A197657 Row sums of A194595.

1, 4, 22, 134, 866, 5812, 40048, 281374, 2006698, 14482064, 105527060, 775113440, 5731756720, 42628923040, 318621793472, 2391808860446, 18023208400634, 136271601087352, 1033449449559724, 7858699302115444, 59906766929537116, 457685157123172664

Offset: 0

Author

Susanne Wienand, Oct 17 2011

Keywords

• nonn
• easy

Comments

Number of meanders of length (n+1)*3 which are composed by arcs of equal length and a central angle of 120 degrees.

Definition of a meander:

A binary curve C is a triple (m, S, dir) such that

(a) S is a list with values in {L,R} which starts with an L,

(b) dir is a list of m different values, each value of S being allocated a value of dir,

(c) consecutive Ls increment the index of dir,

(d) consecutive Rs decrement the index of dir,

(e) the integer m>0 divides the length of S and

(f) C is a meander if each value of dir occurs length(S)/m times.

For this sequence, m = 3.

For 0 <= n <= 16, a(n) = the hypergraph Fuss-Catalan number FC_1^(2,n+1) in the notation of Chavan et al. - see 7.1 in the Appendix. - Peter Bala, Apr 11 2023

Examples

			Some examples of list S and allocated values of dir if n = 4:
Length(S) = (4+1)*3 = 15.
  S: L,L,L,L,L,L,L,L,L,L,L,L,L,L,L
dir: 1,2,0,1,2,0,1,2,0,1,2,0,1,2,0
  S: L,L,L,L,R,L,L,R,L,L,R,L,L,L,L
dir: 1,2,0,1,1,1,2,2,2,0,0,0,1,2,0
  S: L,R,L,L,L,L,L,R,L,L,R,L,R,R,R
dir: 1,1,1,2,0,1,2,2,2,0,0,0,0,2,1
Each value of dir occurs 15/3 = 5 times.
		

Links

• Parth Chavan, Andrew Lee and Karthik Seetharaman, Hypergraph Fuss-Catalan numbers, arXiv:2202.01111 [math.CO]
• Peter Luschny, Meanders and walks on the circle.
• Susanne Wienand, Animation of a meander
• Susanne Wienand, Example of a meander

Crossrefs

Cf. A198060, A198256, A198257, A198258.

Programs

Maple

A197657 := proc(n)
    (A000172(n) + A000172(n+1)) / 3 ;
end proc; # R. J. Mathar, Jul 26 2014
a := n -> 2^n*hypergeom([n + 1, -n/2, -n/2 - 1/2], [1, 1], 1):
seq(simplify(a(n)), n = 0..21); # Peter Luschny, Mar 26 2023

Mathematica

A197657[n_] := Sum[Sum[Sum[(-1)^(j + i)* Binomial[i, j]*Binomial[n, k]^3*(n + 1)^j*(k + 1)^(2 - j)/(k + 1)^2, {i, 0, 2}], {j, 0, 2}], {k, 0, n}]; Table[A197657[n], {n, 0, 16}]  (* Peter Luschny, Nov 02 2011 *)

PARI

A197657(n) = {sum(k=0,n,if(n == 1+2*k,3,(1+k)*(1-((n-k)/(1+k))^3)/(1+2*k-n))*binomial(n,k)^3)} \\ Peter Luschny, Nov 24 2011

SageMath

def A197657(n):
    return 2^n*hypergeometric([n + 1, -n/2, -n/2 - 1/2], [1, 1], 1).simplify_hypergeometric()
for n in (0..21): print(A197657(n)) # Peter Luschny, Mar 26 2023

Formula

a(n) = Sum{k=0..n} Sum{j=0..2} Sum{i=0..2} (-1)^(j+i)*C(i,j)*C(n,k)^3*(n+1)^j*(k+1)^(2-j)/(k+1)^2. - Peter Luschny, Nov 02 2011

a(n) = Sum_{k=0..n} h(n,k)*binomial(n,k)^3, where h(n,k) = (1+k)*(1-((n-k)/(1+k))^3)/(1+2*k-n) if 1+2*k-n <> 0 else h(n,k) = 3. - Peter Luschny, Nov 24 2011

a(n) = A141147(n+1)/2 = A110707(n+1)/6 = (A000172(n)+A000172(n+1))/3. - Max Alekseyev, Jul 15 2014

Conjecture: (n+1)^2*a(n) -3*(n+1)*(2*n+1)*a(n-1) -3*n*(5*n-7)*a(n-2) -8*(n-2)^2*a(n-3)=0. - R. J. Mathar, Jul 26 2014

a(n) = 2^n*hypergeom([n + 1, -n/2, -n/2 - 1/2], [1, 1], 1). - Peter Luschny, Mar 26 2023

a(n) ~ sqrt(3) * 2^(3*n+1) / (Pi*n). - Vaclav Kotesovec, Apr 17 2023