cp's OEIS Frontend

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.

A253194 Number of ways to place non-intersecting diagonals in a convex (n+2)-gon so as to create no pentagons.

Original entry on oeis.org

1, 3, 10, 39, 162, 707, 3190, 14766, 69719, 334481, 1625846, 7989908, 39631204, 198151579, 997623275, 5053274850, 25734158411, 131680565544, 676693557574, 3490897656337, 18071699948492, 93851485181749, 488815126122166
Offset: 1

Views

Author

Michael D. Weiner, Mar 24 2015

Keywords

Examples

			a(3)=10 because the pentagon allows all but the null placement, i.e., 5 placements of 1 diagonal and 5 placements of two diagonals.
		

Crossrefs

Programs

  • Mathematica
    Rest[CoefficientList[(InverseSeries[Series[(y-2*y^2+y^4-y^5)/(1-y),{y,0,24}],x]-x)/x,x]]
  • PARI
    A253194(n)=sum(i=0,(n-1)\3,sum(k=i+1,n-2*i, (-1)^i*binomial(n+k,k)*binomial(k,i)*binomial(n-3*i-1,k-i-1)),if(n%3==0,(-1)^(n/3)*binomial(4*n/3,n/3)))/(n+1) \\ M. F. Hasler, Apr 07 2015

Formula

a(n) = (1/(n+1))*Sum_{i=0..floor(n/3)} Sum_{k=i+1..n-2*i} (-1)^i*binomial(n+k,k)*binomial(k,i)*binomial(n-3*i-1,k-i-1), n !== 0 (mod 3),
a(n) = ((-1)^(n/3)/(n+1))*binomial(4*n/3,n/3) + (1/(n+1))*Sum_{i=0..(n/3)-1} Sum_{k=i+1..n-2*i} (-1)^i*binomial(n+k,k)*binomial(k,i)*binomial(n-3*i-1,k-i-1), n == 0 (mod 3).
Recurrence: 275*(n-2)*(n-1)*n*(n+1)*(13962464*n^5 - 196202616*n^4 + 1069508732*n^3 - 2802358002*n^2 + 3480787751*n - 1597000860)*a(n) = 900*(n-2)*(n-1)*n*(27924928*n^6 - 406367696*n^5 + 2336399896*n^4 - 6678345644*n^3 + 9735406192*n^2 - 6526643891*n + 1424056473)*a(n-1) - 8*(n-2)*(n-1)*(1870970176*n^7 - 30033090896*n^6 + 197840216728*n^5 - 682911269612*n^4 + 1297104157966*n^3 - 1273486799084*n^2 + 486871358313*n + 21712608900)*a(n-2) - 16*(n-2)*(2569093376*n^8 - 47662201536*n^7 + 375012676176*n^6 - 1627682459628*n^5 + 4239503473896*n^4 - 6734585440155*n^3 + 6299789310412*n^2 - 3112752665481*n + 598681926090)*a(n-3) + 16*(2457393664*n^9 - 54190809728*n^8 + 518749193184*n^7 - 2816319789288*n^6 + 9492888047100*n^5 - 20388222826734*n^4 + 27407291375141*n^3 - 21462176121217*n^2 + 8117426803296*n - 745665750648)*a(n-4) - 8*(n-4)*(2*n - 7)*(4*n - 17)*(4*n - 11)*(13962464*n^5 - 126390296*n^4 + 424322908*n^3 - 631422862*n^2 + 369599799*n - 31302531)*a(n-5). - Vaclav Kotesovec, Mar 30 2015