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.

A098746 Number of permutations of [1..n] which avoid 4231 and 42513.

Original entry on oeis.org

1, 1, 2, 6, 23, 102, 495, 2549, 13682, 75714, 428882, 2474573, 14492346, 85926361, 514763279, 3111119358, 18946375767, 116147683902, 716179441293, 4438862153246, 27638747494178, 172805469880497, 1084462349973559, 6828717036765622, 43132158190994223, 273204023401012901
Offset: 0

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Author

N. J. A. Sloane, Oct 30 2004

Keywords

Comments

(a(n)){n>=1} is the INVERT transform of (u(n)){n>=1}:=(1,1,3,12,55,273,...), the ternary numbers A001764. - David Callan, Nov 21 2011
a(n) = number of Dyck paths of semilength 2n for which all descents are of even length (counted by A001764) with no valley vertices at height 1. For example, a(2)=2 counts UUUUDDDD, UUDDUUDD. - David Callan, Nov 21 2011
Conjecture: a(n) is the number of permutations of [1..n] which avoid 1342 and 13254. - Alexander Burstein, Oct 19 2017

Crossrefs

Programs

  • Maple
    1+add( t^n * add( (n-l)*binomial(2*l+n,l)/(2*l+n), l=0..n ), n=1..30);
  • Mathematica
    Flatten[{1,Table[Sum[(n-j)*Binomial[2*j+n,j]/(2*j+n),{j,0,n}],{n,1,20}]}] (* Vaclav Kotesovec, Mar 17 2014 *)
  • PARI
    a(n) = {my(k = 1); if(n > 0, k = sum(j = 0, n, (n-j)*binomial(2*j+n, j)/(2*j+n))); k; } \\ Jinyuan Wang, Aug 03 2019

Formula

G.f.: 1 + Sum_{n>=1} (t^n*Sum_{k=0..n} ((n-k)*binomial(2*k+n,k)/(2*k+n))).
G.f.: sqrt(3)/(sqrt(3)-2*sqrt(x)*sin(asin(3*sqrt(3x)/2)/3)). - Paul Barry, Dec 15 2006
From Gary W. Adamson, Jul 07 2011: (Start)
Let M = the production matrix:
1, 1;
1, 2, 1;
1, 3, 2, 1;
1, 4, 3, 2, 1;
1, 5, 4, 3, 2, 1;
...
a(n) is the upper left term in M^n, with sum of top row terms = a(n+1). Example: top row of M^3 = (6, 11, 5, 1), where a(3) = 6 and a(4) = 23 = (6 + 11 + 5 + 1). (End)
a(n) ~ 3^(3*n+3/2) / (49 * sqrt(Pi) * 4^n * n^(3/2)). - Vaclav Kotesovec, Mar 17 2014
Conjecture: 2*(2*n-1)*(n-1)*a(n) +3*(11*n^2-67*n+76)*a(n-1) +3*(-155*n^2+931*n-1356)*a(n-2) +(469*n^2-2799*n+4070)*a(n-3) -48*(3*n-8)*(3*n-10)*a(n-4)=0. - R. J. Mathar, May 30 2014
G.f: A(x) = 1 + series reversion of x/((1+x)*c(x/(1+x))), where c(x) = (1 - sqrt(1 - 4*x))(2*x) is the g.f. of the Catalan numbers A000108. - Peter Bala, May 05 2024