A198257 Row sums of A197654.

1, 6, 94, 1700, 35466, 795312, 18848992, 464517468, 11801240050, 307073982116, 8147186436324, 219664321959524, 6003343077661216, 165975724832822400, 4634768975107569024, 130553813782898706908, 3705740233107582161538, 105902829964290241990332

Offset

0,2

Comments

Number of meanders of length (n+1)*5 which are composed by arcs of equal length and a central angle of 72 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 = 5.

The terms are proved by brute force for 0 <= n <= 6, but not yet in general. [Susanne Wienand, Oct 29 2011]

Links

Table of n, a(n) for n=0..17.

Peter Luschny, Meanders and walks on the circle.

Project Euler, Robot Walks: Problem 208

Formula

a(n) = Sum{k=0..n} Sum{j=0..4} Sum{i=0..4} (-1)^(j+i)*C(i,j)*C(n,k)^5*(n+1)^j*(k+1)^(4-j)/(k+1)^4. - Peter Luschny, Nov 02 2011

a(n) = Sum_{k=0..n} h(n,k)*binomial(n,k)^5, where h(n,k) = (1+k)*(1-((n-k)/(1+k))^5)/(1+2*k-n) if 1+2*k-n <> 0 else h(n,k) = 5. - Peter Luschny, Nov 24 2011

a(n) ~ sqrt(5) * 2^(5*n+2) / (Pi*n)^2. - Vaclav Kotesovec, Apr 17 2023

Example

Some examples of list S and allocated values of dir if n = 5:
Length(S) = (5+1)*5 = 30.
  S: L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L
dir: 1,2,3,4,0,1,2,3,4,0,1,2,3,4,0,1,2,3,4,0,1,2,3,4,0,1,2,3,4,0
  S: L,L,L,L,L,L,L,L,L,L,L,L,L,R,L,R,R,R,R,R,L,R,L,L,L,L,R,R,R,L
dir: 1,2,3,4,0,1,2,3,4,0,1,2,3,3,3,3,2,1,0,4,4,4,4,0,1,2,2,1,0,0
  S: L,L,L,L,L,R,L,L,L,R,R,L,L,L,L,L,R,R,L,R,R,L,R,R,L,L,L,L,L,R
dir: 1,2,3,4,0,0,0,1,2,2,1,1,2,3,4,0,0,4,4,4,3,3,3,2,2,3,4,0,1,1
Each value of dir occurs 30/5 = 6 times.

Maple

A198257 := proc(n) local i, j, k, pow;
pow := (a, b) -> if a=0 and b=0 then 1 else a^b fi;
add(add(add((-1)^(j+i)*binomial(i, j)*binomial(n, k)^5*pow(n+1, j)*pow(k+1, 4-j)/(k+1)^4, i=0..4), j=0..4), k=0..n) end: seq(A198257(n), n=0..16); # Peter Luschny, Nov 02 2011

Mathematica

Table[Sum[Sum[ Sum[(-1)^(j + i) Binomial[i, j], {i, 0, 4}] Binomial[n, k]^5*(n + 1)^j*(k + 1)^(4 - j), {j, 0, 4}]/(k + 1)^4, {k, 0, n}], {n, 0, 17}] (* Michael De Vlieger, Aug 18 2016 *)

Prog

(PARI)
A198257(n) = {sum(k=0, n, if(n == 1+2*k, 5, (1+k)*(1-((n-k)/(1+k))^5)/(1+2*k-n))*binomial(n, k)^5)} \\ Peter Luschny, Nov 24 2011

Crossrefs

Cf. A198060, A198256, A198258.

Sequence in context: A078103 A221525 A321073 * A296820 A184983 A184980

Adjacent sequences: A198254 A198255 A198256 * A198258 A198259 A198260

Keyword

nonn

Author

Susanne Wienand, Oct 22 2011

Status

approved