A086618 -id:A086618 - cp's OEIS frontend

A228456 Determinant of the (n+1) X (n+1) matrix with (i,j)-entry equal to A086618(i+j) for all i,j = 0,...,n.

1, 3, 41, 2841, 1010845, 1790959059, 15625174448153, 684155685766047665, 153135670584610711281133, 174306862433739623658457865851, 994128440039970508236558371536766105, 28386094077591547319667447446929656332136825

Offset: 0

Zhi-Wei Sun, Aug 22 2013

Conjecture: (i) a(n) is always positive and odd, and not congruent to 7 modulo 8.

(ii) For any odd prime p, if p == 1 (mod 3) and p = x^2 + 3*y^2 with x == 1 (mod 3), then a(p-1) == (-1)^{(p-1)/2}*(2*x-p/(2*x)) (mod p^2); if p == 2 (mod 3) then a(p-1) == (-1)^{(p+1)/2}*3p/binomial((p+1)/2,(p+1)/6) (mod p^2).

			a(0) = 1 since A086618(0) = 1.

Zhi-Wei Sun, Conjectures and results on x^2 mod p^2 with 4p = x^2+d*y^2, in: Number Theory and Related Area (eds., Y. Ouyang, C. Xing, F. Xu and P. Zhang), Higher Education Press & International Press, Beijing and Boston, 2013, pp. 147-195.

Zhi-Wei Sun, Table of n, a(n) for n = 0..25
Zhi-Wei Sun, Conjectures and results on x^2 mod p^2 with 4p = x^2+d*y^2, arXiv:1103.4325 [math.NT], 2011-2014.
Z.-W. Sun, On sums of Apéry polynomials and related congruences, J. Number Theory 132(2012), 2673-2699.
Zhi-Wei Sun, Connections between p = x^2 + 3*y^2 and Franel numbers, J. Number Theory 133(2013), 2914-2928.
Zhi-Wei Sun, On some determinants with Legendre symbol entries, preprint, arXiv:1308.2900 [math.NT], 2013-2018.

Cf. A086618.

f[n_]:=Sum[Binomial[n,k]^2*Binomial[2k,k]/(k+1),{k,0,n}]
a[n_]:=Det[Table[f[i+j],{i,0,n},{j,0,n}]]
Table[a[n],{n,0,10}]

f(n)=sum(k=0,n, binomial(n,k)^2*binomial(2*k,k)/(k+1))
a(n)=my(v=vector(2*n+1,k,f(k-1))); matdet(matrix(n+1,n+1,i,j,v[i+j-1])) \\ Charles R Greathouse IV, Jul 30 2016

A005802 Number of permutations in S_n with longest increasing subsequence of length <= 3 (i.e., 1234-avoiding permutations); vexillary permutations (i.e., 2143-avoiding).

Original entry on oeis.org

1, 1, 2, 6, 23, 103, 513, 2761, 15767, 94359, 586590, 3763290, 24792705, 167078577, 1148208090, 8026793118, 56963722223, 409687815151, 2981863943718, 21937062144834, 162958355218089, 1221225517285209, 9225729232653663, 70209849031116183, 537935616492552297

Offset: 0

N. J. A. Sloane, Simon Plouffe

Also the dimension of SL(3)-invariants in V^n tensor (V^*)^n, where V is the standard 3-dimensional representation of SL(3) and V^* is its dual. - Alec Mihailovs (alec(AT)mihailovs.com), Aug 14 2005
Also the number of doubly-alternating permutations of length 2n with no four-term increasing subsequence (i.e., 1234-avoiding doubly-alternating permutations). The doubly-alternating permutations (counted by sequence A007999) are those permutations w such that both w and w^(-1) have descent set {2, 4, 6, ...}. - Joel B. Lewis, May 21 2009
Any permutation without an increasing subsequence of length 4 has a decreasing subsequence of length >= n/3, where n is the length of the sequence, by the Erdős-Szekeres theorem. - Charles R Greathouse IV, Sep 26 2012
Also the number of permutations of length n simultaneously avoiding patterns 1324 and 3416725 (or 1324 and 3612745). - Alexander Burstein, Jan 31 2014

Eric S. Egge, Defying God: The Stanley-Wilf Conjecture, Stanley-Wilf Limits, and a Two-Generation Explosion of Combinatorics, pp. 65-82 of "A Century of Advancing Mathematics", ed. S. F. Kennedy et al., MAA Press 2015.
S. Kitaev, Patterns in Permutations and Words, Springer-Verlag, 2011. see p. 399 Table A.7.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 7.16(e), p. 453.

Alois P. Heinz, Table of n, a(n) for n = 0..1060
Michael Albert and Mireille Bousquet-Mélou, Permutations sortable by two stacks in parallel and quarter plane walks, arXiv:1312.4487 [math.CO], 2014-2015.
Michael Albert and Mireille Bousquet-Mélou, Permutations sortable by two stacks in parallel and quarter plane walks, European Journal of Combinatorics 43 (2015), 131-164.
Alin Bostan, Andrew Elvey Price, Anthony John Guttmann, and Jean-Marie Maillard, Stieltjes moment sequences for pattern-avoiding permutations, arXiv:2001.00393 [math.CO], 2020.
A. Burstein and J. Pantone, Two examples of unbalanced Wilf-equivalence, arXiv preprint arXiv:1402.3842 [math.CO], 2014.
CombOS - Combinatorial Object Server, Generate pattern-avoiding permutations
Andrew R. Conway and Anthony J. Guttmann, On 1324-avoiding permutations, Advances in Applied Mathematics, Volume 64, March 2015, p. 50-69.
Andrew R. Conway and Anthony J. Guttmann, Counting occurrences of patterns in permutations, arXiv:2306.12682 [math.CO], 2023. See p. 18.
Tom Denton, Algebraic and Affine Pattern Avoidance, arXiv preprint arXiv:1303.3767 [math.CO], 2013.
Eric S. Egge, Defying God: The Stanley-Wilf Conjecture, Stanley-Wilf Limits, and a Two-Generation Explosion of Combinatorics, MAA Focus, August/September 2015, pp. 33-34. [Annotated scanned copy]
Shalosh B. Ekhad, Nathaniel Shar, and Doron Zeilberger, The number of 1...d-avoiding permutations of length d+r for SYMBOLIC d but numeric r, arXiv:1504.02513 [math.CO], 2015.
Steven Finch, Pattern-Avoiding Permutations [Cached copy at the Wayback Machine]
Steven Finch, Pattern-Avoiding Permutations [Cached copy, with permission]
A. L. L. Gao, S. Kitaev, and P. B. Zhang, On pattern avoiding indecomposable permutations, arXiv:1605.05490 [math.CO], 2016.
Ira M. Gessel, Symmetric functions and P-recursiveness, J. Combin. Theory A 53 (1990), 257-285.
O. Guibert and E. Pergola, Enumeration of vexillary involutions which are equal to their mirror/complement, Discrete Math., Vol. 224 (2000), pp. 281-287.
Elizabeth Hartung, Hung Phuc Hoang, Torsten Mütze, and Aaron Williams, Combinatorial generation via permutation languages. I. Fundamentals, arXiv:1906.06069 [cs.DM], 2019.
Eric Marberg, Crossings and nestings in colored set partitions, arXiv preprint arXiv:1203.5738 [math.CO], 2012.
J. Noonan and D. Zeilberger, The Enumeration of Permutations With A Prescribed Number of "Forbidden" Patterns.
J. Noonan and D. Zeilberger, The Enumeration of Permutations With A Prescribed Number of "Forbidden" Patterns, Advances in Applied Mathematics, Vol. 17, 1996, pp. 381-407.
J. Noonan and D. Zeilberger, The Enumeration of Permutations With a Prescribed Number of "Forbidden" Patterns, arXiv:math/9808080 [math.CO], 1998.
Erik Ouchterlony, Pattern avoiding doubly alternating permutations
Permutation Pattern Avoidance Library (PermPAL), 0123
Nathaniel Shar, Experimental methods in permutation patterns and bijective proof, PhD Dissertation, Mathematics Department, Rutgers University, May 2016.
Anders Bjorner and Richard P. Stanley, A combinatorial miscellany
R. P. Stanley, Recent Progress in Algebraic Combinatorics, Bull. Amer. Math. Soc., 40 (2003), 55-68.

A column of A047888. See also A224318, A223034, A223905.
Column k=3 of A214015.
A005802, A022558, A061552 are representatives for the three Wilf classes for length-four avoiding permutations (cf. A099952).

a:= n-> 2*add(binomial(2*k,k)*(binomial(n,k))^2*(3*k^2+2*k+1-n-2*k*n)/ (k+1)^2/(k+2)/(n-k+1),k=0..n);
A005802:=rsolve({a(0) = 1, a(1) = 1, (n^2 + 8*n + 16)*a(n + 2) = (10*n^2 + 42*n + 41)*a(n + 1) - (9*n^2 + 18*n + 9)*a(n)},a(n),makeproc): # Alec Mihailovs (alec(AT)mihailovs.com), Aug 14 2005

a[n_] := 2Sum[Binomial[2k, k]Binomial[n, k]^2(3k^2+2k+1-n-2k*n)/((k+1)^2(k+2)(n-k+1)), {k, 0, n}]
(* Second program:*)
a[0] = a[1] = 1; a[n_] := a[n] = ((10*n^2+2*n-3)*a[n-1] + (-9*n^2+18*n-9)* a[n-2])/(n+2)^2; Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Feb 20 2017 *)
Table[HypergeometricPFQ[{1/2, -1 - n, -n}, {2, 2}, 4] / (n+1), {n, 0, 25}] (* Vaclav Kotesovec, Jun 07 2021 *)

a(n)=2*sum(k=0,n,binomial(2*k,k)*binomial(n,k)^2*(3*k^2+2*k+1-n-2*k*n)/(k+1)^2/(k+2)/(n-k+1)) \\ Charles R Greathouse IV, Sep 26 2012

a(n) = 2 * Sum_{k=0..n} binomial(2*k, k) * (binomial(n, k))^2 * (3*k^2 + 2*k+1 - n - 2*k*n)/((k+1)^2 * (k+2) * (n-k+1)).
(4*n^2 - 2*n + 1)*(n + 2)^2*(n + 1)^2*a(n) = (44*n^3 - 14*n^2 - 11*n + 8)*n*(n + 1)^2*a(n - 1) - (76*n^4 + 42*n^3 - 49*n^2 - 24*n + 24)*(n - 1)^2*a(n - 2) + 9*(4*n^2 + 6*n + 3)*(n - 1)^2*(n - 2)^2*a(n - 3). - Vladeta Jovovic, Jul 16 2004
a(0) = 1, a(1) = 1, (n^2 + 8*n + 16)*a(n + 2) = (10*n^2 + 42*n + 41) a(n + 1) - (9*n^2 + 18*n + 9) a(n). - Alec Mihailovs (alec(AT)mihailovs.com), Aug 14 2005
a(n) = ((18*n+45)*A002893(n) - (7+2*n)*A002893(n+1)) / (6*(n+2)^2). - Mark van Hoeij, Jul 02 2010
G.f.: (1+5*x-(1-9*x)^(3/4)*(1-x)^(1/4)*hypergeom([-1/4, 3/4],[1],64*x/((x-1)*(1-9*x)^3)))/(6*x^2). - Mark van Hoeij, Oct 25 2011
a(n) ~ 3^(2*n+9/2)/(16*Pi*n^4). - Vaclav Kotesovec, Jul 29 2013
a(n) = Sum_{k=0..n} binomial(2k,k)*binomial(n+1,k+1)*binomial(n+2,k+1)/((n+1)^2*(n+2)). [Conway and Guttmann, Adv. Appl. Math. 64 (2015) 50]
For n > 0, (n+2)^2*a(n) - n^2*a(n-1) = 4*A086618(n). - Zhi-Wei Sun, Nov 16 2017
a(n) = hypergeom([1/2, -1 - n, -n], [2, 2], 4) / (n+1). - Vaclav Kotesovec, Jun 07 2021

Additional comments from Emeric Deutsch, Dec 06 2000
More terms from Naohiro Nomoto, Jun 18 2001
Edited by Dean Hickerson, Dec 10 2002
More terms from Alec Mihailovs (alec(AT)mihailovs.com), Aug 14 2005

A246065 a(n) = Sum_{k=0..n}C(n,k)^2C(2k,k)/(2k-1), where C(n,k) denotes the binomial coefficient n!/(k!(n-k)!).

Original entry on oeis.org

-1, 1, 9, 39, 177, 927, 5463, 34857, 234657, 1641471, 11820135, 87080265, 653499135, 4979882385, 38441107305, 300027646647, 2364113123073, 18784242756927, 150351698420247, 1211310469545081, 9816017765368671, 79963826730913809, 654504197331971961, 5380270242617370951

Offset: 0

Zhi-Wei Sun, Aug 24 2014

sign

a(n) is always an integer since (2k-1)|C(2k,k) for any nonnegative integer k.
Conjecture: (i) The sequence a(n+1)/a(n) (n = 2,3,...) is strictly increasing to the limit 9, and the sequence a(n+1)^(1/(n+1))/a(n)^(1/n) (n = 1,2,3,...) is strictly decreasing to the limit 1.
(ii) sum_{k=0}^{n-1}a(k) == 0 (mod n^2) for all n > 0. Moreover, for any prime p we have sum_{k=0}^{p-1}a(k) == -p^2*(1+9*(p/3))/2 (mod p^3), where (p/3) is the Legendre symbol.
We are able to prove n | sum_{k=0}^{n-1}a(k). Note also that sum_{k=0}^{n-1}a(k)*9^(n-1-k) = -n^2*A086618(n-1) for all n > 0 since both sides satisfy the same recurrence via the Zeilberger algorithm.
The congruence (0 mod n^2) in (ii) is true, see the formula for A246138 in terms of A005802. - Mark van Hoeij, Nov 07 2023

			a(2) = 9 since Sum_{k=0,1,2}C(2,k)^2*C(2k,k)/(2k-1) = -1 + 8 + 6/3 = 9.

Zhi-Wei Sun, Table of n, a(n) for n = 0..150
Zhi-Wei Sun, A new kind of numbers and their arithmetic properties, preprint, arXiv:1408.5381 [math.NT], 2014-2018.

Cf. A086618, A245769.

a := n -> -hypergeom([-1/2, -n, -n], [1, 1], 4):
seq(simplify(a(n)), n=0..23); # Peter Luschny, Nov 07 2023
ogf := -(1-9*x)^(1/4)*hypergeom([-1/4, 3/4],[1],64*x^3/((1-9*x)*(x-1)^3))/(1-x)^(5/4);
series(ogf, x=0, 25); # Mark van Hoeij, Nov 12 2023

a[n_]:=Sum[Binomial[n,k]^2*Binomial[2k,k]/(2k-1),{k,0,n}]
Table[a[n],{n,0,20}]

Recurrence (obtained via the Zeilberger algorithm):
9*(n+1)^2*a(n) -(19n^2+58n+63)*a(n+1) + (11n^2+46n+47)*a(n+2)-(n+3)^2*a(n+3) = 0.
a(n) ~ A086618(n)/2 ~ 3^(2*n + 5/2)/(16*Pi*n^2) as n tends to the infinity.
a(n) = (9*(2*n+1)^2*A002893(n) - 4*(n+1)^2*A002893(n+1))/3. - Mark van Hoeij, Nov 07 2023
a(n) = -hypergeom([-1/2, -n, -n], [1, 1], 4). - Peter Luschny, Nov 07 2023

A086617 Symmetric square table of coefficients, read by antidiagonals, where T(n,k) is the coefficient of x^ny^k in f(x,y) that satisfies f(x,y) = 1/((1-x)(1-y)) + xyf(x,y)^2.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 7, 4, 1, 1, 5, 13, 13, 5, 1, 1, 6, 21, 33, 21, 6, 1, 1, 7, 31, 69, 69, 31, 7, 1, 1, 8, 43, 126, 183, 126, 43, 8, 1, 1, 9, 57, 209, 411, 411, 209, 57, 9, 1, 1, 10, 73, 323, 815, 1118, 815, 323, 73, 10, 1, 1, 11, 91, 473, 1471, 2633, 2633, 1471, 473, 91, 11, 1

Offset: 0

Paul D. Hanna, Jul 24 2003

Determinants of upper left n X n matrices results in A003046: {1,1,2,10,140,5880,776160,332972640,476150875200,...}, which is the product of the first n Catalan numbers (A000108).

May also be regarded as a Pascal-Catalan triangle. As a triangle, row sums are A086615, inverse has row sums 0^n.

			Rows begin:
  1, 1,  1,   1,    1,    1,     1,     1, ...
  1, 2,  3,   4,    5,    6,     7,     8, ...
  1, 3,  7,  13,   21,   31,    43,    57, ...
  1, 4, 13,  33,   69,  126,   209,   323, ...
  1, 5, 21,  69,  183,  411,   815,  1471, ...
  1, 6, 31, 126,  411, 1118,  2633,  5538, ...
  1, 7, 43, 209,  815, 2633,  7281, 17739, ...
  1, 8, 57, 323, 1471, 5538, 17739, 49626, ...
As a triangle:
  1;
  1,   1;
  1,   2,   1;
  1,   3,   3,   1;
  1,   4,   7,   4,   1;
  1,   5,  13,  13,   5,   1;
  1,   6,  21,  33,  21,   6,   1;
  1,   7,  31,  69,  69,  31,   7,   1;
  1,   8,  43, 126, 183, 126,  43,   8,   1;

Paul Barry, Riordan arrays, generalized Narayana triangles, and series reversion, Linear Algebra and its Applications, 491 (2016) 343-385.

Cf. A086618 (diagonal), A086615 (antidiagonal sums), A003046 (determinants).

T[n_, k_] := Sum[Binomial[n, j] Binomial[k, j] CatalanNumber[j], {j, 0, n}];
Table[T[n - k, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Oct 02 2019 *)

As a triangle, T(n, k)=sum{j=0..n-k, C(n-k, j)C(k, j)C(j)}; T(n, k)=sum{j=0..n, C(n-k, n-j)C(k, j-k)C(j-k)}; T(n, k)=if(k<=n, sum{j=0..n, C(k, j)C(n-k, n-j)C(k-j)}, 0).
As a square array, T(n, k)=sum{j=0..n, C(n, j)C(k, j)C(j)}; As a square array, T(n, k)=sum{j=0..n+k, C(n, n+k-j)C(k, j-k)C(j-k)}; column k has g.f. sum{j=0..k, C(k, j)C(j)(x/(1-x))^j}x^k/(1-x).
G.f.: (1-sqrt(1-(4*x^2*y)/((1-x)*(1-x*y))))/(2*x^2*y). - Vladimir Kruchinin, Jan 15 2018

Additional comments from Paul Barry, Nov 17 2005

Edited by N. J. A. Sloane, Oct 16 2006

A246459 a(n) = Sum_{k=0..n} C(n,k)^2C(2k,k)(2k+1), where C(n,k) denotes the binomial coefficient n!/(k!*(n-k)!).

Original entry on oeis.org

1, 7, 55, 465, 4047, 35673, 316521, 2819295, 25173855, 225157881, 2016242265, 18070920255, 162071863425, 1454320387575, 13055422263255, 117237213829953, 1053070838993151, 9461217421304505, 85019389336077225, 764113545253570191, 6868417199986308129

Offset: 0

Zhi-Wei Sun, Aug 26 2014

nonn

Zhi-Wei Sun proved that for any n > 0 we have Sum_{k=0..n-1} a(k) = n^2*A086618(n-1), and (Sum_{k=0..n-1}a(k,x))/n is a polynomial with integer coefficients, where a(k,x) = sum_{j=0..k}C(k,j)^2*C(2j,j)*(2j+1)*x^j.

			a(2) = 55 since Sum_{k=0,1,2} C(2,k)^2*C(2k,k)(2k+1) = 1 + 8*3 + 6*5 = 55.

Zhi-Wei Sun, Table of n, a(n) for n = 0..160
Zhi-Wei Sun, Two new kinds of numbers and their arithmetic properties, arXiv:1408.5381 [math.NT], 2014-2018.

Cf. A086618, A245769, A246065, A246138.

A246459:=n->add(binomial(n,k)^2*binomial(2*k,k)*(2*k+1), k=0..n): seq(A246459(n), n=0..20); # Wesley Ivan Hurt, Aug 26 2014

a[n_]:=Sum[Binomial[n,k]^2*Binomial[2k,k](2k+1),{k,0,n}]
Table[a[n],{n,0,20}]

Recurrence (obtained via the Zeilberger algorithm): 9*(n+1)^2*a(n) - (19*n^2+74*n+87)*a(n+1) + (n+3)*(11*n+29)*a(n+2) - (n+3)^2*a(n+3) = 0.
a(n) ~ 3^(2*n+1/2) / Pi. - Vaclav Kotesovec, Aug 27 2014
a(n) = (4*n+3)*A002893(n)/3. - Mark van Hoeij, Nov 12 2023

A238115 Number of states arising in matrix method for enumerating Hamiltonian cycles on a 2n X 2n grid.

Original entry on oeis.org

1, 6, 32, 182, 1117, 7280, 49625, 349998, 2535077, 18758264, 141254654, 1079364104, 8350678169, 65298467486, 515349097712, 4100346740510, 32858696386765, 265001681344568, 2149447880547398, 17524254766905368, 143540915998174577, 1180736721910617182

Offset: 1

N. J. A. Sloane, Mar 05 2014

nonn

Andrew Howroyd, Table of n, a(n) for n = 1..500
Ed Wynn, Enumeration of nonisomorphic Hamiltonian cycles on square grid graphs, arXiv preprint arXiv:1402.0545 [math.CO], 2014.

Cf. A000108, A003763, A086618, A143246, A209077, A222065, A238116, A238117, A238118.

a := n -> hypergeom([1/2, -n, -n], [1, 2], 4) - 1:
seq(simplify(a(n)), n = 1..22);  # Peter Luschny, Dec 13 2024

a(n)=sum(k=1,n,binomial(n,k)^2*binomial(2*k,k)/(k+1)) \\ Andrew Howroyd, Dec 13 2024

From Andrew Howroyd, Dec 13 2024: (Start)
a(n) = Sum_{k=1..n} binomial(n,k)^2 * A000108(k).
a(n) = A086618(n) - 1. (End)

A087457 Number of odd length roads between any adjacent nodes in virtual optimal chordal ring of degree 3 (the length of chord < number of nodes/2).

Original entry on oeis.org

1, 5, 31, 213, 1551, 11723, 90945, 719253, 5773279, 46889355, 384487665, 3177879675, 26442188865, 221278343445, 1860908156031, 15717475208853, 133256583398655, 1133591857814363, 9672323357640129, 82752014457666363, 709719620585186529, 6100394753270329605

Offset: 1

B. Dubalski (dubalski(AT)atr.bydgoszcz.pl), Oct 23 2003

nonn

			a(1)=1; a(2)=9*a(1)-2*2=9-4=5; a(3)=9*5-2*7=31; a(4)=9*31-2*33=213; etc

R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, Addison-Wesley, Reading, MA, 2nd ed. 1998, see page number?

Seiichi Manyama, Table of n, a(n) for n = 1..1052 (terms 1..100 from T. D. Noe)

Cf. A086617, A086618, A002893.

a := 1; s := 0; for k from 1 to 10 do for i from 0 to k do ss := ((2*(i))!/((i)!*(i+1)!))*((k)!/((i)!*(k-i)!))^2; s := s+ss; od; a := (9*a-2*s); s := 0; od;
# Alternative:
a := n -> hypergeom([1/2, -n, -n], [1, 1], 4)/3;
seq(simplify(a(n)), n = 1..22);  # Peter Luschny, Nov 06 2023

Table[Sum[Binomial[n,k]^2*Binomial[2k,k],{k,0,n}]/3,{n,1,20}] (* Vaclav Kotesovec, Oct 14 2012 *)

a(n) = sum(k=0, n, binomial(n,k)^2*binomial(2*k,k))/3; \\ Michel Marcus, May 10 2020

a(1) = 1; a(n) = 9*a(n-1) - 2*A086618(n), where A086618(n) = Sum_{k=0..n} Catalan(n)*binomial(n, k)^2, and Catalan(n) = (2*n)!/(n!*(n+1)!). - Michael Somos
a(n) = A002893(n)/3 = (1/3)*Sum_{k=0..n}binomial(n,k)^2*binomial(2k,k). - Philippe Deléham, Sep 14 2008
Recurrence: n^2*a(n) = (10*n^2-10*n+3)*a(n-1) - 9*(n-1)^2*a(n-2). - Vaclav Kotesovec, Oct 14 2012
a(n) ~ 3^(2*n+1/2)/(4*Pi*n). - Vaclav Kotesovec, Oct 14 2012
G.f.: (hypergeom([1/3, 1/3],[1],-27*x*(x-1)^2/(9*x-1)^2)/(1-9*x)^(2/3)-1)/3.  - Mark van Hoeij, May 14 2013
G.f.: G(0)/(6*x*(1-9*x)^(2/3) ) -1/(3*x), where G(k)= 1 + 1/(1 - 3*(3*k+1)^2*x*(1-x)^2/(3*(3*k+1)^2*x*(1-x)^2 - (k+1)^2*(1-9*x)^2/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 31 2013
a(n) = hypergeom([1/2, -n, -n], [1, 1], 4) / 3. - Peter Luschny, Nov 06 2023

A381676 a(n) = Sum_{k=0..n} binomial(n,k) * ( binomial(n,k) - binomial(n,k-1) )^2.

Original entry on oeis.org

1, 1, 4, 17, 86, 472, 2752, 16753, 105394, 680366, 4484360, 30067160, 204508240, 1408057120, 9796738304, 68786005361, 486845236106, 3470187822754, 24891491746792, 179556655434382, 1301857088258836, 9482632068303296, 69361538748381824, 509303099950899352

Offset: 0

Seiichi Manyama, Mar 25 2025

nonn

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Cf. A000108, A129123, A382433, A382443, A382446.
Cf. A131428, A178824.
Cf. A000172, A086618, A238115.

[ &+[Binomial(n, k)^2 * (Binomial(n, k) - (k gt 0 select Binomial(n, k-1) else 0)) : k in [0..n]] : n in [0..20] ]; // Vincenzo Librandi, Mar 27 2025

Table[Sum[Binomial[n,k]^2*(Binomial[n,k]-Binomial[n,k-1]),{k,0,n}],{n,0,20}] (* Vincenzo Librandi, Mar 27 2025 *)

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

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

a(n) ~ 2^(3*n+3) / (Pi * 3^(3/2) * n^2). - Vaclav Kotesovec, Mar 26 2025

A228511 a(n) = sum_{k=0}^n binomial(n,k)^24^kA000108(k).

Original entry on oeis.org

1, 5, 49, 645, 9921, 167909, 3030705, 57284901, 1120905985, 22531796805, 462793508529, 9674942743365, 205261950829761, 4409503432713765, 95746612458475569, 2098428359692863717, 46366172896708865025, 1031886636204630031493, 23112239140054942651185, 520644236358436868354565, 11789139538117859937032385

Offset: 0

Zhi-Wei Sun, Aug 23 2013

nonn

Conjecture: Let p be any odd prime.
(i) Let A(p) be the p X p determinant with (i,j)-entry equal to a(i+j) for all i,j = 0,...,p-1. Then we have A(p) == (-1)^{(p-1)/2} (mod p).
(ii) Let B(p) be the p X p determinant with (i,j)-entry equal to b(i+j) for all i,j = 0,...,p-1, where b(n) denotes sum_{k=0}^n binomial(n,k)^2*binomial(2k,k)*4^k or sum_{k=0}^n binomial(n,k)^2*binomial(2k,k)*(-2)^(n-k). Then B(p) is congruent to the Legendre symbol (p/3) modulo p.

Zhi-Wei Sun, Table of n, a(n) for n = 0..100
Zhi-Wei Sun, On some determinants with Legendre symbol entries, preprint, arXiv:1308.2900 [math.NT], 2013-2019.

Cf. A000108, A086618.

a[n_]:=Sum[Binomial[n,k]^2*4^k*CatalanNumber[k],{k,0,n}]
Table[a[n],{n,0,20}]

By Zeilberger's algorithm, we have the following recurrence: 225*(12*n+43)*(n+1)^2*(n+2)^2*a(n)
  - (n+2)^2*(3108*n^3+20869*n^2+42172*n+26271)*a(n+1)
  + (n+3)*(420*n^4+4037*n^3+13835*n^2+19872*n+9840)*a(n+2)
= (n+1)*(n+3)*(12*n+31)*(n+4)^2*a(n+3).
a(n) ~ 5^(2*n+5/2)/(32*Pi*n^2). - Vaclav Kotesovec, Aug 25 2013

A228514 a(n) = Sum_{k=0..n} binomial(n,k)^2*binomial(2k,k+1).

Original entry on oeis.org

0, 1, 8, 60, 456, 3535, 27888, 223209, 1807760, 14784759, 121909800, 1012208340, 8454274920, 70975888425, 598536562848, 5067375370380, 43052078886048, 366911053809199, 3135773892098520, 26867522192372988, 230731788606093720

Offset: 0

Zhi-Wei Sun, Aug 24 2013

nonn

Conjecture: Let  p > 3 be a prime.
(i) Let A(p) be the p X p determinant with (i,j)-entry equal to a(i+j) for all i,j = 0,...,p-1. Then A(p) == 0 (mod p^2).
(ii) Let B(p) be the (p-1) X (p-1) determinant with (i,j)-entry equal to a(i+j) for all i,j = 1,...,p-1. If p == 1 (mod 3) then B(p) == (-1)^{(p-1)/2} (mod p); if p == 2 (mod 3), then B(p) == 0 (mod p).

Zhi-Wei Sun, Table of n, a(n) for n = 0..100
Zhi-Wei Sun, On some determinants with Legendre symbol entries, preprint, arXiv:1308.2900 [math.NT], 2013-2019.

Cf. A086618, A228456.

a[n_]:=Sum[Binomial[n,k]^2*Binomial[2k,k+1],{k,0,n}]
Table[a[n],{n,0,20}]

By Zeilberger's algorithm, we have the following recurrence:
(n+2)*(n+4)^2*a(n+3) = (n+3)*(11*n^2+64*n+89)*a(n+2)
  -(n+2)*(19*n^2+94*n+108)*a(n+1)+9*(n+1)^2*(n+3)*a(n).
Recurrence (order 2): (n-1)*(n+1)^2*a(n) = 2*n*(5*n^2-2)*a(n-1) - 9*(n-1)^2*(n+1)*a(n-2). - Vaclav Kotesovec, Aug 25 2013
a(n) ~ 3^(2*n+3/2)/(4*Pi*n). - Vaclav Kotesovec, Aug 25 2013
G.f.: ((1-3*x)*hypergeom([1/3, 1/3],[1],27*(x-1)*x^2/(1-9*x))/(1-9*x)^(1/3)-1)/(6*x). - Mark van Hoeij, Nov 12 2023

cp's OEIS Frontend

A228456 Determinant of the (n+1) X (n+1) matrix with (i,j)-entry equal to A086618(i+j) for all i,j = 0,...,n.

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A005802 Number of permutations in S_n with longest increasing subsequence of length <= 3 (i.e., 1234-avoiding permutations); vexillary permutations (i.e., 2143-avoiding).

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A246065 a(n) = Sum_{k=0..n}C(n,k)^2*C(2k,k)/(2k-1), where C(n,k) denotes the binomial coefficient n!/(k!*(n-k)!).

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A086617 Symmetric square table of coefficients, read by antidiagonals, where T(n,k) is the coefficient of x^n*y^k in f(x,y) that satisfies f(x,y) = 1/((1-x)(1-y)) + xy*f(x,y)^2.

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A246459 a(n) = Sum_{k=0..n} C(n,k)^2*C(2k,k)*(2k+1), where C(n,k) denotes the binomial coefficient n!/(k!*(n-k)!).

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A238115 Number of states arising in matrix method for enumerating Hamiltonian cycles on a 2n X 2n grid.

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A087457 Number of odd length roads between any adjacent nodes in virtual optimal chordal ring of degree 3 (the length of chord < number of nodes/2).

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A246065 a(n) = Sum_{k=0..n}C(n,k)^2C(2k,k)/(2k-1), where C(n,k) denotes the binomial coefficient n!/(k!(n-k)!).

A086617 Symmetric square table of coefficients, read by antidiagonals, where T(n,k) is the coefficient of x^ny^k in f(x,y) that satisfies f(x,y) = 1/((1-x)(1-y)) + xyf(x,y)^2.

A246459 a(n) = Sum_{k=0..n} C(n,k)^2C(2k,k)(2k+1), where C(n,k) denotes the binomial coefficient n!/(k!*(n-k)!).

A228511 a(n) = sum_{k=0}^n binomial(n,k)^24^kA000108(k).