A023427 Number of Dyck n-paths with ascents and descents of length equal to 1 (mod 4).
1, 1, 1, 1, 1, 2, 4, 7, 11, 17, 28, 49, 87, 152, 262, 453, 794, 1408, 2507, 4462, 7943, 14179, 25415, 45713, 82398, 148731, 268859, 486890, 883411, 1605582, 2922259, 5325377, 9716564, 17750332, 32464980, 59443403, 108951953, 199886003, 367052947, 674620772
Offset: 0
Examples
(5)=2; representing unpaired vertices by v and arcs by AA, BB, etc., the 8 (= A004148(5)) secondary structures of size 5 are vvvvv, AvAvv, vvAvA, AvvAv, vAvvA, AvvvA, vAvAv, ABvBA; they have 0,1,1,1,1,1,1,0 stacks of odd length, respectively. - _Emeric Deutsch_, Dec 26 2011
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..1000
- I. L. Hofacker, P. Schuster and P. F. Stadler, Combinatorics of RNA secondary structures, Discrete Appl. Math., 88, 1998, 207-237.
- P. R. Stein and M. S. Waterman, On some new sequences generalizing the Catalan and Motzkin numbers, Discrete Math., 26 (1979), 261-272.
Programs
-
Maple
f := z^4/(1-z^4): eq := G = 1+z*G+f*G*(G-1)/(1+f): G := RootOf(eq, G): Gser := simplify(series(G, z = 0, 42)): seq(coeff(Gser, z, n), n = 0 .. 39); # Emeric Deutsch, Dec 26 2011 a:= proc(n) option remember; `if`(n=0, 1, a(n-1) +add(a(k)*a(n-4-k), k=1..n-4)) end: seq(a(n), n=0..50); # Alois P. Heinz, May 09 2012
-
Mathematica
Clear[ a ]; a[ 0 ]=1; a[ n_Integer ] := a[ n ]=a[ n-1 ]+Sum[ a[ k ]*a[ n-4-k ], {k, 1, n-4} ]; CoefficientList[Series[((1-x+x^4) - Sqrt[(1-x+x^4)^2 - 4*x^4])/(2*x^4), {x, 0, 20}], x] (* Vaclav Kotesovec, Sep 16 2013 *)
-
Maxima
a(n):=if n=0 then 1 else sum(binomial(n-3*q,((q)))*binomial((n-3*q),(q+1))/(n-3*q),q,0,(n-1)/3); /* Vladimir Kruchinin, Jan 21 2019 */
-
PARI
{a(n)=polcoeff(((1-x+x^4) - sqrt((1-x+x^4)^2 - 4*x^4 +x^5*O(x^n)))/(2*x^4), n)} \\ Paul D. Hanna, Oct 29 2012
-
PARI
{a(n)=local(A=1+x+x*O(x^n)); for(i=1, n, A=1 + x*A/(1-x^4*A+x*O(x^n))); polcoeff(A, n)} \\ Paul D. Hanna, Oct 29 2012
Formula
a(n) = A202845(n,0). A(x) satisfies A=1+x*A+f*A*(A-1)/(1+f), where f=x^4/(1-x^4). - Emeric Deutsch, Dec 26 2011
G.f.: A(x) = ((1-x+x^4) - sqrt((1-x+x^4)^2 - 4*x^4))/(2*x^4). - Paul D. Hanna, Oct 29 2012
G.f. satisfies: A(x) = 1 + x*A(x)/(1 - x^4*A(x)). - Paul D. Hanna, Oct 29 2012
G.f.: 1 + x*exp( Sum_{n>=1} x^n/n * Sum_{k=0..n} C(n,k)^2 * x^(3*k) ). - Paul D. Hanna, Oct 29 2012
a(n) = A216116(n-1) for n>0.
Recurrence: (n+4)*a(n) = (2*n+5)*a(n-1) - (n+1)*a(n-2) + 2*(n-2)*a(n-4) + (2*n-7)*a(n-5) - (n-8)*a(n-8). - Vaclav Kotesovec, Sep 16 2013
a(n) ~ sqrt(-4*c^2-3*c^3+4-4*c)*(1+2*c-c^3)^n*(-5*c^3-3*c^2+9*c+10) / (2*n^(3/2)*sqrt(Pi)), where c = 1/2*sqrt((4+(155/2-3*sqrt(849)/2)^(1/3) +(155/2+3*sqrt(849)/2)^(1/3))/3) - 1/2*sqrt(8/3-1/3*(155/2-3*sqrt(849)/2)^(1/3) - 1/3*(155/2+3*sqrt(849)/2)^(1/3) + 2*sqrt(3/(4+(155/2-3*sqrt(849)/2)^(1/3) + (155/2+3*sqrt(849)/2)^(1/3)))) = 0.5248885986564... is the root of the equation c^4-2*c^2-c+1=0. - Vaclav Kotesovec, Sep 16 2013
a(n) = Sum_{k=0..(n-1)/3} C(n-3*k,k)*C(n-3*k,k+1)/(n-3*k), n>0, a(0)=1. - Vladimir Kruchinin, Jan 21 2019
Extensions
New name, using a comment of Alois P. Heinz, from Peter Luschny, Jan 21 2019
Comments