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.

A086622 G.f. A(x) satisfies: A(x) = 1/(1-2*x) + x^2*A(x)^2.

Original entry on oeis.org

1, 2, 5, 12, 30, 76, 197, 520, 1398, 3820, 10594, 29768, 84620, 243000, 704045, 2055760, 6043750, 17875020, 53148310, 158773320, 476311940, 1434313960, 4333867170, 13135533552, 39924668220, 121661345656, 371612931492
Offset: 0

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Author

Paul D. Hanna, Jul 24 2003

Keywords

Comments

a(n) is the number of Motzkin paths of length n having no (1,0)-steps at levels 1,3,5,... and having (1,0)-steps of two colors at levels 2,4,6,... . Example: a(3) = 12 because, denoting U=(1,1), D=(1,-1), and H=(1,0), we have 8 paths of shape HHH, 2 paths of shape HUD, and 2 paths of shape UDH. - Emeric Deutsch, May 02 2011

Crossrefs

Cf. A086620 (table), A086621 (diagonal).

Programs

  • Maple
    A086622 := proc(n)
        option remember;
        if n < 3 then
            op(n+1,[1,2,5]) ;
        else
            4*(-n-1)*procname(n-1) +4*procname(n-2) +4*(2*n-3)*procname(n-3) ;
            -%/(n+2) ;
        end if;
    end proc:
    seq(A086622(n),n=0..20) ; # R. J. Mathar, Nov 02 2021
  • Mathematica
    CoefficientList[Series[(-1+2*x+Sqrt[1-4*x+8*x^3])/(2*(-x^2+2*x^3)), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 13 2014 *)

Formula

Antidiagonal sums of square table A086620.
a(n) = Sum_{k=0..floor(n/2)} C(n-k,k) C(2k,k) 2^(n-2k)/(k+1). - Paul Barry, Nov 13 2004
Hankel transform of a(n) is 1,1,1,....; Hankel transform of a(n+1) is A009531(n+2). - Paul Barry, Nov 06 2007
G.f.: 1/(1-2*x-x^2/(1-x^2/(1-2*x-x^2/(1-x^2/(1-2*x-x^2/..... (continued fraction). - Paul Barry, Dec 21 2008
D-finite with recurrence (n+2)*a(n) +4*(-n-1)*a(n-1) +4*a(n-2) +4*(2*n-3)*a(n-3)=0. - R. J. Mathar, Nov 24 2012
G.f.: (-1+2*x+sqrt(1-4*x+8*x^3))/(2*(-x^2+2*x^3)). - Vaclav Kotesovec, Feb 13 2014
a(n) ~ sqrt(50+22*sqrt(5)) * (sqrt(5)+1)^n / (2 * sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Feb 13 2014. Equivalently, a(n) ~ 5^(1/4) * 2^n * phi^(n + 5/2) / (sqrt(Pi) * n^(3/2)), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Dec 07 2021
a(n) = Sum_{i=0..floor(n/2)}2^(n-2i)*C(i)*binomial(n-i,i), where C(n) is the n-th Catalan number A000108. - José Luis Ramírez Ramírez, Apr 20 2015