cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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).

Original entry on oeis.org

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

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Author

Peter Luschny, Nov 01 2011

Keywords

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

Examples

			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).
		

Crossrefs

Programs

  • 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
  • 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 *)
  • 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.
    

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)