A104709 Triangle read by rows: T(n,k) = Sum_{j=0..n} 2^(n-j)*binomial(j,k) for n >= 0 and 0 <= k <= n; also, Riordan array (1/((1-x)*(1-2*x)), x/(1-x)).

1, 3, 1, 7, 4, 1, 15, 11, 5, 1, 31, 26, 16, 6, 1, 63, 57, 42, 22, 7, 1, 127, 120, 99, 64, 29, 8, 1, 255, 247, 219, 163, 93, 37, 9, 1, 511, 502, 466, 382, 256, 130, 46, 10, 1, 1023, 1013, 968, 848, 638, 386, 176, 56, 11, 1, 2047, 2036, 1981, 1816, 1486, 1024, 562, 232, 67

Offset: 0

Author

Gary W. Adamson, Mar 19 2005

Keywords

• nonn
• tabl

Comments

This array (A104709) is the mirror of the fission, A054143, of the polynomial sequence ((x+1)^n: n >= 0) by the polynomial sequence (q(n,x): n >= 0) given by q(n,x) = x^n + x^(n-1) + ... + x + 1. See A193842 for the definition of fission. - Clark Kimberling, Aug 07 2011

The elements of the matrix inverse appear to be T^(-1)(n,k) = (-1)^(n+k)*A110813(n,k) assuming the same offset in both triangles. - R. J. Mathar, Mar 15 2013

From Paul Curtz, Jun 12 2019: (Start)

Numerators of the triangle [Curtz, page 15, triangle (E)]:

1/2;

3/4, 1/4;

7/8, 4/8, 1/8;

15/16, 11/16, 5/16, 1/16;

31/32, 26/31, 16/32, 6/32, 1/32;

63/64, 57/64, 42/64, 22/64, 7/64, 1/64;

...

Denominators - Numerators: Triangle A054143.

1;

1, 3;

1, 4, 7;

1, 5, 11, 15;

...

(E) is a transform which accelerates the convergence of series.

For log(2) = 1 - 1/2 + 1/3 - 1/4 ... = 0.6931..., we have

1*(1/2) = 1/2,

1*(3/4) - (1/2)*(1/4) = 5/8,

1*(7/8) - (1/2)*(4/8) + (1/3)*(1/8) = 2/3,

1*(15/16) - (1/2)*(11/16) + (1/3)*(5/16) - (1/4)*1/16 = 131/192,

...

This is A068566/A068565. (End)

Examples

			Triangle T(n,k) (with rows n >= 0 and columns k = 0..n) begins:
   1;
   3,  1;
   7,  4,  1;
  15, 11,  5,  1;
  31, 26, 16,  6,  1;
  63, 57, 42, 22,  7,  1;
  ...
		

Links

• Paul Curtz, Accélération de la convergence de certaines séries alternées à l'aide des fonctions de sommation, Thèse de 3ème Cycle d'Analyse Numérique, Faculté des Sciences de l'Université de Paris, 4 mai 1965.
• Clark Kimberling, Fusion, Fission, and Factors, Fib. Q., 52(3) (2014), 195-202.

Crossrefs

Cf. A000124, A000295, A001787 (row sums), A007318, A054143, A055248, A068565, A068566, A104709, A140513.

Programs

Maple

A104709_row := proc(n) add(add(binomial(n,n-i)*x^(n-k-1),i=0..k),k=0..n-1);
coeffs(sort(%)) end; seq(print(A104709_row(n)),n=1..6); # Peter Luschny, Sep 29 2011

Mathematica

z = 10;
p[n_, x_] := (x + 1)^n;
q[0, x_] := 1; q[n_, x_] := x*q[n - 1, x] + 1;
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}]] (* A054143 *)
TableForm[Table[h[n], {n, 0, z}]]
Flatten[Table[h[n], {n, -1, z}]] (* A104709 *)
(* Clark Kimberling, Aug 07 2011 *)

Formula

Begin with A055248 as a triangle, delete leftmost column.

The Riordan array factors as (1/(1-2*x), x)*(1/(1-x), x/(1-x)) - the sequence array for 2^n times Pascal's triangle. - Paul Barry, Aug 05 2005

T(n,k) = Sum_{j=0..n-k} C(n-j, k)*2^j. - Paul Barry, Jan 12 2006

T(n,k) = 3*T(n-1,k) + T(n-1,k-1) - 2*T(n-2,k) - 2*T(n-2,k-1), T(0,0) = 1, T(n,k) = 0 if k < 0 or if k > n. - Philippe Deléham, Nov 30 2013

Working with an offset of 0, we have exp(x) * (e.g.f. for row n) = (e.g.f. for diagonal n). For example, for n = 3 we have exp(x)*(15 + 11*x + 5*x^2/2! + x^3/3!) = 15 + 26*x + 42*x^2/2! + 64*x^3/3! + 93*x^4/4! + .... The same property holds more generally for Riordan arrays of the form (f(x), x/(1 - x)). - Peter Bala, Dec 21 2014

From Petros Hadjicostas, Jun 05 2020: (Start)

Bivariate o.g.f.: A(x,y) = Sum_{n,k >= 0} T(n,k)*x^n*y^k = 1/(1 - 3*x - x*y + 2*x^2 + 2*x^2*y) = 1/((1 - 2*x)*(1 - x*(y+1))).

The o.g.f. of the n-th row is (2^(n+1) - (1 + y)^(n+1))/(1 - y).

Let B(x,y) be the bivariate o.g.f. of triangular array A054143. Because A054143 is the mirror image of the current array, we have A(x,y) = B(x*y, 1/y) and B(x,y) = A(x*y, 1/y). This makes it easy to identify lower diagonals of the array.

For example, if we want to identify the second lower diagonal of the array (i.e., 7, 11, 16, 22, ...), we take the 2nd derivative of B(x,y) with respect to y, set y = 0, and divide by 2!. (Note that columns in A054143 start at k = 0.) We get the g.f. x^2*(7 - 10*x + 4*x^2)/(1 - x)^3.

It is then easy to derive that T(n,n-2) = A000124(n+1) = (n+1)*(n+2)/2 + 1 for n >= 2 (by ignoring the first three terms of A000124). Of course, in the current case, it is much easier to use the formula for T(n,k) to find T(n,n-2). (End)

T(n,0) = 2^(n+1) - 1 for n >= 0; T(n,k) = T(n-1,k) + T(n-1,k-1) for 1 <= k <= n. - Peter Bala, Jan 30 2023

T(n,1) = 2^(n+1) - n - 2 = A000295(n+1) for n >= 1. - Bernard Schott, Feb 22 2023

Extensions

Name edited and offset changed by Petros Hadjicostas, Jun 04 2020