A198258 Row sums of A197655.

1, 7, 190, 5831, 219626, 8976562, 394800204, 18211045575, 873216720082, 43136486269382, 2183722698469676, 112795257850321202, 5925951358743662260, 315869014732813229716, 17048301919464100932440, 930217336628892162216135, 51244644190683748419791970

Offset: 0

Author

Susanne Wienand, Oct 24 2011

Keywords

• nonn

Comments

Number of meanders of length (n+1)*6 which are composed by arcs of equal length and a central angle of 60 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 = 6.

The terms are proved by brute force for 0 <= n <= 5, but not yet in general. - Susanne Wienand, Oct 29 2011

Examples

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

Links

• Peter Luschny, Meanders and walks on the circle.

Crossrefs

Cf. A198060, A198256, A198257.

Programs

Maple

A198258 := 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)^6*pow(n+1,j)*pow(k+1,5-j)/(k+1)^5, i=0..5),j=0..5),k=0..n) end:
seq(A198258(n), n=0..16); # Peter Luschny, Nov 02 2011

Mathematica

T[n_, k_] := (1 + n)(1 + 3 k + 3 k^2 - n - 3 k*n + n^2)(1 + k + k^2 + n - k*n + n^2) Binomial[n, k]^6/(1 + k)^5;
a[n_] := Sum[T[n, k], {k, 0, n}];
Table[a[n], {n, 0, 16}] (* Jean-François Alcover, Jun 29 2019 *)

PARI

A198258(n) = {sum(k=0,n,if(n == 1+2*k,6,(1+k)*(1-((n-k)/(1+k))^6)/(1+2*k-n))*binomial(n,k)^6)} \\ Peter Luschny, Nov 24 2011

Formula

a(n) = Sum_{k=0..n} Sum_{j=0..5} Sum_{i=0..5} (-1)^(j+i)*C(i,j)*C(n,k)^6*(n+1)^j*(k+1)^(5-j)/(k+1)^5. - Peter Luschny, Nov 02 2011

a(n) = Sum_{k=0..n} h(n,k)*binomial(n,k)^6, where h(n,k) = (1+k)*(1-((n-k)/(1+k))^6)/(1+2*k-n) if 1+2*k-n <> 0 else h(n,k) = 6. - Peter Luschny, Nov 24 2011

Conjecture: working with offset 1, that is, a(1) = 1, a(2) = 7, ..., then the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) holds for positive integers n and r and all primes p >= 3. - Peter Bala, Mar 21 2023

a(n) ~ sqrt(3) * 2^(6*n+3) / (Pi*n)^(5/2). - Vaclav Kotesovec, Apr 17 2023