A125558 -id:A125558 - cp's OEIS frontend

A187535 Central Lah numbers: a(n) = A105278(2n,n) = A008297(2n,n).

1, 2, 36, 1200, 58800, 3810240, 307359360, 29682132480, 3339239904000, 428906814336000, 61934143990118400, 9931984545324441600, 1751339941492209868800, 336796142594655744000000, 70149825129001153536000000, 15732267448930658699673600000

Offset: 0

Emanuele Munarini, Mar 11 2011

a(n) is the number of Lah partitions of a set of size 2n with n blocks.

Cf. A008297, A111596, A066667, A187536, A187538, A187539, A187540, A187542 - A187548, A090181, A125558.

A187535:= n -> if n=0 then 1 else binomial(2*n-1,n-1)*(2*n)!/n! fi;
seq(A187535(n),n=0..12);

a[n_]:=If[n==0,1,Binomial[2n-1,n-1](2n)!/n!]
Table[a[n],{n,0,12}]
(* Alternative: *)
a[n_] := Binomial[2*n, n] FactorialPower[2*n - 1, n];
Table[a[n], {n, 0, 15}] (* Peter Luschny, Jun 15 2022 *)

a(n) := if n=0 then 1 else binomial(2*n-1,n-1)*(2*n)!/n!;
makelist(a(n),n,0,12);

[catalan_number(n)*binomial(2*n-1,n)*factorial(n+1) for n in range(15)] # Peter Luschny, Oct 07 2014

a(n) = binomial(2n-1,n-1)*(2n)!/n! (for n>0).
D-finite with recurrence (n+1)*a(n+1) = 4*(2n+1)^2*a(n) - delta(n,0).
a(n) ~ 2^(4*n)*n^n*exp(-n)/sqrt(2*n*Pi).
a(n)*a(n+2) - a(n+1)^2 is >= 0 and is a multiple of 2^(n+3) for all nonnegative  n.
a(n) == 0 (mod 10) for n>3.
E.g.f.: 1/2 + K(16x)/Pi, where K(z) is the complete elliptic integral of the first kind, which can also be written as a Legendre function of the second kind.
a(n) = Catalan(n)*C(2*n-1,n)*(n+1)!. - Peter Luschny, Oct 07 2014
a(n) = A125558(n)*(n+1)! = A090181(2*n,n)*(n+1)!. - Peter Luschny, Oct 07 2014
a(n) = (2/n)*(Gamma(2*n)^2/Gamma(n)^3) for n>0. - Peter Luschny, Oct 17 2014

A000888 a(n) = (2n)!^2 / ((n+1)!n!^3).

Original entry on oeis.org

1, 2, 12, 100, 980, 10584, 121968, 1472328, 18404100, 236390440, 3103161776, 41469525552, 562496897872, 7726605740000, 107289439704000, 1503840313184400, 21252802073091300, 302539888334593800, 4334635827016110000, 62464383654579522000, 904841214653480504400

Offset: 0

N. J. A. Sloane

a(n) is the number of walks of 2n unit steps North, East, South, or West, starting at the origin, bounded above by y=x, below by y=-x and terminating on the ray y = x >= 0. Example: a(1) counts EN, EW; a(2) counts ESNN, ESNW, ENSN, ENSW, ENEN, ENEW, EENN, EENW, EEWN, EEWW, EWEN, EWEW. - David Callan, Oct 11 2005
Bijective proof: given such an NESW walk, construct a pair (P_1, P_2) of lattice paths of upsteps U=(1,1) and downsteps D=(1,-1) as follows. To get P_1, replace each E and S with U and each W and N with D. To get P_2, replace each N and E with U and each S and W with D. For example, EENSNW -> (UUDUDD, UUUDUD). This mapping is 1-to-1 and its range is the Cartesian product of the set of Dyck n-paths and the set of nonnegative paths of length 2n. The Dyck paths are counted by the Catalan number C_n (A000108) and the nonnegative paths are counted (see for example the Callan link) by the central binomial coefficient binomial(2n,n) (A000984). So this is a bijection from these NESW walks to a set of size C_n*binomial(2n,n) = a(n). - David Callan, Sep 18 2007
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^3))^{2n}]. - Andrew V. Sutherland, Apr 01 2008
Number of walks within N^2 (the first quadrant of Z^2) starting at (0,0), ending on the vertical axis and consisting of 2 n steps taken from {(-1,-1), (-1,1), (1,-1), (1,1)}. - Manuel Kauers, Nov 18 2008
a(n) is equal to the n-th moment of the following positive function defined on x in (0,16), in Maple notation: (EllipticK(sqrt(1-x/16)) - EllipticE(sqrt(1-x/16)))/(Pi^2*sqrt(x)). This is the solution of the Hausdorff moment problem and thus it is unique. - Karol A. Penson, Feb 11 2011
The partial sums of a(n)/A013709(n) absolutely converge to 1/Pi. - Ralf Steiner, Jan 21 2016

			G.f.: 1 + 2*x + 12*x^2 + 100*x^3 + 980*x^4 + 10584*x^5 + 121968*x^6 + ...

E. R. Hansen, A Table of Series and Products, Prentice-Hall, Englewood Cliffs, NJ, 1975, p. 93.
T. M. MacRobert, Functions of a Complex Variable, 4th ed., Macmillan & Co., London, 1958, p. 177.

Vincenzo Librandi, Table of n, a(n) for n = 0..100
Marco S. Bianchi, Protected and uniformly transcendental, arXiv:2306.06239 [hep-th], 2023.
M. Bousquet-Mélou and M. Mishna, Walks with small steps in the quarter plane, arXiv:0810.4387 [math.CO], 2008-2009.
David Callan, Bijections for the identity 4^n = ... .
Kiran S. Kedlaya and Andrew V. Sutherland, Hyperelliptic curves, L-polynomials and random matrices, arXiv:0803.4462 [math.NT], 2008-2010.
Helmut Prodinger, Two New Identities Involving the Catalan Numbers: A classical approach, arXiv:1911.07604 [math.CO], 2019.
Ralf Steiner, Beispiele zur modifizierten Wallis-Lambert-Reihe, 2016.

Cf. A000108, A002894, A088538, A089835, A125558.

[(Factorial(2*n))^2/(Factorial(n))^4/(n+1): n in [0..20]]; // Vincenzo Librandi, Aug 15 2011

[seq(binomial(2*n,n)^2/(n+1),n=0..17)]; # Zerinvary Lajos, May 27 2006

f[n_] := Binomial[2 n, n]^2/(n + 1); Array[f, 18, 0]  (* Robert G. Wilson v *)
a[ n_] := SeriesCoefficient[ (1/8) (EllipticE[ 16 x] - (1 - 16 x) EllipticK[ 16 x]) / (Pi/2), {x, 0, n + 1}]; (* Michael Somos, Jan 23 2012 *)

{a(n) = if( n<0, 0, (2*n)!^2 / n!^4 / (n+1))}; /* Michael Somos, Sep 11 2005 */

G.f.: 1/4*((16*x-1)*EllipticK(4*x^(1/2)) + EllipticE(4*x^(1/2)))/x/Pi. - Vladeta Jovovic, Oct 12 2003
Given G.f. A(x), y = x*A(x) satisfies y = y'' * (1 - 16*x) * x/4. - Michael Somos, Sep 11 2005
a(n) = binomial(2*n,n)^2/(n+1). - Zerinvary Lajos, May 27 2006
G.f.: 2F1(1/2,1/2;2;16*x). - Paul Barry, Sep 03 2008
a(n) = 2*A125558(n) (n >= 1). - Olivier Gérard, Feb 16 2011
A002894(n) = (n+1) * a(n). A001246(n) = a(n) / (n+1). A089835(n) = n! * a(n). - Michael Somos, May 12 2012
G.f.: 1 + 4*x/(G(0)-4*x) where G(k) = 4*x*(2*k+1)^2 + (k+1)*(k+2) - 4*x*(k+1)*(k+2)*(2*k+3)^2/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Jul 30 2012
D-finite with recurrence: (n+1)*(n+2)*a(n+1) = 4*(2*n+1)^2*a(n). - Vaclav Kotesovec, Sep 11 2012
a(n) = C(n)*binomial(2*n,n) = Sum_{k=0..2*n} binomial(2*n,k)*C(k)*C(2*n-k) where C(k) are Catalan numbers (A000108), see Prodinger. - Michel Marcus, Nov 19 2019
Sum_{n>=0} a(n)/16^n = 4/Pi (A088538). - Amiram Eldar, May 06 2023

A131198 Triangle T(n,k), 0 <= k <= n, read by rows, given by [1,0,1,0,1,0,1,0,...] DELTA [0,1,0,1,0,1,0,1,...] where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 3, 1, 0, 1, 6, 6, 1, 0, 1, 10, 20, 10, 1, 0, 1, 15, 50, 50, 15, 1, 0, 1, 21, 105, 175, 105, 21, 1, 0, 1, 28, 196, 490, 490, 196, 28, 1, 0, 1, 36, 336, 1176, 1764, 1176, 336, 36, 1, 0, 1, 45, 540, 2520, 5292, 5292, 2520, 540, 45, 1, 0

Offset: 0

Philippe Deléham, Oct 20 2007

Mirror image of triangle A090181, another version of triangle of Narayana (A001263).

Equals A133336*A130595 as infinite lower triangular matrices. - Philippe Deléham, Oct 23 2007

			Triangle begins:
  1;
  1,  0;
  1,  1,   0;
  1,  3,   1,   0;
  1,  6,   6,   1,   0;
  1, 10,  20,  10,   1,   0;
  1, 15,  50,  50,  15,   1,  0;
  1, 21, 105, 175, 105,  21,  1, 0;
  1, 28, 196, 490, 490, 196, 28, 1, 0; ...

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Paul Barry, Continued fractions and transformations of integer sequences, JIS 12 (2009), Article 09.7.6.
Paul Barry, On a Generalization of the Narayana Triangle, J. Int. Seq. 14 (2011), Article 11.4.5.
Paul Barry, On a transformation of Riordan moment sequences, arXiv:1802.03443 [math.CO], 2018.
Paul Barry and A. Hennessy, A Note on Narayana Triangles and Related Polynomials, Riordan Arrays, and MIMO Capacity Calculations, J. Int. Seq. 14 (2011), Article 11.3.8.
FindStat - Combinatorial Statistic Finder, The number of peaks of a Dyck path., The number of double rises of a Dyck path., The number of valleys of a Dyck path., The number of left oriented leafs except the first one of a binary tree., The number of left tunnels of a Dyck path.
Aoife Hennessy, A Study of Riordan Arrays with Applications to Continued Fractions, Orthogonal Polynomials and Lattice Paths, Ph. D. Thesis, Waterford Institute of Technology, Oct. 2011.

Cf. A000217, A002415, A006542, A006857.

[[n le 0 select 1 else (n-k)*Binomial(n,k)^2/(n*(k+1)): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Feb 06 2018

T := (n,k) -> `if`(n=0, 0^n, binomial(n,k)^2*(n-k)/(n*(k+1)));
seq(print(seq(T(n,k), k=0..n)), n=0..5); # Peter Luschny, Jun 08 2014
R := n -> simplify(hypergeom([1 - n, -n], [2], x)):
Trow := n -> seq(coeff(R(n, x), x, k), k = 0..n):
seq(print(Trow(n)), n = 0..9); # Peter Luschny, Apr 26 2022

Table[If[n == 0, 1, (n-k)*Binomial[n,k]^2/(n*(k+1))], {n,0,10}, {k,0,n}] //Flatten (* G. C. Greubel, Feb 06 2018 *)

for(n=0,10, for(k=0,n, print1(if(n==0,1, (n-k)*binomial(n,k)^2/(n* (k+1))), ", "))) \\ G. C. Greubel, Feb 06 2018

Sum_{k=0..n} T(n,k)*x^k = A000012(n), A000108(n), A001003(n), A007564(n), A059231(n), A078009(n), A078018(n), A081178(n), A082147(n), A082181(n), A082148(n), A082173(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 respectively.
Sum_{k=0..n} T(n,k)*x^(n-k) = A000007(n), A000108(n), A006318(n), A047891(n+1), A082298(n), A082301(n), A082302(n), A082305(n), A082366(n), A082367(n), for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively. - Philippe Deléham, Oct 23 2007
Sum_{k=0..floor(n/2)} T(n-k,k) = A004148(n). - Philippe Deléham, Nov 06 2007
T(2*n,n) = A125558(n). - Philippe Deléham, Nov 16 2011
T(n, k) = [x^k] hypergeom([1 - n, -n], [2], x). - Peter Luschny, Apr 26 2022

A281000 Triangle read by rows: T(n,k) = binomial(2n+1, 2k+1)binomial(2n-2*k, n-k)/(n+1-k) for 0 <= k <= n.

Original entry on oeis.org

1, 3, 1, 10, 10, 1, 35, 70, 21, 1, 126, 420, 252, 36, 1, 462, 2310, 2310, 660, 55, 1, 1716, 12012, 18018, 8580, 1430, 78, 1, 6435, 60060, 126126, 90090, 25025, 2730, 105, 1, 24310, 291720, 816816, 816816, 340340, 61880, 4760, 136, 1, 92378, 1385670, 4988412, 6651216, 3879876, 1058148, 135660, 7752, 171, 1

Offset: 0

Werner Schulte, Jan 12 2017

			Triangle begins:
n\k:      0       1       2       3       4      5     6    7  8  . . .
  0:      1
  1:      3       1
  2:     10      10       1
  3:     35      70      21       1
  4:    126     420     252      36       1
  5:    462    2310    2310     660      55      1
  6:   1716   12012   18018    8580    1430     78     1
  7:   6435   60060  126126   90090   25025   2730   105    1
  8:  24310  291720  816816  816816  340340  61880  4760  136  1
  etc.
T(3,2) = binomial(7,5) * binomial(2,1) / (3+1-2) = 21 * 2 / 2 = 21. - _Indranil Ghosh_, Feb 15 2017

Indranil Ghosh, Rows 0..100 of triangle, flattened

Row sums are A099250.
Triangle related to A000108, A097610, A280580.
Cf. A000007, A001003, A001006, A002426, A103365, A125558, A273055.

Table[Binomial[2n+1,2k+1] Binomial[2n-2k,n-k]/(n+1-k),{n,0,10},{k,0,n}]// Flatten (* Harvey P. Dale, Nov 25 2018 *)

T(n,k) = A097610(2*n+1, 2*k+1) = binomial(2*n+1, 2*k+1)*A000108(n-k) = A280580(n,k)*(2*n+1)/(2*k+1) for 0 <= k <= n.
Recurrences: T(n,0) = (2*n+1)*A000108(n) and
  (1) T(n,k) = T(n,k-1)*(n+1-k)*(n+2-k)/(2*k*(2*k+1)) for 0 < k <= n,
  (2) T(n,k) = T(n-1, k-1)*n*(2*n+1)/(k*(2*k+1)).
The row polynomials p(n,x) = Sum_{k=0..n} T(n,k)*x^(2*k+1) satisfy the recurrence equation p"(n,x) = (2*n+1)*2*n*p(n-1,x) with initial value p(0,x) = x (n > 0, p" is the second derivative of p), and Sum_{n>=0} p(n,x)*t^(2*n+1)/ ((2*n+1)!) = sinh(x*t)*(Sum_{n>=0} A000108(n)*t^(2*n)/((2*n)!)).
Conjectures:
  (1) Antidiagonal sums equal A001003(n+1);
  (2) Sum_{k=0..n} (-1)^k*T(n,k)*(2*k+1)*A000108(k)*A103365(k) = A000007(n);
  (3) Matrix inverse equals T(n,k)*A103365(n+1-k).
Sum_{k=0..n} (n+1-k)*T(n,k) = A002426(2*n+1) = A273055(n).
Sum_{k=0..n} T(n,k)*(2*k+1)*A000108(k) = (2*n+1)*A000108(n)*A000108(n+1) = A125558(n+1).
Matrix product: Sum_{i=0..n} T(n,i)*T(i,k) = T(n,k)*A000108(n+1-k) for 0<=k<=n.

cp's OEIS Frontend

A187535 Central Lah numbers: a(n) = A105278(2*n,n) = A008297(2*n,n).

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A000888 a(n) = (2*n)!^2 / ((n+1)!*n!^3).

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A131198 Triangle T(n,k), 0 <= k <= n, read by rows, given by [1,0,1,0,1,0,1,0,...] DELTA [0,1,0,1,0,1,0,1,...] where DELTA is the operator defined in A084938.

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A281000 Triangle read by rows: T(n,k) = binomial(2*n+1, 2*k+1)*binomial(2*n-2*k, n-k)/(n+1-k) for 0 <= k <= n.

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A187535 Central Lah numbers: a(n) = A105278(2n,n) = A008297(2n,n).

A000888 a(n) = (2n)!^2 / ((n+1)!n!^3).

A281000 Triangle read by rows: T(n,k) = binomial(2n+1, 2k+1)binomial(2n-2*k, n-k)/(n+1-k) for 0 <= k <= n.