A193845 Mirror of the triangle A193844.

1, 3, 1, 7, 5, 1, 15, 17, 7, 1, 31, 49, 31, 9, 1, 63, 129, 111, 49, 11, 1, 127, 321, 351, 209, 71, 13, 1, 255, 769, 1023, 769, 351, 97, 15, 1, 511, 1793, 2815, 2561, 1471, 545, 127, 17, 1, 1023, 4097, 7423, 7937, 5503, 2561, 799, 161, 19, 1

Offset: 0

Author

Clark Kimberling, Aug 07 2011

Keywords

• nonn
• tabl

Comments

This triangle is obtained by reversing the rows of the triangle A193844.

From Philippe Deléham, Jan 17 2014: (Start)

Subtriangle of the triangle in A112857.

T(n,0) = A000225(n+1).

T(n,1) = A000337(n).

T(n+2,2) = A055580(n).

T(n+3,3) = A027608(n).

T(n+4,4) = A211386(n).

T(n+5,5) = A211388(n).

T(n,n) = A000012(n).

T(n+1,n) = A005408(n).

T(n+2,n) = A056220(n+2).

T(n+3,n) = A199899(n+1).

Row sums are A003462(n+1).

Diagonal sums are A048739(n).

Riordan array (1/((1-2*x)*(1-x)), x/(1-2*x)). (End)

Consider the transformation 1 + x + x^2 + x^3 + ... + x^n = A_0*(x-2)^0 + A_1*(x-2)^1 + A_2*(x-2)^2 + ... + A_n*(x-2)^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

The n-th row lists the coefficients of the polynomial sum_{k=0..n} (X+2)^k, in order of increasing powers. - M. F. Hasler, Oct 15 2014

Examples

			First six rows:
1
3....1
7....5....1
15...17...7....1
31...49...31...9...1
63...129..111..49..11..1
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 0..11475 (rows 0 <= n <= 150, flattened)
• Russell Jay Hendel, A Method for Uniformly Proving a Family of Identities, arXiv:2107.03549 [math.CO], 2021.

Crossrefs

Cf. A193844.

Cf. Columns: A000225, A000337, A055580, A027608, A211386, A211388

Cf. Diagonals: A000012, A005408, A056220, A199899

Programs

Mathematica

z = 10;
p[n_, x_] := (x + 1)^n;
q[n_, x_] := (x + 1)^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}]]  (* A193844 *)
TableForm[Table[h[n], {n, 0, z}]]
Flatten[Table[h[n], {n, -1, z}]]  (* A193845 *)
Table[2^k*Binomial[n + 1, k]*Hypergeometric2F1[1, -k, -k + n + 2, 1/2], {n, 0, 9}, {k, n, 0, -1}] // Flatten (* Michael De Vlieger, Nov 09 2021 *)

PARI

for(n=0,20,for(k=0,n,print1(1/k!*sum(i=0,n,(2^(i-k)*prod(j=0,k-1,i-j))),", "))) \\ Derek Orr, Oct 14 2014

Formula

T(n,k) = A193844(n,n-k).

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