0,2

a(n-2)+1 = number of (3412,1243)-, (3412,2134)-, and (3412,1324)-avoiding involutions in S_n, n>1. - Ralf Stephan (ralf(AT)ark.in-berlin.de), Jul 06 2003

The number of terms in all ordered partitions of (n+1) using only ones and twos. For example, a(3)=15 because there are 15 terms in 1+1+1+1;2+1+1;1+2+1;1+1+2;2+2 - Geoffrey Critzer (critzer.geoffrey(AT)usd443.org), Apr 07 2008

a(n)=number of n-matchings in the graph obtained by a zig-zag triangulation of a convex (2n+1)-gon. Example: a(2)=7 because in the triangulation of the convex pentagon ABCDEA with diagonals AD and AC we have 7 2-matchings: {AB,CD},{AB,DE},{BC,AD},{BC,DE},{BC,EA},{CD,EA}, and {DE,AC}. - Emeric Deutsch (deutsch(AT)duke.poly.edu), Dec 25 2004

Partial sums of A029907. First differences of A002940. - Peter Bala (pbala(AT)toucansurf.com), Oct 24 2007

E. S. Egge, Restricted 3412-Avoiding Involutions: Continued Fractions, Chebyshev Polynomials and Enumerations, sec. 8

O.g.f.: (x+1)/(1-x-x^2)^2. - Len Smiley (smiley(AT)math.uaa.alaska.edu), Dec 11 2001

a(n) = 1/5*((n+2)F(n+4)+(n-1)F(n+2)), with F(n)=A000047(n). - Ralf Stephan (ralf(AT)ark.in-berlin.de), Jul 06 2003

a(n)=sum{k=0..n+1, (n-k+1)*C(n-k+1, k)} - Paul Barry (pbarry(AT)wit.ie), Nov 05 2005

Recurrence: a(n+2) = a(n+1) + a(n) + Fib(n+4), n >= 0. For n >= 2, a(n-2) = (-1)^n*((-2n+3)*Fib(-n) - (-n)*Fib(-n-1))/5 = (-1)^n*A010049(-n), the second-order Fibonacci numbers of negative index, where Fib(-n) = (-1)^(n+1)*Fib(n). - Peter Bala (pbala(AT)toucansurf.com), Oct 24 2007

a(n) = (n+1)F(n+2) - A001629(n+1) where F(n) is the Fibbonacci sequence - Geoffrey Critzer (critzer.geoffrey(AT)usd443.org), Apr 07 2008

Column 1 of triangle A063967.

Cf. A002940, A010049, A029907.

Adjacent sequences: A023607 A023608 A023609 this_sequence A023611 A023612 A023613

Sequence in context: A002545 A055795 A058695 this_sequence A062544 A120411 A069112

nonn

Clark Kimberling (ck6(AT)evansville.edu)

More terms from Christian G. Bower (bowerc(AT)usa.net), Jan 29 2004