A089849 Number of fixed points in range [A014137(n-1)..A014138(n-1)] of permutation A069772.

1, 1, 2, 1, 6, 2, 20, 5, 70, 14, 252, 42, 924, 132, 3432, 429, 12870, 1430, 48620, 4862, 184756, 16796, 705432, 58786, 2704156, 208012, 10400600, 742900, 40116600, 2674440, 155117520, 9694845, 601080390, 35357670, 2333606220, 129644790

Offset: 0

Author

Antti Karttunen, Nov 29 2003

Keywords

• nonn
• easy

Comments

The number of n-node binary trees fixed by the corresponding automorphism(s). A000984 interleaved with A000108.

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• Antti Karttunen, C-program for computing the initial terms of this sequence.

Crossrefs

Cf. A014137, A014138, A069772, A089880.

Cf. A000984 interleaved with A000108.

Programs

Magma

A089849:= func< n | n eq 0 select 1 else ((1+(-1)^n)*Binomial(n,Floor(n/2))+(1-(-1)^n)*Catalan(Floor((n-1)/2)))/2 >;
[A089849(n): n in [0..50]]; // G. C. Greubel, Feb 22 2025

Mathematica

a[n_] := If[EvenQ[n], Binomial[n, n/2], CatalanNumber[(n-1)/2]];
Table[a[n], {n, 0, 35}] (* Jean-François Alcover, Mar 09 2023 *)

SageMath

def A089849(n): return binomial(n, n//2) if n%2==0 else catalan_number((n-1)//2)
print([A089849(n) for n in range(51)]) # G. C. Greubel, Feb 22 2025

Scheme

(define (A089849 n) (if (even? n) (A000984 (/ n 2)) (A000108 (/ (- n 1) 2))))

Formula

a(2n) = A000984(n), a(2n+1) = A000108(n).

a(n) = Sum_{k=0..floor(n/2)} binomial(k+1,n-k)*Catalan(k). - Paul Barry, Feb 23 2005

From Paul Barry, Jan 23 2006: (Start)

a(n+1) = Jacobi_P(n, 2, 0, 0)*2^n*(cos(Pi*n/2)+sin(Pi*n/2)).

a(n+1) = (Sum_{k=0..n} C(n,k)*C(n+2,k)*(-1)^k)*(cos(Pi*n/2)+sin(Pi*n/2)). (End)

From Sergei N. Gladkovskii, Dec 18 2012 (Start)

E.g.f.: 1 + integral(G(0)) dx where G(k) = 1 + 2*x/(1 - 2*x/(2*x + (2*k+2)*(2*k+4)/G(k+1) )); (continued fraction).

E.g.f.: 1 + x*G(0) where G(k) = 1 + x*(2*k+1)/(k+1 - x*(k+1)/(x + (k+2)*(2*k+3)/G(k+1) )); (continued fraction).

E.g.f.: E(x) = integral( (1/x + 2)*BesselI(1,2*x) ) dx. (End)

G.f.: G(0), where G(k) = 1 + x/(k+1 - (k+1)*(4*k+2)*x/((4*k+2)*x + 1/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 19 2013

From Amiram Eldar, Mar 12 2023: (Start)

Sum_{n>=0} 1/a(n) = 10/3 + 2*Pi/(3*sqrt(3)).

Sum_{n>=0} (-1)^n/a(n) = 2/3 + 2*Pi/(9*sqrt(3)). (End)