1,2

A034910 occurs in connection with the Vandermonde permanent of (1,3,5,7,9,...); see the Mathematica section of A203516. - Clark Kimberling, Jan 03 2012

Table of n, a(n) for n=1..14.

Index entries for sequences related to factorial numbers

4*a(n) = (8*n-4)(!^8) := product(8*j-4, j=1..n) = 4^n*A001147(n) = 2^n*(2*n)!/n!, A001147(n) = (2*n-1)!!; E.g.f. (-1+(1-8*x)^(-1/2))/4.

a(n) = A090802(2n-1, n). - Ross La Haye, Oct 18 2005

a(n) = (2*n)!/(n)!*2^(n-2). - Zerinvary Lajos, Sep 25 2006

G.f.: x/(1-12x/(1-8x/(1-20x/(1-16x/(1-28x/(1-24x/(1-36x/(1-32x/(1-... (continued fraction). - Philippe Deléham, Jan 07 2011

From Peter Bala, Feb 01 2015: (Start)

Recurrence equation: a(n) = (7*n - 3)*a(n-1) + 4*(n - 1)*(2*n - 3)*a(n-2).

The sequence b(n) := a(n)* Sum {k = 0..n-1} (-1)^k/( 2^k*(2*k + 1)*binomial(2*k,k) ) beginning [1, 11, 222, 6210, 223584, ...] satisfies the same recurrence. This leads to the finite continued fraction expansion b(n)/a(n) = 1/(1 + 1/(11 + 24/(18 + 60/(25 + ... + 4*(n - 1)*(2*n - 3)/(7*n - 3) )))) for n >= 3.

Letting n tend to infinity gives the continued fraction expansion Sum {k = 0..inf} (-1)^k/( 2^k*(2*k + 1)*binomial(2*k,k) ) = 4/3*log(2) = 1/(1 + 1/(11 + 24/(18 + 60/(25 + ... + 4*(n - 1)*(2*n - 3)/((7*n - 3) + ... ))))). (End)

From Peter Bala, Feb 03 2015: (Start)

This sequence satisfies several other second order recurrence equations leading to some continued fraction expansions.

1) a(n) = (9*n + 4)*a(n-1) - 4*n*(2*n - 1)*a(n-2).

This recurrence is also satisfied by the (integer) sequence c(n) := a(n)*Sum {k = 0..n} 1/( 2^k*(2*k + 1)*binomial(2*k,k) ). From this we can obtain the continued fraction expansion Sum {k >= 0} 1/( 2^k*(2*k + 1)*binomial(2*k,k) ) = 8/sqrt(7)*arctan(sqrt(7)/7) = 8/sqrt(7)*A195699 = 1 + 1/(12 - 24/(22 - 60/(31 - ... - 4*n*(2*n - 1)/((9*n + 4) - ... )))).

2) a(n) = (12*n + 2)*a(n-1) - 8*(2*n - 1)^2*a(n-2).

This recurrence is also satisfied by the (integer) sequence d(n) := a(n)*Sum {k = 0..n} 1/( (2*k + 1)*2^k ). From this we can obtain the continued fraction expansion Sum {k >= 0} 1/( (2*k + 1)*2^k ) = 1/sqrt(2)*log(3 + 2*sqrt(2)) = 1 + 2/(12 - 8*3^2/(26 - 8*5^2/(38 - ... - 8*(2*n - 1)^2/((12*n + 2) - ... )))). Cf. A002391.

3) a(n) = (4*n + 6)*a(n-1) + 8*(2*n - 1)^2*a(n-2).

This recurrence is also satisfied by the (integer) sequence e(n) := a(n)*Sum {k = 0..n} (-1)^k/( (2*k + 1)*2^k ). From this we can obtain the continued fraction expansion Sum {k >= 0} (-1)^k/( (2*k + 1)*2^k ) = 1/sqrt(2)*arctan(sqrt(2)/2) = 1 - 2/(12 + 8*3^2/(14 + 8*5^2/(18 + ... + 8*(2*n - 1)^2/((4*n + 6) + ... )))). Cf. A073000. (End)

a(n) = (-1)^n / (16*a(-n)) for all n in Z. - Michael Somos, Feb 04 2015

G.f. = x + 12*x^2 + 240*x^3 + 6720*x^4 + 241920*x^5 + 10644480*x^6 + ...

[seq((2*n)!/(n)!*2^(n-2), n=1..14)]; # Zerinvary Lajos, Sep 25 2006

s=1; lst={s}; Do[s+=n*s; AppendTo[lst, s], {n, 11, 5!, 8}]; lst (* Vladimir Joseph Stephan Orlovsky, Nov 08 2008 *)

0[ n_] := Pochhammer[ 1/2, n] 8^n / 4; (* Michael Somos, Feb 04 2015 *)

(PARI) {a(n) = if( n==1, 1, n>1, a(n-1) * (8*n - 4), a(n+1) / (8*n + 4))}; /* Michael Somos, Feb 04 2015 */

Cf. A001147, A045755, A034908, A034909, A034911, A034912, A203516.

Sequence in context: A009150 A009080 A002166 * A332958 A091745 A012344

Adjacent sequences:  A034907 A034908 A034909 * A034911 A034912 A034913

easy,nonn

Wolfdieter Lang

approved