A091593 Reversion of Jacobsthal numbers A001045.

1, -1, -1, 5, -3, -21, 51, 41, -391, 407, 1927, -6227, -2507, 49347, -71109, -236079, 966129, 9519, -7408497, 13685205, 32079981, -167077221, 60639939, 1209248505, -2761755543, -4457338681, 30629783831, -22124857219, -206064020315, 572040039283, 590258340811

Offset: 0

Author

Paul Barry, Jan 23 2004

Keywords

• easy
• sign

Comments

Hankel transform is (-2)^C(n+1,2). - Paul Barry, Apr 28 2009

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..1000
• Index entries for reversions of series

Crossrefs

Cf. A001263, A154825.

Programs

Maple

a := n -> hypergeom([-n,-n-1], [2], -2);
seq(simplify(a(n)), n=0..30); # Peter Luschny, Sep 22 2014
# Using function CompInv from A357588.
CompInv(25, n -> (2^n - (-1)^n)/3 ); # Peter Luschny, Oct 07 2022

Mathematica

a[n_] := Hypergeometric2F1[-n - 1, -n - 1, 2, -2] + (n + 1)*Hypergeometric2F1[-n, -n, 3, -2]; Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Oct 03 2016, after Vladimir Kruchinin *)

Maxima

a(n) := hypergeometric([ -n - 1, -n - 1 ], [ 2 ], -2) + (n + 1) * hypergeometric([ -n, -n ], [ 3 ], -2); /* Vladimir Kruchinin, Oct 12 2011 */

Sage

# Algorithm of L. Seidel (1877)
def A091593_list(n) :
    D = [0]*(n+2); D[1] = 1
    R = []; b = false; h = 1
    for i in range(2*n) :
        if b :
            for k in range(1, h, 1) : D[k] += -2*D[k+1]
            R.append(D[1])
        else :
            for k in range(h, 0, -1) : D[k] += D[k-1]
            h += 1
        b = not b
    return R
A091593_list(30)  # Peter Luschny, Oct 19 2012

Formula

G.f.: (-(1+x)+sqrt(1+2*x+9*x^2))/(4*x^2). - Corrected by Seiichi Manyama, May 12 2019

a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2*k)*C(k)*(-1)^(n-k)*2^k, where C(n) is A000108(n). - Paul Barry, May 16 2005

G.f.: 1/(1+x+2x^2/(1+x+2x^2/(1+x+2x^2/(1+x+2x^2/(1+ ... (continued fraction). - Paul Barry, Apr 28 2009

a(n) = Sum_{i=0..n} (2^(i)*(-1)^(n-i)*binomial(n+1,i)^2*(n-i+1)/(i+1))/(n+1). - Vladimir Kruchinin, Oct 12 2011

Conjecture: (n+2)*a(n) +(2*n+1)*a(n-1) +9*(n-1)*a(n-2)=0. - R. J. Mathar, Nov 26 2012

a(n) = (-1)^n*hypergeom([-n/2, (1-n)/2], [2], -8). - Peter Luschny, May 28 2014

R. J. Mathar's conjecture confirmed by Maple using this hypergeom form. - Robert Israel, Sep 22 2014

a(n) = Sum_{k = 0..n} (-2)^k * (1/(n+1))*binomial(n+1, k)*binomial(n+1, k+1) = Sum_{k = 0..n} (-2)^k * N(n+1,k+1), where N(n,k) = A001263(n,k) are the Narayana numbers. - Peter Bala, Sep 01 2023