A143565 Triangle T(n,k), n>=1, 1<=k<=n, where the e.g.f. for column k satisfies: A_k(x) = exp(x*A_k(x^k/k!)).
1, 3, 1, 16, 4, 1, 125, 13, 5, 1, 1296, 46, 21, 6, 1, 16807, 241, 61, 31, 7, 1, 262144, 1471, 211, 106, 43, 8, 1, 4782969, 9409, 1401, 281, 169, 57, 9, 1, 100000000, 67348, 8065, 946, 505, 253, 73, 10, 1, 2357947691, 564841, 37241, 7561, 1261, 841, 361, 91, 11, 1
Offset: 1
Examples
Triangle begins:
1;
3, 1;
16, 4, 1;
125, 13, 5, 1;
1296, 46, 21, 6, 1;
16807, 241, 61, 31, 7, 1;
262144, 1471, 211, 106, 43, 8, 1;
...
Links
- Alois P. Heinz, Rows n = 1..100, flattened
Crossrefs
Programs
-
Maple
A:= proc(n,k::posint) option remember; if n<=0 then 1 else unapply( convert(series(exp(x*A(n-k,k)(x^k/k!)), x,n+1), polynom),x) fi end: T:= (n,k)-> coeff(A(n,k)(x), x,n)*n!: seq(seq(T(n,k), k=1..n), n=1..12); -
Mathematica
a[n_, k_] := a[n, k] = If[n <= 0, 1&, Function[x, Series[E^(x*a[n - k, k][x^k/k!]), {x, 0, n+1}] // Normal // Evaluate]]; t[n_, k_] := Coefficient[a[n, k][x], x, n]*n!; Flatten[Table[Table[t[n, k], {k, 1, n}], {n, 1, 12}]] (* Jean-François Alcover, Dec 16 2013, translated from Maple *)
Formula
E.g.f. for column k satisfies: A_k(x) = exp(x*A_k(x^k/k!)).
T(0,k) = 1; T(n,k) = (n-1)! * Sum_{j=0..floor((n-1)/k)} (k*j+1) * T(j,k) * T(n-1-k*j,k) / (k!^j * j! * (n-1-k*j)!). - Seiichi Manyama, Nov 28 2023