A000736 Boustrophedon transform of Catalan numbers 1, 1, 1, 2, 5, 14, ...

1, 2, 4, 10, 32, 120, 513, 2455, 13040, 76440, 492231, 3465163, 26530503, 219754535, 1959181266, 18710532565, 190588702776, 2062664376064, 23636408157551, 285900639990875, 3640199365715769, 48665876423760247

Offset: 0

Author

N. J. A. Sloane, Simon Plouffe

Keywords

• nonn

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..400
• Peter Luschny, An old operation on sequences: the Seidel transform
• J. Millar, N. J. A. Sloane and N. E. Young, A new operation on sequences: the Boustrophedon transform, J. Combin. Theory, 17A 44-54 1996 (Abstract, pdf, ps).
• N. J. A. Sloane, Transforms
• Wikipedia, Boustrophedon transform
• Index entries for sequences related to boustrophedon transform

Crossrefs

Cf. A000108, A000753, A109449.

Programs

Haskell

a000736 n = sum $ zipWith (*) (a109449_row n) (1 : a000108_list)
-- Reinhard Zumkeller, Nov 05 2013

Maple

egf := (sec(x/2)+tan(x/2))*(exp(x)*((x-1/2)*BesselI(0,x)-x*BesselI(1,x))+3/2);
s := n -> 2^n*n!*coeff(series(egf,x,n+2),x,n); seq(s(n), n=0..22); # Peter Luschny, Oct 30 2014, after Sergei N. Gladkovskii

Mathematica

CoefficientList[Series[1/2*(3 + E^(2*x)*((4*x-1)*BesselI[0, 2*x] - 4*x*BesselI[1, 2*x]))*(Sec[x] + Tan[x]), {x, 0, 20}], x] * Range[0, 20]! (* Vaclav Kotesovec, Oct 30 2014, after Peter Luschny *)
t[n_, 0] := If[n == 0, 1, CatalanNumber[n - 1]]; t[n_, k_] := t[n, k] = t[n, k-1] + t[n-1, n-k]; a[n_] := t[n, n]; Array[a, 30, 0] (* Jean-François Alcover, Feb 12 2016 *)

Python

from itertools import accumulate, count, islice
def A000736_gen(): # generator of terms
    yield 1
    blist, c = (1,), 1
    for i in count(0):
        yield (blist := tuple(accumulate(reversed(blist),initial=c)))[-1]
        c = c*(4*i+2)//(i+2)
A000736_list = list(islice(A000736_gen(),40)) # Chai Wah Wu, Jun 12 2022

Formula

E.g.f.: (sec(x) + tan(x))*(integral(exp(2*x)*(BesselI(0,2*x)-BesselI(1,2*x)),x)+1). - Sergei N. Gladkovskii, Oct 30 2014

a(n) ~ n! * (6/Pi+2*exp(Pi)*((2-1/Pi)*BesselI(0,Pi)-2*BesselI(1,Pi))) * 2^n / Pi^n. - Vaclav Kotesovec, Oct 30 2014