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

0,2

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

Links

Table of n, a(n) for n=0..19.

Peter Luschny, Meanders and walks on the circle.

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

Example

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.

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 *)

Prog

(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
(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

Crossrefs

Cf. A198060, A198257, A198258.

Sequence in context: A309185 A169963 A371744 * A232972 A127304 A112029

Adjacent sequences: A198253 A198254 A198255 * A198257 A198258 A198259

Keyword

nonn

Author

Susanne Wienand, Oct 22 2011

Status

approved