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T(3, 43) = 1 = card(1111).

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Array A(m, k) starts:

.

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 a the representation of meanders as multisets as given in multiset permutations (see A361043) and generated by the Julia program section:below.

T(3, 0) = 8 = card(1000, 1100, 1010, 1001, 1110, 1101, 1011, 1111).

AT(3, 1, 2) = 10 = card(110000, 100100, 100001, 111100, 111001, 110110, 110011, 101101, 100111, 111111). Note that this is a subset of the 32 multiset permutations A361043(2, 4).

T(3, 2) = 4 = card(111000, 110001, 100011, 111111).

T(3, 4) = 1 = card(1111).

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:

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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)^(m-j)/(k+1)^m.

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)^(m-j)/(k+1)^m, , 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;

for m from 0 to 6 do seq(A198060(m, n), n=0..6) od;

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for n in range(1, 7): print([n], Arow(n, 7)) # _Peter Luschny_, Mar 24 2023

(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))

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