A352680 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).

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

Offset: 0

Author

Peter Luschny, Mar 27 2022

Keywords

• nonn
• tabl

Examples

			Array starts:
n\k 0, 1,  2,  3,  4,   5,   6,    7,    8,     9, ...
------------------------------------------------------
[0] 1, 0,  1,  3,  9,  28,  90,  297, 1001,  3432, ... A071724
[1] 1, 1,  2,  5, 14,  42, 132,  429, 1430,  4862, ... A000108
[2] 1, 2,  3,  7, 19,  56, 174,  561, 1859,  6292, ... A071716
[3] 1, 3,  4,  9, 24,  70, 216,  693, 2288,  7722, ... A038629
[4] 1, 4,  5, 11, 29,  84, 258,  825, 2717,  9152, ... A352681
[5] 1, 5,  6, 13, 34,  98, 300,  957, 3146, 10582, ...
[6] 1, 6,  7, 15, 39, 112, 342, 1089, 3575, 12012, ...
[7] 1, 7,  8, 17, 44, 126, 384, 1221, 4004, 13442, ...
[8] 1, 8,  9, 19, 49, 140, 426, 1353, 4433, 14872, ...
[9] 1, 9, 10, 21, 54, 154, 468, 1485, 4862, 16302, ...
.
Seen as a triangle:
[0] 1;
[1] 1, 0;
[1] 1, 1, 1;
[2] 1, 2, 2,  3;
[3] 1, 3, 3,  5,  9;
[4] 1, 4, 4,  7, 14, 28;
[5] 1, 5, 5,  9, 19, 42,  90;
[6] 1, 6, 6, 11, 24, 56, 132, 297;
		

Crossrefs

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

Diagonals: A077587 (main), A271823.

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

Programs

Julia

# Compare with the Julia function A352686Row.
function A352680Row(n, len)
    a = BigInt(n)
    P = BigInt[1]; T = BigInt[1]
    for k in 0:len-1
        T = push!(T, a)
        P = cumsum(push!(P, a))
        a = P[end]
    end
T end
for n in 0:9 println(A352680Row(n, 9)) end

Maple

for n from 0 to 9 do
    ogf := ((n - 1)*x + 1)*(1 - sqrt(1 - 4*x))/(2*x);
    ser := series(ogf, x, 12):
    print(seq(coeff(ser, x, k), k = 0..9)); od:
# Alternative:
alias(PS = ListTools:-PartialSums):
CatalanRow := proc(n, len) local a, k, P, R;
a := n; P := [1]; R := [1];
for k from 0 to len-1 do
    R := [op(R), a]; P := PS([op(P), a]); a := P[-1] od;
R end: seq(lprint(CatalanRow(n, 9)), n = 0..9);
# Recurrence:
A := proc(n, k) option remember: if k < 3 then [1, n, n + 1][k + 1] else
A(n, k-1)*((6 - 4*k)*(n - 3 + k*(3 + n)))/((1 + k)*(6 - k*(3 + n))) fi end:
seq(print(seq(A(n, k), k = 0..9)), n = 0..9);

Mathematica

T[n_, 0] := 1;
T[n_, k_] := (n - 1) CatalanNumber[k - 1] + CatalanNumber[k];
Table[T[n, k], {n, 0, 9}, {k, 0, 9}] // TableForm

Formula

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

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.

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:

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

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

Return R.