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.

Showing 1-2 of 2 results.

A108447 Number of paths from (0,0) to (3n,0) that stay in the first quadrant (but may touch the horizontal axis), consisting of steps u=(2,1),U=(1,2), or d=(1,-1) and have no peaks of the form ud.

Original entry on oeis.org

1, 1, 4, 20, 113, 688, 4404, 29219, 199140, 1385904, 9807820, 70364704, 510609620, 3741212535, 27639233548, 205660399220, 1539916433473, 11594310041792, 87725707127600, 666681174728724, 5086601816592432, 38948589882247968
Offset: 0

Views

Author

Emeric Deutsch, Jun 10 2005

Keywords

Comments

Column 0 of A108446.

Examples

			a(2)=4 because we have uUddd, UddUdd, UdUddd and UUdddd.
		

Crossrefs

Programs

  • Maple
    a:=n->(1/n)*sum(binomial(n,j)*binomial(n+2*j,j-1),j=0..n): 1, seq(a(n),n=1..25);
    a := n -> `if`(n=0,1,simplify(hypergeom([1-n,(n+3)/2,(n+4)/2],[2, n+3],-4))): seq(a(n), n=0..21); # Peter Luschny, Oct 30 2015
  • Mathematica
    Flatten[{1, Table[Sum[Binomial[n, j]*Binomial[n + 2*j, j-1], {j, 0, n}]/n, {n, 1, 20}]}] (* Vaclav Kotesovec, Nov 27 2017 *)
    terms = 22; g[] = 1; Do[g[x] = 1+x*g[x]*(g[x]^2+g[x]-1) + O[x]^terms // Normal, {terms}]; CoefficientList[g[x], x] (* Jean-François Alcover, Jul 19 2018 *)

Formula

a(n) = (1/n) * Sum_{j=0..n} binomial(n, j)*binomial(n+2j, j-1) (n>=1); a(0)=1.
G.f.: G satisfies G = 1 + z*G*(G^2+G-1).
a(n) = hypergeom([1-n,(n+3)/2,(n+4)/2],[2,n+3],-4) for n>=1. - Peter Luschny, Oct 30 2015
a(n) ~ sqrt((s-1) / (Pi*(1 + 3*s))) / (2*n^(3/2) * r^(n + 1/2)), where r = 0.1215851068721183026145063923222031450327682505108... and s = 1.451605962955776643742608112028547116887657025022... are real roots of the system of equations 1 + r*s*(-1 + s + s^2) = s, r*(-1 + 2*s + 3*s^2) = 1. - Vaclav Kotesovec, Nov 27 2017
O.g.f.: A(x) = (1/x) * Revert( x/c(x/(1 - x)) ), where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. of the Catalan numbers A000108. - Peter Bala, Mar 08 2020
D-finite with recurrence 8*n*(2*n+1)*a(n) -6*(2*n-1)*(13*n-10)*a(n-1) +24*(4*n-7)*(2*n-5)*a(n-2) +4*(19*n-40)*(n-3)*a(n-3) -35*(n-3)*(n-4)*a(n-4)=0. - R. J. Mathar, Jul 26 2022

A108448 Number of peaks of the form ud in all paths from (0,0) to (3n,0) that stay in the first quadrant (but may touch the horizontal axis), consisting of steps u=(2,1), U=(1,2), or d=(1,-1).

Original entry on oeis.org

1, 7, 61, 575, 5641, 56695, 579125, 5984767, 62390545, 654862247, 6911195501, 73265596607, 779594526361, 8321683861015, 89070157349221, 955598531432447, 10273391096237089, 110647714508386375, 1193641560393864605
Offset: 1

Views

Author

Emeric Deutsch, Jun 10 2005

Keywords

Comments

a(n) = Sum_{k=1..n} k*A108446(n,k). Example: a(3) = 1*32 + 2*13 + 3*1 = 61.

Examples

			a(2) = 7 because in the ten paths (ud)(ud), (ud)Udd, u(ud)d, uUddd, Udd(ud), UddUdd, Ud(ud)d, UdUddd, U(ud)dd and UUdddd (see A027307) we have 7 ud's (shown between parentheses).
		

Crossrefs

Programs

  • Maple
    A:=(2/3)*sqrt((z+3)/z)*sin((1/3)*arcsin(sqrt(z)*(z+18)/(z+3)^(3/2)))-1/3: G:=z*A/(1-2*z*A-3*z*A^2): Gser:=series(G,z=0,25): seq(coeff(Gser,z^n),n=1..23);
  • Mathematica
    RecurrenceTable[{(n-1)*(2*n-1)*a[n]==(18*n^2-26*n+1)*a[n-1] +(46*n^2-225*n+276)*a[n-2]+2*(n-3)*(2*n-5)*a[n-3], a[1]==1, a[2]==7, a[3]==61},a,{n,20}] (* Vaclav Kotesovec, Oct 18 2012 *)

Formula

G.f.: z*A/(1-2*z*A-3*z*A^2), where A=1+z*A^2+z*A^3 or, equivalently, A=(2/3)*sqrt((z+3)/z)*sin((1/3)*arcsin(sqrt(z)*(z+18)/(z+3)^(3/2)))-1/3 (the g.f. of A027307).
Recurrence: (n-1)*(2*n-1)*a(n) = (18*n^2-26*n+1)*a(n-1) + (46*n^2-225*n+276)*a(n-2) + 2*(n-3)*(2*n-5)*a(n-3). - Vaclav Kotesovec, Oct 18 2012
a(n) ~ sqrt(70*sqrt(5)-150)*((11+5*sqrt(5))/2)^n/(20*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 18 2012. Equivalently, a(n) ~ phi^(5*n - 2) / (2 * 5^(1/4) * sqrt(Pi*n)), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Dec 07 2021
a(n) = Sum_{k=1..n} k*C(n-1,k-1)*C(2*n+k-1,n)/(n+k). - Vladimir Kruchinin, Mar 03 2014
a(n) = P(n-1,n,0,3), where P is the Jacobi Polynomial. - Richard Turk, Jun 27 2018
From Peter Bala, Feb 08 2024: (Start)
a(n) = Sum_{k = 0..n-1} binomial(2*n-1, k)*binomial(n-1, k)*2^k.
(n - 1)*(2*n - 1)*(10*n - 17)*a(n) = (220*n^3 - 814*n^2 + 950*n - 341)*a(n-1) + (n - 2)*(2*n - 3)*(10*n - 7)*a(n-2) with a(1) = 1 and a(2) = 7.. (End)
Showing 1-2 of 2 results.