^n HypergeometricPFQ[{1/2, -2 n, 2 n + 1}, {1, 1}, 1];
Table[a[n], {n, 0, 19}] (* Peter Luschny, Mar 14 2018 *)

PARI
```
{a(n) = binomial(2*n, n)^2};
```

{a(n) = if( n<0, 0, polcoeff( polcoeff( polcoeff( 1 / (1 - x * (y + z + 1/y + 1/z)) + x * O(x^(2*n)), 2*n), 0), 0))}; /* Michael Somos, Jun 12 2004 */

[binomial(2*n, n)**2 for n in range(17)]  # Zerinvary Lajos, Apr 21 2009

D-finite with recurrence: (n+1)^2*a(n+1) = 4*(2*n + 1)^2*a(n). - Matthijs Coster, Apr 28 2004
a(n) ~ Pi^(-1)*n^(-1)*2^(4*n). - Joe Keane (jgk(AT)jgk.org), Jun 06 2002
G.f.: F(1/2, 1/2; 1; 16*x) = 1 / AGM(1, (1 - 16*x)^(1/2)) = K(4*sqrt(x)) / (Pi/2), where AGM(x, y) is the arithmetic-geometric mean of Gauss and Legendre. - Michael Somos, Mar 04 2003
G.f.: 2*EllipticK(4*sqrt(x))/Pi, using Maple's convention for elliptic integrals.
E.g.f.: Sum_{n>=0} a(n)*x^(2*n)/(2*n)! = BesselI(0, 2x)^2.
a(n) = A000984(n)^2. - Jonathan Vos Post, Jun 17 2007
E.g.f.: (BesselI(0, 2*x))^2 = 1+2*x^2/(U(0)-2*x^2); U(k) = 2*x^2*(2*k+1)+(k+1)^3-2*x^2*(2*k+3)*(k+1)^3/U(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 23 2011
In generally, for (BesselI(b, 2x))^2=((x^(2*b))/(GAMMA(b+1))^2)*(1+(2*x^2)*(2*b+1)/(Q(0)-(2*x^2)*(2*b+1)); Q(k)=(2*x^2)*(2*k+2*b+1)+(k+1)*(k+b+1)*(k+2*b+1)-(2*x^2)*(k+1)*(k+b+1)*(k+2*b+1)*(2*k+2*b+3)/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, Nov 23 2011
G.f.: G(0)/2, where G(k)= 1 + 1/(1 - 4*(2*k+1)^2*x*(1+4*x)^2/(4*(2*k+1)^2*x*(1+4*x)^2 + (k+1)^2*(1+4*x)^2/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 01 2013
0 = +a(n)*(+393216*a(n+2) -119040*a(n+3) +6860*a(n+4)) +a(n+1)*(-16128*a(n+2) +6928*a(n+3) -465*a(n+4)) +a(n+2)*(+36*a(n+2) -63*a(n+3) +6*a(n+4)) for all n in Z. - Michael Somos, Aug 06 2014
Integral representation as the n-th moment of a positive function W(x) on (0,16), in Maple notation, W(x) = EllipticK(sqrt(1-x/16))/(2*Pi^2*sqrt(x)); a(n) = Integral_{x=0..16} x^n*W(x) dx, n>=0. The function W(x) is singular at x=0 and W(16) = 1/(16*Pi). This representation is unique since W(x) is the solution of the Hausdorff moment problem. - Stanley Smith and Karol A. Penson, Jun 19 2015
a(n) ~ 16^n*(2-2/(8*n+2)^2+21/(8*n+2)^4-671/(8*n+2)^6+45081/(8*n+2)^8)^2/((4*n+1)* Pi). - Peter Luschny, Oct 14 2015
a(n) = binomial(2*n,n)*binomial(2*n,n) = ( [x^n](1 + x)^(2*n) ) *( [x^n](1 + x)^(2*n) ) = [x^n](F(x)^(4*n)), where F(x) = 1 + x + x^2 + 4*x^3 + 20*x^4 + 120*x^5 + 798*x^6 + 5697*x^7 + ... appears to have integer coefficients. For similar results see A000897, A002897, A006480, A008977, A186420 and A188662. - Peter Bala, Jul 14 2016
a(n) = Sum_{k = 0..n} binomial(2*n + k,k)*binomial(n,k)^2. Cf. A005258(n) = Sum_{k = 0..n} binomial(n + k,k)*binomial(n,k)^2. - Peter Bala, Jul 27 2016
a(n) = A241530(2*n), n >= 0. - Wolfdieter Lang, Sep 06 2016
E.g.f.: 2F2(1/2,1/2; 1,1; 16*x). - Ilya Gutkovskiy, Jan 23 2018
a(n) = 16^n*hypergeom([1/2, -2*n, 2*n + 1], [1, 1], 1). - Peter Luschny, Mar 14 2018
The right-hand side of the binomial coefficient identity Sum_{k = 0..n} C(n,k)*C(n+k,k)*C(2*n+2*k,n+k)*(-4)^(n-k) = a(n). - Peter Bala, Mar 16 2018
a(n) = [x^n] (1 - x)^(2*n) * P(2*n,(1 + x)/(1 - x)), where P(n,x) denotes the n-th Legendre polynomial. Compare with A245086(n) = [x^n] (1 - x)^(2*n) * P(n,(1 + x)/(1 - x)). - Peter Bala, Mar 23 2022
a(n) = Sum_{k=0..n} multinomial(2n [k k (n-k) (n-k)]), which is another way to count random walks on Z^2, with steps of (0,+-1) or (+-1,0), that return to the point of origin after 2n steps (not necessarily for the first time), as is C(2n,n)^2. - Shel Kaphan, Jan 12 2023
0 = a(n)*(+393216*a(n+2) -119040*a(n+3) +6860*a(n+4)) +a(n+1)*(-16128*a(n+2) +6928*a(n+3) -465*a(n+4)) +a(n+2)*(+36*a(n+2) -63*a(n+3) +6*a(n+4)) for n>=0. - Michael Somos, May 30 2023
From Peter Bala, Sep 12 2023: (Start)
Right-hand side of the binomial coefficient identities
1) Sum_{k = 0..n} (-1)^(n+k) * C(n,k)*C(n+k,n)*C(2*n+k,n) = a(n).
2) 2*Sum_{k = 0..n} (-1)^(n+k) * C(n,k)*C(n+k-1,n)*C(2*n+k-1,n) = a(n) for n >= 1.
3) (4/3)*Sum_{k = 0..n} (-1)^(n+k) * C(n,k)*C(n+k,n)*C(2*n+k-1,n) = a(n) for n >= 1. (End)

Edited by N. J. A. Sloane, Feb 18 2016

A005249 Determinant of inverse Hilbert matrix.

Original entry on oeis.org

1, 1, 12, 2160, 6048000, 266716800000, 186313420339200000, 2067909047925770649600000, 365356847125734485878112256000000, 1028781784378569697887052962909388800000000, 46206893947914691316295628839036278726983680000000000

Offset: 0

N. J. A. Sloane

a(n) = 1/determinant of M(n)*(-1)^floor(n/2) where M(n) is the n X n matrix m(i,j)=1/(i-j+n).

For n>=2, a(n) = Product k=1...(n-1) (2k+1) * C(2k,k)^2. This is a special case of the Cauchy determinant formula. A similar formula exists also for A067689. - Sharon Sela (sharonsela(AT)hotmail.com), Mar 23 2002

			The matrix begins:
  1    1/2  1/3  1/4  1/5  1/6  1/7  1/8  ...
  1/2  1/3  1/4  1/5  1/6  1/7  1/8  1/9  ...
  1/3  1/4  1/5  1/6  1/7  1/8  1/9  1/10 ...
  1/4  1/5  1/6  1/7  1/8  1/9  1/10 1/11 ...
  1/5  1/6  1/7  1/8  1/9  1/10 1/11 1/12 ...
  1/6  1/7  1/8  1/9  1/10 1/11 1/12 1/13 ...

Philip J. Davis, Interpolation and Approximation, Dover Publications, 1975, p. 288.
Jerry Glynn and Theodore Gray, "The Beginner's Guide to Mathematica Version 4," Cambridge University Press, Cambridge UK, 2000, page 76.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..25
Man-Duen Choi, Tricks or treats with the Hilbert matrix, Amer. Math. Monthly, 90 (1983), 301-312.
Richard K. Guy, Letter to N. J. A. Sloane, Sep 1986.
John E. Lauer, Letter to N. J. A. Sloane, Dec 1980.
Sajad Salami, On special matrices related to Cauchy and Toeplitz matrices, Instítuto da Matemática e Estatística, Universidade Estadual do Rio de Janeiro (Brazil, 2019).
Eric Weisstein's World of Mathematics, Hilbert Matrix.

Cf. A000178, A000515, A067689, A060739, A069945, A056040, A163085, A074962.

List([0..10],n->Product([1..n-1],k->(2*k+1)*Binomial(2*k,k)^2)); # Muniru A Asiru, Jul 07 2018

H=: % @: >: @: (+/~) @: i.
det=: -/ .* NB. Roger Hui, Oct 12 2005

with(linalg): A005249 := n-> 1/det(hilbert(n));

Table[ 1 / Det[ Table[ 1 / (i + j), {i, 1, n}, {j, 0, n - 1} ]], {n, 1, 10} ]
Table[Denominator[Det[HilbertMatrix[n]]], {n, 0, 12}]//Quiet (* L. Edson Jeffery, Aug 05 2014 *)
Table[BarnesG[2 n + 1]/BarnesG[n + 1]^4, {n, 0, 10}] (* Jan Mangaldan, Sep 22 2021 *)

a(n)=n^n*prod(k=1,n-1,(n^2-k^2)^(n-k))/prod(k=0,n-1,k!^2)

PARI
```
a(n)=if(n<0,0,1/matdet(mathilbert(n)))
```

a(n)=if(n<0,0,prod(k=0,n-1,(2*k)!*(2*k+1)!/k!^4))

def A005249(n):
    swing = lambda n: factorial(n)/factorial(n//2)^2
    return mul(swing(i) for i in (1..2*n-1))
[A005249(i) for i in (0..10)] # Peter Luschny, Sep 18 2012

a(n) = n^n*(Product_{k=1..n-1} (n^2 - k^2)^(n-k))/Product_{k=0..n-1} k!^2. - Benoit Cloitre, Jan 15 2003
The reciprocal of the determinant of an n X n matrix whose element at T(i, j) is 1/(i+j-1).
a(n+1) = a(n)*A000515(n) = a(n)*(2*n+1)*binomial(2n,n)^2. - Enrique Pérez Herrero, Mar 31 2010 [In other words, the partial products of sequence A000515. - N. J. A. Sloane, Jul 10 2015]
a(n) = n!*Product_{i=1..2n-1} binomial(i,floor(i/2)) = n!*|A069945(n)|. - Peter Luschny, Sep 18 2012
a(n) = Product_{i=1..2n-1} A056040(i) = A163085(2*n-1). - Peter Luschny, Sep 18 2012
a(n) ~ A^3 * 2^(2*n^2 - n - 1/12) * n^(1/4) / (exp(1/4) * Pi^n), where A = A074962 = 1.2824271291... is the Glaisher-Kinkelin constant. - Vaclav Kotesovec, May 01 2015
a(n) = A000178(2*n-1)/A000178(n-1)^4, for n >= 1. - Amiram Eldar, Oct 20 2022

1 more term from Jud McCranie, Jul 16 2000

Additional comments from Robert G. Wilson v, Feb 06 2002

A228329 a(n) = Sum_{k=0..n} (k+1)^2*T(n,k)^2 where T(n,k) is the Catalan triangle A039598.

Original entry on oeis.org

1, 8, 98, 1320, 18590, 268736, 3952228, 58837680, 883941750, 13373883600, 203487733020, 3110407163760, 47726453450988, 734694122886080, 11341161925265480, 175489379096245984, 2721169178975361702, 42273090191785999728, 657788911222324942060, 10250564041646388681200

Offset: 0

N. J. A. Sloane, Aug 26 2013

nonn

Let h(m) denote the sequence whose n-th term is Sum_{k=0..n} (k+1)^m*T(n,k)^2, where T(n,k) is the Catalan triangle A039598. This is h(2).

G. C. Greubel, Table of n, a(n) for n = 0..825
Pedro J. Miana and Natalia Romero, Moments of combinatorial and Catalan numbers, Journal of Number Theory, Volume 130, Issue 8, August 2010, Pages 1876-1887. See Remark 3 p. 1882. Omega2(n) = a(n-1).
Yidong Sun and Fei Ma, Four transformations on the Catalan triangle, arXiv:1305.2017 [math.CO], 2013.
Yidong Sun and Fei Ma, Some new binomial sums related to the Catalan triangle, Electronic Journal of Combinatorics 21(1) (2014), #P1.33.

Cf. A039598, A000108, A024492 (h(0)), A000894 (h(1)), A000515 (h(3)), A228330 (h(4)), A228331 (h(5)), A228332 (h(6)), A228333 (h(7)).

Cf. A000142, A007318, A051960.

B:=(n,k)->binomial(2*n, n-k) - binomial(2*n, n-k-2); #A039598
Omega:=(m,n)->add((k+1)^m*B(n,k)^2,k=0..n);
h:=m->[seq(Omega(m,n),n=0..20)];
h(2);
# Second solution:
h := n -> I*HeunG(8/5,0,-1/4,1/4,3/2,1/2,16*x)/sqrt(16*x-1);
seq(coeff(series(h(x),x,n+2),x,n),n=0..19); # Peter Luschny, Nov 26 2013

a[n_] := Binomial[4n, 2n] (3n+1)/(2n+1);
Table[a[n], {n, 0, 19}] (* Jean-François Alcover, Jul 30 2018, after Philippe Deléham *)

@CachedFunction
def A228329(n):
    return A228329(n-1)*(6*n+2)*(4*n-3)*(4*n-1)/(n*(2*n+1)*(3*n-2)) if n>0 else 1
[A228329(n) for n in (0..19)]  # Peter Luschny, Nov 26 2013

Conjecture: n*(2*n+1)*a(n) + 2*(-26*n^2+25*n-11)*a(n-1) + 20*(4*n-5)*(4*n-7)*a(n-2) = 0. - R. J. Mathar, Sep 08 2013
a(n) = ((4n)!*(3n+1))/((2n)!^2*(2n+1)) = binomial(4n,2n)*(3n+1)/(2n+1). - Philippe Deléham, Nov 25 2013
Therefore a(n) = A051960(2*n) / 2. - F. Chapoton, Jun 14 2024
From Peter Luschny, Nov 26 2013: (Start)
a(n) = 16^n*(3*n+1)*gamma(2*n+1/2)/(sqrt(Pi)*gamma(2*n+2)).
a(n) = a(n-1)*(6*n+2)*(4*n-3)*(4*n-1)/(n*(2*n+1)*(3*n-2)) if n > 0 else 1.
a(n) = [x^n] I*HeunG(8/5,0,-1/4,1/4,3/2,1/2,16*x)/sqrt(16*x-1) where [x^n] f(x) is the coefficient of x^n in f(x) and HeunG is the Heun general function. (End)

A228330 Let h(m) denote the sequence whose n-th term is Sum_{k=0..n} (k+1)^m*T(n,k)^2, where T(n,k) is the Catalan triangle A039598. This is h(4).

Original entry on oeis.org

1, 20, 362, 6504, 114686, 1992536, 34231540, 583027920, 9862508790, 165918037560, 2778642667020, 46358257249200, 770951008563372, 12785838603285104, 211540243555702376, 3492587812271418784, 57557091526140668070, 946970607665938615032, 15557339429900195819164, 255246113991506558429936

Offset: 0

N. J. A. Sloane, Aug 26 2013

nonn

G. C. Greubel, Table of n, a(n) for n = 0..825
Pedro J. Miana and Natalia Romero, Moments of combinatorial and Catalan numbers, Journal of Number Theory, Volume 130, Issue 8, August 2010, Pages 1876-1887. See Remark 3 p. 1882. Omega4(n) = a(n-1).
Yidong Sun and Fei Ma, Four transformations on the Catalan triangle, arXiv:1305.2017 [math.CO], 2013.
Yidong Sun and Fei Ma, Some new binomial sums related to the Catalan triangle, Electronic Journal of Combinatorics 21(1) (2014), #P1.33.

Cf. A000108, A039598, A024492 (h(0)), A000894 (h(1)), A228329 (h(2)), A000515 (h(3)), this sequence (h(4)), A228331 (h(5)), A228332 (h(6)), A228333 (h(7)).

List([0..20], n-> Binomial(4*n,2*n)*(15*n^3+15*n^2+n-1)/((2*n+1)*(4*n-1))); # G. C. Greubel, Mar 02 2019

[Binomial(4*n,2*n)*(15*n^3+15*n^2+n-1)/((2*n+1)*(4*n-1)): n in [0..20]]; // G. C. Greubel, Mar 02 2019

Table[4*Sum[(k+1)^6*(Binomial[2n+1, n-k]/(n+k+2))^2, {k,0,n}], {n,0,20}] (* Vaclav Kotesovec, Dec 08 2013 *)

vector(20, n, n--; binomial(4*n,2*n)*(15*n^3+15*n^2+n-1)/((2*n+1)*(4*n-1))) \\ G. C. Greubel, Mar 02 2019

[binomial(4*n,2*n)*(15*n^3+15*n^2+n-1)/((2*n+1)*(4*n-1)) for n in (0..20)] # G. C. Greubel, Mar 02 2019

Conjecture: n*(2*n+1)*(3467*n-4029)*a(n) + 8*(-36721*n^3 + 109040*n^2 - 137926*n + 69822)*a(n-1) + 4*(4*n-9)*(45706*n-7907)*(4*n-7)*a(n-2) = 0. - R. J. Mathar, Sep 08 2013
Recurrence: n*(2*n+1)*(15*n^3 - 30*n^2 + 16*n - 2)*a(n) = 2*(4*n-5)*(4*n-3)*(15*n^3 + 15*n^2 + n - 1)*a(n-1). - Vaclav Kotesovec, Dec 08 2013
From Vaclav Kotesovec, Dec 08 2013: (Start)
a(n) = binomial(4*n,2*n) * (15*n^3+15*n^2+n-1)/((2*n+1)*(4*n-1)).
a(n) = 4*Sum_{k=0..n} (k+1)^6*(binomial(2*n+1, n-k)/(n+k+2))^2. (End)

A228331 Let h(m) denote the sequence whose n-th term is Sum__{k=0..n} (k+1)^m*T(n,k)^2, where T(n,k) is the Catalan triangle A039598. This is h(5).

Original entry on oeis.org

1, 36, 780, 16240, 321300, 6131664, 114017904, 2079380160, 37356642180, 663144710800, 11657925495216, 203295462691776, 3521108298744400, 60632838691387200, 1038859802556120000, 17721669103065158400, 301147406355880764900, 5099997408534884394000, 86106549929771707182000

Offset: 0

N. J. A. Sloane, Aug 26 2013

nonn

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Pedro J. Miana and Natalia Romero, Moments of combinatorial and Catalan numbers, Journal of Number Theory, Volume 130, Issue 8, August 2010, Pages 1876-1887. See Omega5. Remark 3 p. 1882.
Yidong Sun and Fei Ma, Four transformations on the Catalan triangle, arXiv preprint arXiv:1305.2017, 2013
Yidong Sun and Fei Ma, Some new binomial sums related to the Catalan triangle, Electronic Journal of Combinatorics 21(1) (2014), #P1.33

Cf. A000108, A039598, A024492, A000894, A228329, A000515, A228330, A228332, A228333.

Table[Sum[(k+1)^5*(Binomial[2n+1, n-k]*2*(k+1)/(n+k+2))^2,{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Dec 08 2013 *)

Conjecture: n^2*a(n) +4*(2*n^2-22*n+11)*a(n-1) +16*(-7*n^2-54*n+166)*a(n-2) -1088*(2*n-3)*(2*n-7)*a(n-3)=0. - R. J. Mathar, Sep 08 2013
Recurrence: n^2*(3*n^2 - 5*n + 1)*a(n) = 4*(2*n-3)*(2*n+1)*(3*n^2 + n - 1)*a(n-1). - Vaclav Kotesovec, Dec 08 2013
a(n) = binomial(2*n,n)^2 * (2*n+1)*(3*n^2+n-1)/(2*n-1). - Vaclav Kotesovec, Dec 08 2013
G.f.: ((28*x+3)*hypergeom([1/2, 5/2],[1],16*x)+20*x*(1-16*x)*hypergeom([3/2, 7/2],[2],16*x))/3. - Mark van Hoeij, Apr 12 2014

A228333 Let h(m) denote the sequence whose n-th term is Sum__{k=0..n} (k+1)^m*T(n,k)^2, where T(n,k) is the Catalan triangle A039598. This is h(7).

Original entry on oeis.org

1, 132, 4260, 120400, 3017700, 69776784, 1524611088, 31951782720, 648578888100, 12837530477200, 248966505964176, 4747739344525632, 89267646282614800, 1658349027407016000, 30489930211792680000, 555544747397829254400, 10042477557290424843300, 180267292319119226298000, 3215718323211443887530000

Offset: 0

N. J. A. Sloane, Aug 26 2013

nonn

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Pedro J. Miana and Natalia Romero, Moments of combinatorial and Catalan numbers, Journal of Number Theory, Volume 130, Issue 8, August 2010, Pages 1876-1887. See Omega7. Remark 3 p. 1882.
Yidong Sun and Fei Ma, Four transformations on the Catalan triangle, arXiv preprint arXiv:1305.2017 [math.CO], 2013.
Yidong Sun and Fei Ma, Some new binomial sums related to the Catalan triangle, Electronic Journal of Combinatorics 21(1) (2014), #P1.33.

Cf. A000108, A039598, A024492, A000894, A228329, A000515, A228330, A228331, A228332.

Table[Sum[(k+1)^7*(Binomial[2n+1, n-k]*2*(k+1)/(n+k+2))^2,{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Dec 08 2013 *)

Conjecture: n^2*(304*n-411)*a(n) + 4*(-1814*n^3+2554*n^2-4776*n+7567)*a(n-1) + 32*(2*n-5)*(2*n-1)*(299*n-176)*a(n-2) = 0. - R. J. Mathar, Dec 04 2013
Recurrence: n^2*(6*n^3 - 12*n^2 + 6*n - 1)*a(n) = 4*(2*n-3)*(2*n+1)*(6*n^3 + 6*n^2 - 1)*a(n-1). - Vaclav Kotesovec, Dec 08 2013
a(n) = binomial(2*n,n)^2 * (2*n+1)*(6*n^3+6*n^2-1)/(2*n-1). - Vaclav Kotesovec, Dec 08 2013
G.f.: ((256*x+3)*hypergeom([1/2, 5/2],[1],16*x)+80*(38*x+1)*x*hypergeom([3/2, 7/2],[2],16*x))/3.  - Mark van Hoeij, Apr 12 2014

A189765 Triangle in which row n has the n(n+1)/2 elements of the lower triangular part of the inverse of the n-th order Hilbert matrix.

Original entry on oeis.org

1, 4, -6, 12, 9, -36, 192, 30, -180, 180, 16, -120, 1200, 240, -2700, 6480, -140, 1680, -4200, 2800, 25, -300, 4800, 1050, -18900, 79380, -1400, 26880, -117600, 179200, 630, -12600, 56700, -88200, 44100, 36, -630, 14700, 3360, -88200, 564480, -7560, 211680

Offset: 1

T. D. Noe, May 02 2011

The n-th order Hilbert matrix has elements h(i,j) = 1/(i+j-1) for 1 <= i,j <=n. Only the lower triangular matrix is shown because the Hilbert matrix and its inverse are symmetric. The n-th row begins with n^2 and ends with A000515(n+1).
The sums of select rows of the inverse matrix are sequences A002457, A002736, A002738, A007531, and A054559.
The largest magnitude in the matrix is A210356(n). - T. D. Noe, Mar 28 2012
The sum of the elements of the n-th matrix is n^2. - T. D. Noe, Apr 02 2012

			Row 3 is 9, -36, 192, 30, -180, 180 which corresponds to the inverse
  9  -36   30
-36  192 -180
30 -180  180

T. D. Noe, Rows n = 1..25, flattened
Eric W. Weisstein, MathWorld: Hilbert Matrix

Cf. A002457, A002736, A002738, A005249 (determinant), A007531, A054559, A189766 (trace).

lowerTri[m_List] := Module[{n = Length[m]}, Flatten[Table[Take[m[[i]], i], {i, n}]]]; Flatten[Table[lowerTri[Inverse[HilbertMatrix[n]]], {n, 6}]]

a(n,i,j) = (-1)^(i+j) (i+j-1) binomial(n+i-1, n-j) binomial(n+j-1, n-i) binomial(i+j-2, i-1)^2 is the (i,j) element of the inverse of the n-th Hilbert matrix.

A228332 Let h(m) denote the sequence whose n-th term is Sum_{k=0..n} (k+1)^m*T(n,k)^2, where T(n,k) is the Catalan triangle A039598. This is h(6).

Original entry on oeis.org

1, 68, 1778, 43080, 958430, 20119736, 405350788, 7921691280, 151231519350, 2834134359000, 52320693313020, 953960351550960, 17212782834351468, 307826474156801840, 5462948893700675720, 96303960593503261984, 1687752152779483045542, 29424712141610821296408, 510621541414656188646220

Offset: 0

N. J. A. Sloane, Aug 26 2013

nonn

Pedro J. Miana and Natalia Romero, Moments of combinatorial and Catalan numbers, Journal of Number Theory, Volume 130, Issue 8, August 2010, Pages 1876-1887. See Remark 3 p. 1882. Omega6(n) = a(n-1).
Yidong Sun and Fei Ma, Four transformations on the Catalan triangle, arXiv preprint arXiv:1305.2017 [math.CO], 2013.
Yidong Sun and Fei Ma, Some new binomial sums related to the Catalan triangle, Electronic Journal of Combinatorics 21(1) (2014), #P1.33.

Cf. A000108, A039598, A024492, A000894, A228329, A000515, A228330, A228331, A228333.

Table[Sum[(k+1)^6*(Binomial[2n+1, n-k]*2*(k+1)/(n+k+2))^2,{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Dec 08 2013 *)

Recurrence: n*(2*n+1)*(105*n^5 - 420*n^4 + 588*n^3 - 356*n^2 + 96*n - 10)*a(n) = 2*(4*n-7)*(4*n-5)*(105*n^5 + 105*n^4 - 42*n^3 - 62*n^2 - 7*n + 3)*a(n-1). - Vaclav Kotesovec, Dec 08 2013

a(n) = binomial(4*n,2*n) * (105*n^5 + 105*n^4 - 42*n^3 - 62*n^2 - 7*n + 3) / ((2*n+1)*(4*n-3)*(4*n-1)). - Vaclav Kotesovec, Dec 08 2013

A061065 For n <= 6, entry of maximal modulus in the inverse of the n-th Hilbert matrix. For n >= 3, this is the (n-1,n-1)-th entry.

Original entry on oeis.org

1, 12, 192, 6480, 179200, 4410000, 100590336, 2175421248, 45229916160, 912328045200, 17965673440000, 346945899203904, 6592659294154752, 123580568462478400, 2289795064260480000, 42003815644116000000

Offset: 1

Roger Cuculière, May 28 2001

Incorrect version of the largest element in the inverse of Hilbert's matrix. See A210356 for the correct version. See A210357 for the location of the maximal value. - T. D. Noe and Clark Kimberling, Mar 28 2012

Harry J. Smith, Table of n, a(n) for n = 1..100

Cf. A005249, A000515, A210356, A210357.

{ for (n=1, 100, if (n>2, a=((2*n-2)^2)*(2*n-3)*binomial(2*n-4, n-2)^2, if (n==1, a=1, a=12)); write("b061065.txt", n, " ", a) ) } \\ Harry J. Smith, Jul 17 2009

For n >= 3, a(n) = ((2n-2)^2)*(2n-3)*C(2n-4, n-2)^2. - David Wasserman, Jun 08 2002

More terms from David Wasserman, Jun 08 2002

A144630 Triangle read by rows: T(n,k) (1 <= k <= n) is the sum of the entries in the lower right k X k submatrix of the n X n inverse Hilbert matrix.

Original entry on oeis.org

1, 12, 4, 180, 12, 9, 2800, 880, 40, 16, 44100, 46900, 4480, 40, 25, 698544, 1615824, 411264, 13104, 84, 36, 11099088, 45094896, 23653476, 2268756, 36036, 84, 49, 176679360, 1115345088, 1017615456, 207193536, 9660816, 79776, 144, 64

Offset: 1

Daniel McLaury and Ben Golub, Dec 23 2008

The initial entries in each row form A000515. The second entries give A144631. The final entries are the squares (A000290).

Row sums are A144632. The penultimate entries in each row appear to be 4*A014105. - N. J. A. Sloane, Jan 20 2009

			The first three inverse Hilbert matrices are:
--------------
[ 1 ]
--------------
[4 -6 ]
[-6 12]
--------------
[ 9 -36 30 ]
[-36 192 -180]
[30 -180 180]
--------------
Triangle begins:
1,
12, 4,
180, 12, 9,
2800, 880, 40, 16,
44100, 46900, 4480, 40, 25,
698544, 1615824, 411264, 13104, 84, 36

Klaus Brockhaus, Table of n, a(n) for n=1..1830 (rows 1 - 60)
Wikipedia, Hilbert matrix (gives inverse Hilbert matric explicitly).

Cf. A000290, A000515, A144631.

invhilb(1), invhilb(2), invhilb(3), etc.

&cat[ [ &+[I[i][j]: i,j in [k..n] ]: k in [n..1 by -1] ] where I:=H^-1 where H:=Matrix(Rationals(), n, n, [ < i, j, 1/(i+j-1) >: i, j in [1..n] ] ): n in [1..8] ]; // Klaus Brockhaus, Jan 21 2009

invH := proc(n,i,j) (-1)^(i+j)*(i+j-1)*binomial(n+i-1,n-j)*binomial(n+j-1,n-i)* (binomial(i+j-2,i-1))^2 ; end: A144630 := proc(n,k) local T,i,j ; T := 0 ; for i from n-k+1 to n do for j from n-k+1 to n do T := T+invH(n,i,j) ; od; od; RETURN(T) ; end: for n from 1 to 10 do for k from 1 to n do printf("%a,", A144630(n,k)) : od: od: # R. J. Mathar, Jan 21 2009

inverseHilbert[n_, i_, j_] := (-1)^(i+j)*(i+j-1) * Binomial[n+i-1, n-j] * Binomial[n+j-1, n-i] * Binomial[i+j-2, i-1]^2; inverseHilbert[n_, k_] := Table[ inverseHilbert[n, i, j], {i, n-k+1, n}, {j, n-k+1, n}]; t[n_, k_] := Tr[ Flatten[ inverseHilbert[n, k]]]; Flatten[ Table[t[n, k], {n, 1, 8}, {k, 1, n}]] (* Jean-François Alcover, Jul 16 2012 *)

More terms from R. J. Mathar and Klaus Brockhaus, Jan 21 2009

cp's OEIS Frontend

A002894 a(n) = binomial(2n, n)^2.

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A005249 Determinant of inverse Hilbert matrix.

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A228329 a(n) = Sum_{k=0..n} (k+1)^2*T(n,k)^2 where T(n,k) is the Catalan triangle A039598.

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A228330 Let h(m) denote the sequence whose n-th term is Sum_{k=0..n} (k+1)^m*T(n,k)^2, where T(n,k) is the Catalan triangle A039598. This is h(4).

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A228331 Let h(m) denote the sequence whose n-th term is Sum__{k=0..n} (k+1)^m*T(n,k)^2, where T(n,k) is the Catalan triangle A039598. This is h(5).

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A228333 Let h(m) denote the sequence whose n-th term is Sum__{k=0..n} (k+1)^m*T(n,k)^2, where T(n,k) is the Catalan triangle A039598. This is h(7).

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