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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)).

%I A104709 #94 Jan 05 2025 19:51:38
%S A104709 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,
%T A104709 8,1,255,247,219,163,93,37,9,1,511,502,466,382,256,130,46,10,1,1023,
%U A104709 1013,968,848,638,386,176,56,11,1,2047,2036,1981,1816,1486,1024,562,232,67
%N 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)).
%C A104709 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
%C A104709 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
%C A104709 From _Paul Curtz_, Jun 12 2019: (Start)
%C A104709 Numerators of the triangle [Curtz, page 15, triangle (E)]:
%C A104709      1/2;
%C A104709      3/4,  1/4;
%C A104709      7/8,  4/8,    1/8;
%C A104709    15/16, 11/16,  5/16,  1/16;
%C A104709    31/32, 26/31, 16/32,  6/32, 1/32;
%C A104709    63/64, 57/64, 42/64, 22/64, 7/64, 1/64;
%C A104709    ...
%C A104709 Denominators - Numerators: Triangle A054143.
%C A104709   1;
%C A104709   1, 3;
%C A104709   1, 4, 7;
%C A104709   1, 5, 11, 15;
%C A104709   ...
%C A104709 (E) is a transform which accelerates the convergence of series.
%C A104709 For log(2) = 1 - 1/2 + 1/3 - 1/4 ... = 0.6931..., we have
%C A104709   1*(1/2) = 1/2,
%C A104709   1*(3/4) - (1/2)*(1/4) = 5/8,
%C A104709   1*(7/8) - (1/2)*(4/8) + (1/3)*(1/8) = 2/3,
%C A104709   1*(15/16) - (1/2)*(11/16) + (1/3)*(5/16) - (1/4)*1/16 = 131/192,
%C A104709   ...
%C A104709 This is A068566/A068565. (End)
%H A104709 Paul Curtz, <a href="http://paul.curtz.free.fr/wp-content/uploads/These_Curtz_1965.pdf">Accélération de la convergence de certaines séries alternées à l'aide des fonctions de sommation</a>, Thèse de 3ème Cycle d'Analyse Numérique, Faculté des Sciences de l'Université de Paris, 4 mai 1965.
%H A104709 Clark Kimberling, <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Papers1/52-3/Kimberling11132013.pdf">Fusion, Fission, and Factors</a>, Fib. Q., 52(3) (2014), 195-202.
%F A104709 Begin with A055248 as a triangle, delete leftmost column.
%F A104709 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
%F A104709 T(n,k) = Sum_{j=0..n-k} C(n-j, k)*2^j. - _Paul Barry_, Jan 12 2006
%F A104709 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
%F A104709 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
%F A104709 From _Petros Hadjicostas_, Jun 05 2020: (Start)
%F A104709 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))).
%F A104709 The o.g.f. of the n-th row is (2^(n+1) - (1 + y)^(n+1))/(1 - y).
%F A104709 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.
%F A104709 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.
%F A104709 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)
%F A104709 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
%F A104709 T(n,1) = 2^(n+1) - n - 2 = A000295(n+1) for n >= 1. - _Bernard Schott_, Feb 22 2023
%e A104709 Triangle T(n,k) (with rows n >= 0 and columns k = 0..n) begins:
%e A104709    1;
%e A104709    3,  1;
%e A104709    7,  4,  1;
%e A104709   15, 11,  5,  1;
%e A104709   31, 26, 16,  6,  1;
%e A104709   63, 57, 42, 22,  7,  1;
%e A104709   ...
%p A104709 A104709_row := proc(n) add(add(binomial(n,n-i)*x^(n-k-1),i=0..k),k=0..n-1);
%p A104709 coeffs(sort(%)) end; seq(print(A104709_row(n)),n=1..6); # _Peter Luschny_, Sep 29 2011
%t A104709 z = 10;
%t A104709 p[n_, x_] := (x + 1)^n;
%t A104709 q[0, x_] := 1; q[n_, x_] := x*q[n - 1, x] + 1;
%t A104709 p1[n_, k_] := Coefficient[p[n, x], x^k];
%t A104709 p1[n_, 0] := p[n, x] /. x -> 0;
%t A104709 d[n_, x_] := Sum[p1[n, k]*q[n - 1 - k, x], {k, 0, n - 1}]
%t A104709 h[n_] := CoefficientList[d[n, x], {x}]
%t A104709 TableForm[Table[Reverse[h[n]], {n, 0, z}]]
%t A104709 Flatten[Table[Reverse[h[n]], {n, -1, z}]] (* A054143 *)
%t A104709 TableForm[Table[h[n], {n, 0, z}]]
%t A104709 Flatten[Table[h[n], {n, -1, z}]] (* A104709 *)
%t A104709 (* _Clark Kimberling_, Aug 07 2011 *)
%Y A104709 Cf. A000124, A000295, A001787 (row sums), A007318, A054143, A055248, A068565, A068566, A104709, A140513.
%K A104709 nonn,tabl
%O A104709 0,2
%A A104709 _Gary W. Adamson_, Mar 19 2005
%E A104709 Name edited and offset changed by _Petros Hadjicostas_, Jun 04 2020