A000888 -id:A000888 - cp's OEIS frontend

A186418 a(n) = binomial(2*n,n)^4/(n + 1)^2.

1, 4, 144, 10000, 960400, 112021056, 14876193024, 2167749739584, 338710896810000, 55880440123393600, 9629613008027474176, 1719721549507980904704, 316402760115623198128384, 59700436261400947600000000

Offset: 0

Emanuele Munarini, Feb 21 2011

Cf. A000108, A002894, A000888, A002897, A186414, A186415, A186416.

Table[Binomial[2n,n]^4/(n+1)^2,{n,0,40}]

makelist(binomial(2*n,n)^4/(n+1)^2,n,0,40);

G.f.: 4F3({1/2,1/2,1/2,1/2},{1,2,2},256x), where 4F3 is a hypergeometric series.

A186266 Expansion of 2F1( 1/2, 3/2; 4; 16*x ).

Original entry on oeis.org

1, 3, 18, 140, 1260, 12474, 132132, 1472328, 17065620, 204155380, 2506399896, 31443925968, 401783498480, 5215458874500, 68633685693000, 914099013896400, 12304253831789700, 167193096184907100, 2291164651422801000, 31637804708163654000, 439903041116118980400

Offset: 0

Olivier Gérard, Feb 16 2011

Combinatorial interpretation welcome.

Could involve planar maps, lattice walks, and interpretations of Catalan numbers.

Indranil Ghosh, Table of n, a(n) for n = 0..800
H. Franzen and T. Weist, The Value of the Kac Polynomial at One, arXiv preprint arXiv:1608.03419 [math.RT], 2016.

Formula close to A000257, A000888, A172392.

Cf. A000108.

CoefficientList[
Series[HypergeometricPFQ[{1/2, 3/2}, {4}, 16*x], {x, 0, 20}], x]
Table[3 CatalanNumber[n] CatalanNumber[n+1] * (n+1) / (n+3), {n, 0, 20}] (* Indranil Ghosh, Mar 05 2017 *)

c(n) = binomial(2*n,n) / (n+1);
a(n) = 3 * c(n) * c(n+1) *(n+1) / (n+3); \\ Indranil Ghosh, Mar 05 2017

import math
f=math.factorial
def C(n,r): return f(n) / f(r) / f(n-r)
def Catalan(n): return C(2*n, n) / (n+1)
def A186266(n): return 3 * Catalan(n) * Catalan(n+1) * (n+1) / (n+3) # Indranil Ghosh, Mar 05 2017

a(n) = 3*A000108(n)*A000108(n+1)*(n+1)/(n+3). - David Scambler, Aug 18 2012

D-finite with recurrence n*(n+3)*a(n) -4*(2*n-1)*(2*n+1)*a(n-1)=0. - R. J. Mathar, Jun 17 2016

A186419 a(n) = binomial(2*n,n)^4/(n + 1).

Original entry on oeis.org

1, 8, 432, 40000, 4802000, 672126336, 104133351168, 17341997916672, 3048398071290000, 558804401233936000, 105925743088302215936, 20636658594095770856448, 4113235881503101575668992, 835806107659613266400000000, 172665358079973774114240000000

Offset: 0

Emanuele Munarini, Feb 21 2011

Cf. A000108, A002894, A000888, A002897, A186414, A186415, A186416, A186418.

Mathematica
```
Table[Binomial[2n,n]^4/(n+1),{n,0,40}]
```

makelist(binomial(2*n,n)^4/(n+1),n,0,12);

G.f.: 4F3({1/2,1/2,1/2,1/2},{1,1,2},256x), where 4F3 is a hypergeometric series.

A275607 a(n) = 212^nGamma(n+1/2)(n+1)/(sqrt(Pi)Gamma(n+3)).

Original entry on oeis.org

1, 4, 27, 216, 1890, 17496, 168399, 1667952, 16888014, 173997720, 1818276174, 19225409616, 205299909828, 2210922105840, 23984556773175, 261854925711840, 2874948871877910, 31722346066169880, 351589335566716170, 3912422681494285200, 43694647856506630620, 489597172255515289680

Offset: 0

Karol A. Penson, Nov 14 2016

nonn

In reference of K. Szymanski et al. the function g(x) from the Eq.(4.6) satisfies the equality g(x/4)/4 = W(x) where W(x) is the weight function of the integral representation, see below.

G. C. Greubel, Table of n, a(n) for n = 0..925
Simeon T. Stefanov, Counting fixed points free vector fields on B^2, arXiv:1807.03714 [math.GT], 2018.
K. Szymanski, B. Collins, T. Szarek and K. Zyczkowski, Convex set of quantum states with positive partial transpose analysed by hit and run algorithm, arXiv:1611.01194 [quant-ph], 2016.

Cf. A000108, A002293, A002894, A000168, A000139, A000257, A004987, A005568, A000888, A004981.

a := n -> (2^(2*n+1)*3^n*(n+1)*GAMMA(n+1/2))/(sqrt(Pi)*GAMMA(n+3)):
seq(a(n), n=0..21); # Peter Luschny, Nov 14 2016

g[z_] :=  E^z (BesselI[0,z] - (1-1/z) BesselI[1,z])
Table[CoefficientList[2/3 Series[g[6z], {z,0,21}],z]] Range[0, 21]! //Flatten (* Peter Luschny, Nov 14 2016 *)
Table[ 2*12^n*(n + 1)*Gamma[n + 1/2]/(Sqrt[Pi]*Gamma[n + 3]), {n,0,100}] (* G. C. Greubel, Jan 13 2017 *)

a(n)=2*12^n*gamma(n+1/2)*(n+1)\/(sqrt(Pi)*(n+2)!) \\ Charles R Greathouse IV, Nov 14 2016

a(n)=2*3^n*binomial(2*n+1,n-1)*(n+1)/(2*n+1)/n \\ Charles R Greathouse IV, Nov 14 2016

O.g.f: (1/54)*(1-(6*z+1)*sqrt(1-12*z))/z^2;
E.g.f.(in Maple notation): (1/9)*exp(6*z)*(6*z*(BesselI(0,6*z)-BesselI(1,6*z))+ BesselI(1,6*z))/z;
Recurrence: (-12*n^2-54*n-54)*a(n+1)+(n^2+6*n+8)*a(n+2)=0, n=0,1..., for the initial values a(0)=1, a(1)=4.
Integral representation as the n-th Hausdorff moment of the positive function W(x) on the segment x=(0,12), i.e., a(n) = Integral_{x=0..12} x^n*W(x) dx, where W(x) = (1/27)*sqrt(12-x)*(3+(1/2)*x)/(Pi*sqrt(x)). This representation is unique.
a(n) ~ 2^(2*n+1)*3^n/(sqrt(Pi)*n^(3/2)). - Ilya Gutkovskiy, Nov 14 2016
a(n) = 2*3^n*binomial(2n+1, n-1)*(n+1)/(2n^2+n). - Charles R Greathouse IV, Nov 14 2016

A379100 Triangle read by rows: T(n, k) = binomial(2k, k) binomial(2*n, n) / (n + 1).

Original entry on oeis.org

1, 1, 2, 2, 4, 12, 5, 10, 30, 100, 14, 28, 84, 280, 980, 42, 84, 252, 840, 2940, 10584, 132, 264, 792, 2640, 9240, 33264, 121968, 429, 858, 2574, 8580, 30030, 108108, 396396, 1472328, 1430, 2860, 8580, 28600, 100100, 360360, 1321320, 4907760, 18404100

Offset: 0

Peter Luschny, Dec 15 2024

			  [0]   1;
  [1]   1,   2;
  [2]   2,   4,   12;
  [3]   5,  10,   30,  100;
  [4]  14,  28,   84,  280,   980;
  [5]  42,  84,  252,  840,  2940,  10584;
  [6] 132, 264,  792, 2640,  9240,  33264, 121968;
  [7] 429, 858, 2574, 8580, 30030, 108108, 396396, 1472328;

Cf. A002894, A000888, A270577, A379099 (row sums).

T := (n, k) -> binomial(2*k, k) * binomial(2*n, n) / (n + 1):
seq(seq(T(n, k), k = 0..n), n = 0..8);

T(n, k) = ((2*k)! * (2*n)!) / ((k! * n!)^2 * (n+1)).

A189942 Table, read by rows, of the number of quivers of type Ã_(n-1) according to the parameter k (n >= 2, 1 <= k <= [n/2]).

Original entry on oeis.org

1, 2, 5, 4, 14, 12, 42, 36, 22, 132, 108, 100, 429, 349, 315, 172, 1430, 1144, 1028, 980, 4862, 3868, 3432, 3240, 1651, 16796, 13260, 11700, 10920, 10584, 58786, 46210, 40520, 37556, 36036, 18028, 208012, 162792, 142120, 130900, 124740, 121968

Offset: 2

Jonathan Vos Post, May 01 2011

Table 1, p. 15 of Bastian.

There is a bijection with dissections of an annulus [Hermund André Torkildsen]. - N. J. A. Sloane, Jan 31 2013

			The table begins
===================================
n  |  r=1 | r=2 | r=3 | r=4 | r=5 |
===================================
n=2     1
n=3     2
n=4     5      4
n=5    14     12
n=6    42     36    22
n=7   132    108   100
n=8   429    349   315   172
n=9  1430   1144  1028   980
n=10 4862   3868  3432  3240  1651
===================================

Francois Bergeron, Gilbert Labelle and Pierre Leroux, Combinatorial species and tree-like structures, Encyclopedia of Mathematics and its Applications, vol. 67, Cambridge University Press, Cambridge, 1998, Translated from the 1994 French original by Margaret Readdy, With a foreword by Gian-Carlo Rota.

Ibrahim Assem, Thomas Brustle, Gabrielle Charbonneau-Jodoin and Pierre-Guy Plamondon, Gentle algebras arising from surface triangulations, Algebra & Number Theory 4 (2010), no. 2, 201-229; arXiv:0903.3347 [math.RT], 2009.
Janine Bastian, Thomas Prellberg, Martin Rubey and Christian Stump, Counting the number of elements in the mutation classes of Ã_n-quivers, arXiv:0906.0487 [math.CO], 2009-2011.
Janine Bastian, Mutation classes of Ã_n-quivers and derived equivalence classification of cluster tilted algebras of type Ã_n, Algebra Number Theory 5 (2011), no. 5, 567-594; arXiv:0901.1515 [math.RT], 2009-2012.
Hermund André Torkildsen, A geometric realization of the m-cluster category of type Ã, arXiv 1208.2138 [math.RT], 2012. - From _N. J. A. Sloane_, Jan 31 2013

Cf. A000108 (r=1), A000888 (n=2r+1).

a[r_, r_] := 1/2 (Binomial[2 r, r]/2 + Sum[EulerPhi[k]/(4 r) Binomial[2 r/k, r/k]^2, {k, Divisors@r}]);
a[r_, s_] := 1/2 Sum[EulerPhi[k]/(r + s) Binomial[2 r/k, r/k] Binomial[2 s/k, s/k], {k, Intersection[Divisors@r, Divisors@s]}];
Table[a[r, n - r], {n, 2, 10}, {r, n/2}] // TableForm
(* Andrey Zabolotskiy, Jan 19 2022 *)

Rows 11-13 added by Andrey Zabolotskiy, Jan 19 2022

A318614 Scaled g.f. S(u) = Sum_{n>0} a(n)16(u/16)^n satisfies T(u) = d/du S(u), with T(u) as defined by A318417; sequence gives a(n).

Original entry on oeis.org

1, 6, 76, 1260, 24276, 515592, 11721072, 280020312, 6945369860, 177358000248, 4635276570288, 123449340098448, 3339525750984528, 91535631253610400, 2537277723600799680, 71015600640006437040, 2004523477053308685540, 57003431104378084982040

Offset: 1

Bradley Klee, Aug 30 2018

nonn

Area interior to the central loop of u = 2*H = x^2 + y^2 - (1/2)*(x^4 + y^4) equals to Pi*S(u), when u in [0,1/2].

			Singular Value: S(1/2) = 1/sqrt(2).
N=4, h=1/sqrt(2) Quantization: S(u) = (n+1/2)*h/N.
  n  |                  u
==================================================
  0  |  0.08544689553344134756293807606337...
  1  |  0.23840989875904155311088418238272...
  2  |  0.36638282702449450473835851051425...
  3  |  0.46595506694324457665483887176081...

E. Heller, The Semiclassical Way to Dynamics and Spectroscopy, Princeton University Press, 2018, page 204.

Eric Weisstein's World of Mathematics, Plane Division by Ellipses.
Eric Weisstein's World of Mathematics, Circle Ellipse Intersection.

Cf. A318417, A010503, A247719, A000888.

a:=[1,6];; for n in [3..20] do a[n]:=(1/(n*(n-1)^2))*(12*(n-1)*(2*n-3)^2*a[n-1]-(128*(n-2)*(2*n-5)*(2*n-3)*a[n-2])); od; a; # Muniru A Asiru, Sep 24 2018

RecurrenceTable[{(n-1)^2*n*a[n] - 12*(n-1)*(2*n-3)^2*a[n-1] + 128*(n-2)*(2*n-5)*(2*n-3)*a[n-2] == 0, a[1] == 1, a[2] == 6}, a, {n, 1, 1000}]

(n-1)^2*n*a(n) - 12*(n-1)*(2*n-3)^2*a(n-1) + 128*(n-2)*(2*n-5)*(2*n-3)*a(n-2) == 0.

a(n) = A000108(n-1)*A098410(n-1).

cp's OEIS Frontend

A186418 a(n) = binomial(2*n,n)^4/(n + 1)^2.

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A186266 Expansion of 2F1( 1/2, 3/2; 4; 16*x ).

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A186419 a(n) = binomial(2*n,n)^4/(n + 1).

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A275607 a(n) = 2*12^n*Gamma(n+1/2)*(n+1)/(sqrt(Pi)*Gamma(n+3)).

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A379100 Triangle read by rows: T(n, k) = binomial(2*k, k) * binomial(2*n, n) / (n + 1).

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A189942 Table, read by rows, of the number of quivers of type Ã_(n-1) according to the parameter k (n >= 2, 1 <= k <= [n/2]).

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A318614 Scaled g.f. S(u) = Sum_{n>0} a(n)*16*(u/16)^n satisfies T(u) = d/du S(u), with T(u) as defined by A318417; sequence gives a(n).

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A275607 a(n) = 212^nGamma(n+1/2)(n+1)/(sqrt(Pi)Gamma(n+3)).

A379100 Triangle read by rows: T(n, k) = binomial(2k, k) binomial(2*n, n) / (n + 1).

A318614 Scaled g.f. S(u) = Sum_{n>0} a(n)16(u/16)^n satisfies T(u) = d/du S(u), with T(u) as defined by A318417; sequence gives a(n).