A198256 Row sums of A197653.

1, 5, 46, 485, 5626, 69062, 882540, 11614437, 156343330, 2142556130, 29791689148, 419260001030, 5960334608788, 85469709312860, 1234797737654296, 17955907741675749, 262607675818816050, 3860239468267647914, 57002176852356800700, 845159480056345448610

Offset: 0

Author

Susanne Wienand, Oct 22 2011

Keywords

• nonn

Comments

Number of meanders of length (n+1)*4 which are composed by arcs of equal length and a central angle of 90 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 = 4.

The terms are proved by brute force for 0 <= n <= 8, 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)*4 = 20.
  S: L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L
dir: 1,2,3,0,1,2,3,0,1,2,3,0,1,2,3,0,1,2,3,0
  S: L,L,L,L,R,L,R,R,L,R,R,R,L,R,L,L,R,L,L,L
dir: 1,2,3,0,0,0,0,3,3,3,2,1,1,1,1,2,2,2,3,0
  S: L,L,R,L,L,L,L,R,R,L,R,R,L,R,L,L,L,L,R,R
dir: 1,2,2,2,3,0,1,1,0,0,0,3,3,3,3,0,1,2,2,1
Each value of dir occurs 20/4 = 5 times.
		

Links

• Peter Luschny, Meanders and walks on the circle.

Crossrefs

Cf. A197653, A198060, A198257, A198258.

Cf. A005260, A181069.

Cf. A181067, A374614.

Programs

Mathematica

A198256[n_] := Sum[Sum[Sum[(-1)^(j + i)* Binomial[i, j]*Binomial[n, k]^4*(n + 1)^j*(k + 1)^(3 - j)/(k + 1)^3, {i, 0, 3}], {j, 0, 3}], {k, 0, n}]; Table[A198256[n], {n, 0, 16}] (* Peter Luschny, Nov 02 2011 *)

PARI

A198256(n) = {sum(k=0,n,if(n == 1+2*k,4,(1+k)*(1-((n-k)/(1+k))^4)/(1+2*k-n))*binomial(n,k)^4)} \\ Peter Luschny, Nov 24 2011

Sage

from mpmath import mp, hyper
def A198256(n) : return hyper([1-n, 1-n, 1-n, 1-n], [3, 3, 3], 1)*(n^4-n^6)/4 + hyper([-n, -n, -n, -n], [2, 2, 2], 1)*(1+n+n^2+n^3) + hyper([2, 1-n, 1-n, 1-n, 1-n], [1, 3, 3, 3], 1)*(n^4+n^5)/4
mp.dps = 32
for n in (0..19) : print(int(A198256(n)))  # Peter Luschny, Oct 24 2011

Formula

a(n) = Sum_{k=0..n} Sum_{j=0..3} Sum_{i=0..3} (-1)^(j+i)*C(i,j)*C(n,k)^4*(n+1)^j*(k+1)^(3-j)/(k+1)^3. - Peter Luschny, Nov 02 2011

a(n) = Sum_{k=0..n} h(n,k)*binomial(n,k)^4, where h(n,k) = (1+k)*(1-((n-k)/(1+k))^4)/(1+2*k-n) if 1+2*k-n <> 0 else h(n,k) = 4. - Peter Luschny, Nov 24 2011

From Peter Bala, Mar 21 2023: (Start)

Conjecture 1: a(n) = Sum_{k = 0..n} binomial(n+1,k)^2*binomial(n,k)^2.

If true, then we have the third-order recurrence equation

n^2*(n + 1)^3*P(n-1)*a(n) = 2*n^2*(400*n^8 - 1260*n^7 + 20*n^6 + 3020*n^5 - 1646*n^4 - 1951*n^3 + 1142*n^2 + 465*n - 290)*a(n-1) + 4*(n - 1)*(2800*n^9 - 15420*n^8 + 30620*n^7 - 23710*n^6 + 808*n^5 + 6863*n^4 - 1309*n^3 - 1218*n^2 + 496*n - 60)*a(n-2) + 8*(n - 2)^2*(2*n - 3)*(4*n - 5)*(4*n - 7)*P(n)*a(n-3) with a(0) = 1, a(1) = 5 and a(2) = 46 and where P(n) = 100*n^5 - 65*n^4 - 35*n^3 + 25*n^2 + 6*n - 5.

Conjecture 2: working with offset 1, that is, a(1) = 1, a(2) = 5, ..., 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 >= 5. (End)

a(n) ~ 2^(4*n + 5/2) / (Pi*n)^(3/2). - Vaclav Kotesovec, Apr 17 2023

Peter Bala's conjecture 1 can equivalently written: a(n) = hypergeom([-n - 1, -n - 1, -n, -n], [1, 1, 1], 1). - Detlef Meya, May 28 2024

a(n) = Sum_{k=0..n+1} (k/(n+1))^2 * binomial(n+1,k)^4. - Seiichi Manyama, Jul 14 2024