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.

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

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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)