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%I #27 Nov 09 2021 18:39:33
%S 1,3,1,7,5,1,15,17,7,1,31,49,31,9,1,63,129,111,49,11,1,127,321,351,
%T 209,71,13,1,255,769,1023,769,351,97,15,1,511,1793,2815,2561,1471,545,
%U 127,17,1,1023,4097,7423,7937,5503,2561,799,161,19,1
%N Mirror of the triangle A193844.
%C This triangle is obtained by reversing the rows of the triangle A193844.
%C From _Philippe Deléham_, Jan 17 2014: (Start)
%C Subtriangle of the triangle in A112857.
%C T(n,0) = A000225(n+1).
%C T(n,1) = A000337(n).
%C T(n+2,2) = A055580(n).
%C T(n+3,3) = A027608(n).
%C T(n+4,4) = A211386(n).
%C T(n+5,5) = A211388(n).
%C T(n,n) = A000012(n).
%C T(n+1,n) = A005408(n).
%C T(n+2,n) = A056220(n+2).
%C T(n+3,n) = A199899(n+1).
%C Row sums are A003462(n+1).
%C Diagonal sums are A048739(n).
%C Riordan array (1/((1-2*x)*(1-x), x/(1-2*x)). (End)
%C 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
%C 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
%H Michael De Vlieger, <a href="/A193845/b193845.txt">Table of n, a(n) for n = 0..11475</a> (rows 0 <= n <= 150, flattened)
%H Russell Jay Hendel, <a href="https://arxiv.org/abs/2107.03549">A Method for Uniformly Proving a Family of Identities</a>, arXiv:2107.03549 [math.CO], 2021.
%F T(n,k) = A193844(n,n-k).
%F 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
%e First six rows:
%e 1
%e 3....1
%e 7....5....1
%e 15...17...7....1
%e 31...49...31...9...1
%e 63...129..111..49..11..1
%t z = 10;
%t p[n_, x_] := (x + 1)^n;
%t q[n_, x_] := (x + 1)^n
%t p1[n_, k_] := Coefficient[p[n, x], x^k];
%t p1[n_, 0] := p[n, x] /. x -> 0;
%t d[n_, x_] := Sum[p1[n, k]*q[n - 1 - k, x], {k, 0, n - 1}]
%t h[n_] := CoefficientList[d[n, x], {x}]
%t TableForm[Table[Reverse[h[n]], {n, 0, z}]]
%t Flatten[Table[Reverse[h[n]], {n, -1, z}]] (* A193844 *)
%t TableForm[Table[h[n], {n, 0, z}]]
%t Flatten[Table[h[n], {n, -1, z}]] (* A193845 *)
%t 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 *)
%o (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
%Y Cf. A193844.
%Y Cf. Columns: A000225, A000337, A055580, A027608, A211386, A211388
%Y Cf. Diagonals: A000012, A005408, A056220, A199899
%K nonn,tabl
%O 0,2
%A _Clark Kimberling_, Aug 07 2011
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