a(n) = numerator(n!/2) and
A141044(n) = denominator(n!/2).
D-finite with recurrence: a(0) = a(1) = a(2) = 1; a(n) = n*a(n-1) for n>2. -
Chad Brewbaker, Jan 31 2003 [Corrected by
N. J. A. Sloane, Jul 25 2008]
a(0) = 0, a(1) = 1; a(n) = Sum_{k=1..n-1} k*a(k). -
Amarnath Murthy, Oct 29 2002
Stirling transform of a(n+1) = [1, 3, 12, 160, ...] is
A083410(n) = [1, 4, 22, 154, ...]. -
Michael Somos, Mar 04 2004
a(n) = 0^n + Sum_{k=0..n} (-1)^(n-k-1)*T(n-1, k)*cos(Pi*(n-k-1)/2)^2.
E.g.f.: (2 - x^2)/(2 - 2*x).
E.g.f. of a(n+2), n>=0, is 1/(1-x)^3.
a(n+1) = (-1)^n *
A136656(n,1), n>=1.
a(n) = n!/2 for n>=2 (proof from the e.g.f). -
Wolfdieter Lang, Apr 30 2010
a(n) = (n-2)! * t(n-1), n>1, where t(n) is the n-th triangular number (
A000217). -
Gary Detlefs, May 21 2010
O.g.f.: 1 + x*Sum_{n>=0} n^n*x^n/(1 + n*x)^n. -
Paul D. Hanna, Sep 13 2011
a(n) = if n < 2 then 1, otherwise Pochhammer(n,n)/binomial(2*n,n). -
Peter Luschny, Nov 07 2011
a(n) = Sum_{k=0..floor(n/2)} s(n,n-2*k) where s(n,k) are Stirling number of the first kind,
A048994. -
Mircea Merca, Apr 07 2012
a(n-1), n>=3, is M_1([2,1^(n-2)])/n = (n-1)!/2, with the M_1 multinomial numbers for the given n-1 part partition of n. See the second to last entry in line n>=3 of
A036038, and the above necklace comment by W. Lang. -
Wolfdieter Lang, Jun 26 2012
G.f.: A(x) = 1 + x + x^2/(G(0)-2*x) where G(k) = 1 - (k+1)*x/(1 - x*(k+3)/G(k+1)); (continued fraction). -
Sergei N. Gladkovskii, Dec 26 2012.
G.f.: 1 + x + (Q(0)-1)*x^2/(2*(sqrt(x)+x)), where Q(k) = 1 + (k+2)*sqrt(x)/(1 - sqrt(x)/(sqrt(x) + 1/Q(k+1))); (continued fraction). -
Sergei N. Gladkovskii, May 15 2013
G.f.: 1 + x + (x*Q(x)-x^2)/(2*(sqrt(x)+x)), where Q(x) = Sum_{n>=0} (n+1)!*x^n*sqrt(x)*(sqrt(x) + x*(n+2)). -
Sergei N. Gladkovskii, May 15 2013
G.f.: 1 + x/2 + (Q(0)-1)*x/(2*(sqrt(x)+x)), where Q(k) = 1 + (k+1)*sqrt(x)/(1 - sqrt(x)/(sqrt(x) + 1/Q(k+1))); (continued fraction). -
Sergei N. Gladkovskii, May 15 2013
G.f.: 1 + x + x^2*G(0)/2, where G(k) = 1 + 1/(1 - x/(x + 1/(k+3)/G(k+1))); (continued fraction). -
Sergei N. Gladkovskii, Jun 01 2013
G.f.: 1+x + x^2*W(0), where W(k) = 1 - x*(k+3)/( x*(k+3) - 1/(1 - x*(k+1)/( x*(k+1) - 1/W(k+1) ))); (continued fraction). -
Sergei N. Gladkovskii, Aug 26 2013
a(0)=a(1)=1; after which, for even n: a(n) = (n/2) * (n-1)!, and for odd n: a(n) = (n-1)/2 * ((n-1)! + (n-2)!). [The formula was empirically found after viewing these numbers in factorial base,
A007623, and is easily proved by considering formulas from Lang (Apr 30 2010) and Detlefs (May 21 2010) shown above.]
For n >= 1, a(2*n+1) = a(2*n) +
A153880(a(2*n)). [Follows from above.] (End)
The o.g.f. A(x) satisfies the Riccati equation x^2*A'(x) + (x - 1)*A(x) + 1 - x^2 = 0.
G.f.: A(x) = 1 + x + x^2/(1 - 3*x/(1 - x/(1 - 4*x/(1 - 2*x/(1 - 5*x/(1 - 3*x/(1 - ... - (n + 2)*x/(1 - n*x/(1 - ... ))))))))) (apply Stokes, 1982).
A(x) = 1 + x + x^2/(1 - 2*x - x/(1 - 3*x/(1 - 2*x/(1 - 4*x/(1 - 3*x/(1 - 5*x/(1 - ... - n*x/(1 - (n+2)*x/(1 - ... ))))))))). (End)
H(x) = (1 - (1 + x)^(-2)) / 2 = x - 3*x^2/2! + 12*x^3/3! - ..., an e.g.f. for the signed sequence here (n!/2!), ignoring the first two terms, is the compositional inverse of G(x) = (1 - 2*x)^(-1/2) - 1 = x + 3*x^2/2! + 15*x^3/3! + ..., an e.g.f. for
A001147. Cf.
A094638. H(x) is the e.g.f. for the sequence (-1)^m * m!/2 for m = 2,3,4,... . Cf.
A001715 for n!/3! and
A001720 for n!/4!. Cf. columns of
A094587,
A173333, and
A213936 and rows of
A138533. -
Tom Copeland, Dec 27 2019
Sum_{n>=0} 1/a(n) = 2*(e-1).
Sum_{n>=0} (-1)^n/a(n) = 2/e. (End)
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