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
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).
Links
- Peter Luschny, Meanders and walks on the circle.
Crossrefs
Programs
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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;
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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)
Comments