A193843 Mirror image of the triangle A193842.

1, 4, 1, 13, 7, 1, 40, 34, 10, 1, 121, 142, 64, 13, 1, 364, 547, 334, 103, 16, 1, 1093, 2005, 1549, 643, 151, 19, 1, 3280, 7108, 6652, 3478, 1096, 208, 22, 1, 9841, 24604, 27064, 17086, 6766, 1720, 274, 25, 1, 29524, 83653, 105796, 78322, 37384, 11926

Offset: 0

Author

Clark Kimberling, Aug 07 2011

Keywords

• nonn
• tabl
• easy

Comments

A193843 is obtained by reversing the rows of the triangle A193842.

Consider the transformation 1 + x + x^2 + x^3 + ... + x^n = A_0*(x-3)^0 + A_1*(x-3)^1 + A_2*(x-3)^2 + ... + A_n*(x-3)^n. This sequence gives A_0, ... A_n as the entries in the n-th row of this triangle, starting at n = 0. - Derek Orr, Oct 14 2014

Examples

			First six rows:
    1;
    4,   1;
   13,   7,   1;
   40,  34,  10,   1;
  121, 142,  64,  13,  1;
  364, 547, 334, 103, 16, 1;
		

Links

• E. Neuwirth, Recursively defined combinatorial functions: Extending Galton's board, Discrete Math. 239 (2001) 33-51.

Crossrefs

Cf. A193842.

Cf. A000225 (alt.row sums), A002450 (row sums), A014916 (weighted sums).

Cf. A003462 (first col.), A082574 (anti-diag.sums).

Programs

Maple

T := proc(n,k) option remember;
if k<0 or k>n then 0 elif n=k then 1 elif n=1 and k=0 then 4
else 4*T(n-1,k) + T(n-1,k-1) -3*T(n-2,k) - T(n-2,k-1) fi end;
seq(seq(T(n,k), k=0..n), n=0..9); # Peter Luschny, Jan 18 2014

Mathematica

z = 10;
p[n_, x_] := (x + 1)^n;
q[n_, x_] := (x + 2)^n
p1[n_, k_] := Coefficient[p[n, x], x^k];
p1[n_, 0] := p[n, x] /. x -> 0;
d[n_, x_] := Sum[p1[n, k]*q[n - 1 - k, x], {k, 0, n - 1}]
h[n_] := CoefficientList[d[n, x], {x}]
TableForm[Table[Reverse[h[n]], {n, 0, z}]]
Flatten[Table[Reverse[h[n]], {n, -1, z}]]  (* A193842 *)
TableForm[Table[h[n], {n, 0, z}]]  (* A193843 *)
Flatten[Table[h[n], {n, -1, z}]]

PARI

for(n=0,20,for(k=0,n,print1(1/k!*sum(i=0,n,(3^(i-k)*prod(j=0,k-1,i-j))),", "))) \\ Derek Orr, Oct 14 2014

Formula

Write w(n,k) for the triangle at A193842. The triangle at A193843 is then given by w(n,n-k).

From Peter Bala, Jul 31 2012: (Start)

Matrix product of the shifted Pascal triangle {C(n+1,k+1)}n,k>=0 and the square of the Pascal triangle {2^(n-k)*C(n,k)}n,k>=0. Thus the triangle is the product of two triangular Galton arrays and so is also a Galton array (Neuwirth, Theorem 10).

T(n,k) = Sum_{i = 0..n} C(n+1,i+1)*C(i,k)*2^(i-k).

Riordan array (1/((1 - x)*(1 - 3*x)), x/(1 - 3*x)).

O.g.f.: 1/((1 - x)*(1 - (3 + t)*x)) = 1 + (4 + t)*x + (13 + 7*t + t^2)*x^2 + ....

First column A003462. Row sums A002450. Alternating row sums A000225.

Antidiagonal sums (Sum_{k} T(n-k,k)) A082574. Weighted sums (Sum_{k} k*T(n,k)) A014916. (End)

T(n,k) = 4*T(n-1,k) + T(n-1,k-1) - 3*T(n-2,k) - T(n-2,k-1), T(0,0) = T(1,1) = 1, T(1,0) = 4, T(n,k) = 0 if k < 0 or if k > n. - Philippe Deléham, Jan 17 2014