A007172 Erroneous version of A027610.
1, 1, 1, 3, 7, 24, 93, 434, 2110, 10957, 58713, 321576, 1792133, 10131027, 57949430, 334970205, 1953890318, 11489753730, 68054102361, 405715557048
Offset: 1
This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
a(2) = 3 because the only dissections with 5 edges are given by a square dissected by any of the two diagonals and the pentagon with no dissecting diagonal. G.f. = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + 273*x^5 + 1428*x^6 + 7752*x^7 + 43263*x^8 + ...
List([0..25],n->Binomial(3*n,n)/(2*n+1)); # Muniru A Asiru, Oct 31 2018
a001764 n = a001764_list !! n a001764_list = 1 : [a258708 (2 * n) n | n <- [1..]] -- Reinhard Zumkeller, Jun 23 2015
[Binomial(3*n,n)/(2*n+1): n in [0..30]]; // Vincenzo Librandi, Sep 04 2014
A001764 := n->binomial(3*n,n)/(2*n+1): seq(A001764(n), n=0..25); with(combstruct): BB:=[T,{T=Prod(Z,F),F=Sequence(B),B=Prod(F,Z,F)}, unlabeled]:seq(count(BB,size=i),i=0..22); # Zerinvary Lajos, Apr 22 2007 with(combstruct):BB:=[S, {B = Prod(S,S,Z), S = Sequence(B)}, labelled]: seq(count(BB, size=n)/n!, n=0..21); # Zerinvary Lajos, Apr 25 2008 n:=30:G:=series(RootOf(g = 1+x*g^3, g),x=0,n+1):seq(coeff(G,x,k),k=0..n); # Robert FERREOL, Apr 03 2015 alias(PS=ListTools:-PartialSums): A001764List := proc(m) local A, P, n; A := [1,1]; P := [1]; for n from 1 to m - 2 do P := PS(PS([op(P), P[-1]])); A := [op(A), P[-1]] od; A end: A001764List(25); # Peter Luschny, Mar 26 2022
InverseSeries[Series[y-y^3, {y, 0, 24}], x] (* then a(n)=y(2n+1)=ways to place non-crossing diagonals in convex (2n+4)-gon so as to create only quadrilateral tiles *) (* Len Smiley, Apr 08 2000 *) Table[Binomial[3n,n]/(2n+1),{n,0,25}] (* Harvey P. Dale, Jul 24 2011 *)
{a(n) = if( n<0, 0, (3*n)! / n! / (2*n + 1)!)};
{a(n) = if( n<0, 0, polcoeff( serreverse( x - x^3 + O(x^(2*n + 2))), 2*n + 1))};
{a(n) = my(A); if( n<0, 0, A = 1 + O(x); for( m=1, n, A = 1 + x * A^3); polcoeff(A, n))};
b=vector(22);b[1]=1;for(n=2,22,for(i=1,n-1,for(j=1,n-1,for(k=1,n-1,if((i-1)+(j-1)+(k-1)-(n-2),NULL,b[n]=b[n]+b[i]*b[j]*b[k])))));a(n)=b[n+1]; print1(a(0));for(n=1,21,print1(", ",a(n))) \\ Gerald McGarvey, Oct 08 2008
Vec(1 + serreverse(x / (1+x)^3 + O(x^30))) \\ Gheorghe Coserea, Aug 05 2015
from math import comb def A001764(n): return comb(3*n,n)//(2*n+1) # Chai Wah Wu, Nov 10 2022
def A001764_list(n) : D = [0]*(n+1); D[1] = 1 R = []; b = false; h = 1 for i in range(2*n) : for k in (1..h) : D[k] += D[k-1] if not b : R.append(D[h]) else : h += 1 b = not b return R A001764_list(22) # Peter Luschny, May 03 2012
Table[Binomial[3 n, n]/(3 (2 n + 1) (2 n + 2)) + If[OddQ[n], Binomial[3 (n - 1)/2 + 1, n]/(n + 1), Binomial[3 n/2, n]/(n + 1)]/2 + 2 Switch[Mod[n, 3], 0, 0, 1, Binomial[n, (n - 1)/3]/n, 2, Binomial[n, (n - 2)/3]/n]/3, {n, 1, 30}] (* Robert A. Russell, Apr 11 2012 *)
Table[Boole[1==n]+Switch[Mod[n,24],5,12Binomial[(n-5)/8,(n-5)/12],17,24Binomial[(n-9)/8,(n-17)/24],,0]/(n+7),{n,60}] (* _Robert A. Russell, Mar 22 2024 *)
Table[(If[OddQ[n],3Binomial[(3n-1)/2,n],2Binomial[3n/2,n]]+If[1==Mod[n,4],3Binomial[(3n-3)/4,(n-1)/2],0]+If[2==Mod[n,6],3Binomial[n/2-1,(n-2)/3],0])/(3n+3),{n,30}]
Table[Switch[Mod[n,8],1,4Binomial[(3n-3)/8,(n-1)/8]/(n+3)-If[17==Mod[n,24],24Binomial[(n-9)/8,(n-17)/24]/(n+7),0],5,4Binomial[(3n-7)/8,(n+3)/8]/(n-1)-If[5==Mod[n,24],12Binomial[(n-5)/8,(n-5)/12]/(n+7),0],,0]-Boole[1==n],{n,50}] (* _Robert A. Russell, Mar 22 2024 *)
Table[Switch[Mod[n,12],2,6Binomial[(n-2)/4,(n-2)/12],8,12Binomial[(n-4)/4,(n-2)/6],,0]/(n+4),{n,50}] (* _Robert A. Russell, Mar 22 2024 *)
Table[Switch[Mod[n,3],1,Binomial[n,(n-1)/3],2,Binomial[n,(n-2)/3],_,0]/(3n)+(Binomial[3n,n]/(6n+3)-If[OddQ[n],Binomial[3(n-1)/2+1,n],Binomial[3n/2,n]/3]-2If[1==Mod[n,4],Binomial[(3n-3)/4,(n-1)/2],0]-2If[2==Mod[n,6],Binomial[n/2-1,n/3-2/3],0])/(4n+4),{n,30}]
Table[Switch[Mod[n,12],2,3Binomial[(n-2)/2,(n-2)/6]/(2n+2)-3Binomial[(n-2)/4,(n-2)/12]/(n+4),8,3Binomial[(n-2)/2,(n-2)/6]/(2n+2)-6Binomial[(n-4)/4,(n-2)/6]/(n+4),,0],{n,50}] (* _Robert A. Russell, Mar 22 2024 *)
Table[Switch[Mod[n,24],5,6Binomial[(n-5)/4,(n-5)/12]/(n+1)-12Binomial[(n-5)/8,(n-5)/12]/(n+7),17,6Binomial[(n-5)/4,(n-5)/12]/(n+1)-24Binomial[(n-9)/8,(n-17)/24]/(n+7),,0]/2,{n,60}] (* _Robert A. Russell, Mar 22 2024 *)
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