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Array read by ascending antidiagonals. A family of Catalan-like sequences. A(n, k) = [x^k] ((n - 1)*x + 1)*(1 - sqrt(1 - 4*x))/(2*x).
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%I #20 Mar 31 2022 18:12:41

%S 1,1,0,1,1,1,1,2,2,3,1,3,3,5,9,1,4,4,7,14,28,1,5,5,9,19,42,90,1,6,6,

%T 11,24,56,132,297,1,7,7,13,29,70,174,429,1001,1,8,8,15,34,84,216,561,

%U 1430,3432,1,9,9,17,39,98,258,693,1859,4862,11934,1,10,10,19,44,112,300,825,2288,6292,16796,41990

%N Array read by ascending antidiagonals. A family of Catalan-like sequences. A(n, k) = [x^k] ((n - 1)*x + 1)*(1 - sqrt(1 - 4*x))/(2*x).

%F A(n, k) = (n-1)*CatalanNumber(k-1) + CatalanNumber(k) for n >= 0 and k >= 1, A(n, 0) = 1. (Cf. A352682.)

%F D-finite with recurrence: A(n, k) = A(n, k-1)*((6 - 4*k)*(n - 3 + k*(3 + n)))/((1 + k)*(6 - k*(3 + n))) for k >= 3, otherwise 1, n, n + 1 for k = 0, 1, 2.

%F Given a list T let PS(T) denote the list of partial sums of T. Given two list S and T let [S, T] denote the concatenation of the lists. Further let P[end] denote the last element of the list P. Row n of the array A with length k can be computed by the following procedure:

%F A = [n], P = [1], R = [1];

%F Repeat k times: R = [R, A], P = PS([P, A]), A = [P[end]];

%F Return R.

%e Array starts:

%e n\k 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ...

%e ------------------------------------------------------

%e [0] 1, 0, 1, 3, 9, 28, 90, 297, 1001, 3432, ... A071724

%e [1] 1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, ... A000108

%e [2] 1, 2, 3, 7, 19, 56, 174, 561, 1859, 6292, ... A071716

%e [3] 1, 3, 4, 9, 24, 70, 216, 693, 2288, 7722, ... A038629

%e [4] 1, 4, 5, 11, 29, 84, 258, 825, 2717, 9152, ... A352681

%e [5] 1, 5, 6, 13, 34, 98, 300, 957, 3146, 10582, ...

%e [6] 1, 6, 7, 15, 39, 112, 342, 1089, 3575, 12012, ...

%e [7] 1, 7, 8, 17, 44, 126, 384, 1221, 4004, 13442, ...

%e [8] 1, 8, 9, 19, 49, 140, 426, 1353, 4433, 14872, ...

%e [9] 1, 9, 10, 21, 54, 154, 468, 1485, 4862, 16302, ...

%e .

%e Seen as a triangle:

%e [0] 1;

%e [1] 1, 0;

%e [1] 1, 1, 1;

%e [2] 1, 2, 2, 3;

%e [3] 1, 3, 3, 5, 9;

%e [4] 1, 4, 4, 7, 14, 28;

%e [5] 1, 5, 5, 9, 19, 42, 90;

%e [6] 1, 6, 6, 11, 24, 56, 132, 297;

%p for n from 0 to 9 do

%p ogf := ((n - 1)*x + 1)*(1 - sqrt(1 - 4*x))/(2*x);

%p ser := series(ogf, x, 12):

%p print(seq(coeff(ser, x, k), k = 0..9)); od:

%p # Alternative:

%p alias(PS = ListTools:-PartialSums):

%p CatalanRow := proc(n, len) local a, k, P, R;

%p a := n; P := [1]; R := [1];

%p for k from 0 to len-1 do

%p R := [op(R), a]; P := PS([op(P), a]); a := P[-1] od;

%p R end: seq(lprint(CatalanRow(n, 9)), n = 0..9);

%p # Recurrence:

%p A := proc(n, k) option remember: if k < 3 then [1, n, n + 1][k + 1] else

%p A(n, k-1)*((6 - 4*k)*(n - 3 + k*(3 + n)))/((1 + k)*(6 - k*(3 + n))) fi end:

%p seq(print(seq(A(n, k), k = 0..9)), n = 0..9);

%t T[n_, 0] := 1;

%t T[n_, k_] := (n - 1) CatalanNumber[k - 1] + CatalanNumber[k];

%t Table[T[n, k], {n, 0, 9}, {k, 0, 9}] // TableForm

%o (Julia) # Compare with the Julia function A352686Row.

%o function A352680Row(n, len)

%o a = BigInt(n)

%o P = BigInt[1]; T = BigInt[1]

%o for k in 0:len-1

%o T = push!(T, a)

%o P = cumsum(push!(P, a))

%o a = P[end]

%o end

%o T end

%o for n in 0:9 println(A352680Row(n, 9)) end

%Y Rows: A071724, A000108, A071716, A038629, A352681.

%Y Diagonals: A077587 (main), A271823.

%Y Compare A352682 for a similar array based on the Bell numbers.

%K nonn,tabl

%O 0,8

%A _Peter Luschny_, Mar 27 2022