A198060 Array read by antidiagonals, m>=0, n>=0, A(m,n) = Sum_{k=0..n} Sum_{j=0..m} Sum_{i=0..m} (-1)^(j+i)*C(i,j)*C(n,k)^(m+1)*(n+1)^j*(k+1)^(-j).

1, 1, 2, 1, 3, 4, 1, 4, 10, 8, 1, 5, 22, 35, 16, 1, 6, 46, 134, 126, 32, 1, 7, 94, 485, 866, 462, 64, 1, 8, 190, 1700, 5626, 5812, 1716, 128, 1, 9, 382, 5831, 35466, 69062, 40048, 6435, 256, 1, 10, 766, 19682, 219626, 795312, 882540, 281374, 24310, 512

Offset

0,3

Comments

We repeat the definition of a meander as given in the link below and used in the sequences in the cross-references:

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.

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

The rows of the array A(m, n) show the number of meanders of length n and central angle 360/m as specified by the columns of a table in the given link. - Peter Luschny, Mar 20 2023

Links

Table of n, a(n) for n=0..54.

Peter Luschny, Meanders and walks on the circle.

Formula

From Peter Bala, Apr 22 2022: (Start)

Conjectures:

1) the m-th row entries satisfy the Gauss congruences T(m, n*p^r - 1) == T(m, n*p^(r-1) - 1) (mod p^r)) for primes p >= 3 and positive integers n and r.

2) for m even, the m-th row entries satisfy the congruences T(m, p^r - 1) == 2^(p^r - 1) (mod p^2) for primes p >= 3 and positive integers r.

3) for m odd, the m-th row entries satisfy the supercongruences T(m,n*p^r - 1) == T(m,n*p*(r-1) - 1) (mod p^(3*r)) for primes p >= 5 and positive integers n and r. (End)

Example

Array A(m, k) starts:
m\n  [0] [1]  [2]   [3]     [4]     [5]        [6]
--------------------------------------------------
[0]   1   2    4     8      16       32         64   A000079
[1]   1   3   10    35     126      462       1716   A001700
[2]   1   4   22   134     866     5812      40048   A197657
[3]   1   5   46   485    5626    69062     882540   A198256
[4]   1   6   94  1700   35466   795312   18848992   A198257
[5]   1   7  190  5831  219626  8976562  394800204   A198258
.
Triangle T(m, k) starts:
[0]   1;
[1]   2,    1;
[2]   4,    3,    1;
[3]   8,   10,    4,    1;
[4]  16,   35,   22,    5,    1;
[5]  32,  126,  134,   46,    6,   1;
[6]  64,  462,  866,  485,   94,   7, 1;
[7] 128, 1716, 5812, 5626, 1700, 190, 8, 1;
.
Using the representation of meanders as multiset permutations (see A361043) and generated by the Julia program below.
T(3, 0) =  8 = card(1000, 1100, 1010, 1001, 1110, 1101, 1011, 1111).
T(3, 1) = 10 = card(110000, 100100, 100001, 111100, 111001, 110110, 110011, 101101, 100111, 111111).
T(3, 2) =  4 = card(111000, 110001, 100011, 111111).
T(3, 3) =  1 = card(1111).

Maple

A198060 := proc(m, n) local i, j, k; add(add(add((-1)^(j+i)*binomial(i, j)* binomial(n, k)^(m+1)*(n+1)^j*(k+1)^(-j), i=0..m), j=0..m), k=0..n) end:
for m from 0 to 6 do seq(A198060(m, n), n=0..6) od;

Mathematica

a[m_, n_] := Sum[ Sum[ Sum[(-1)^(j + i)*Binomial[i, j]*Binomial[n, k]^(m+1)*(n+1)^j*(k+1)^(m-j)/(k+1)^m, {i, 0, m}], {j, 0, m}], {k, 0, n}]; Table[ a[m-n, n], {m, 0, 9}, {n, 0, m}] // Flatten (* Jean-François Alcover, Jun 27 2013 *)

Prog

(SageMath) # This function assumes an offset (1, 1).
def A(m: int, n: int) -> int:
    S = sum(
            sum(
                sum((
                    (-1) ** (j + i)
                    * binomial(i, j)
                    * binomial(n - 1, k) ** m
                    * n ** j )
                    // (k + 1) ** j
                for i in range(m) )
            for j in range(m) )
        for k in range(n) )
    return S
def Arow(n: int, size: int) -> list[int]:
    return [A(n, k) for k in range(1, size + 1)]
for n in range(1, 7): print([n], Arow(n, 7)) # Peter Luschny, Mar 24 2023
# These functions compute the number of meanders by generating and counting.
# Their primary purpose is to illustrate that meanders are a special class of
# multiset permutations. They are not suitable for numerical calculation.
(Julia)
using Combinatorics
function isMeander(m::Int, c::Vector{Bool})::Bool
    l = length(c)
    (l == 0 || c[1] != true) && return false
    vec = fill(Int(0), m)
    max = div(l, m)
    dir = Int(1)
    ob = c[1]
    for b in c
        if b && ob
            dir += 1
        elseif !b && !ob
            dir -= 1
        end
        dir = mod(dir, m)
        v = vec[dir + 1] + 1
        vec[dir + 1] = v
        if v > max
            return false
        end
        ob = b
    end
true end
function CountMeanders(n, k)
    n == 0 && return k + 1
    count = 0
    size = n * k
    for a in range(0, stop=size; step=n)
        S = [(i <= a) for i in 1:size]
        count += sum(1 for c in multiset_permutations(S, size)
                     if isMeander(n, c); init = 0)
    end
count end
A198060ByCount(m, n) = CountMeanders(m + 1, n + 1)
for n in 0:4
    [A198060ByCount(n, k) for k in 0:4] |> println
end
# Peter Luschny, Mar 20 2023

Crossrefs

T(n, 2) = A033484(n+1).

Cf. A033484, A000079, A001700, A197657, A198256, A198257, A198258, A198061.

Cf. A361043.

Sequence in context: A104711 A133112 A247239 * A327084 A159856 A137649

Adjacent sequences: A198057 A198058 A198059 * A198061 A198062 A198063

Keyword

nonn,tabl

Author

Peter Luschny, Nov 01 2011

Status

approved