A193862 - OEIS

%I #12 Apr 10 2020 12:38:07

%S 1,2,2,3,6,4,4,12,16,8,5,20,40,40,16,6,30,80,120,96,32,7,42,140,280,

%T 336,224,64,8,56,224,560,896,896,512,128,9,72,336,1008,2016,2688,2304,

%U 1152,256,10,90,480,1680,4032,6720,7680,5760,2560,512,11,110,660

%N Mirror of the triangle A115068.

%C A193862 is obtained by reversing the rows of the triangle A115068.

%C Riordan array (1/(1-x)^2, 2*x/(1-x)). - _Philippe Deléham_, Jan 29 2014

%C Let P(n, x) := Sum_{k=1..n} T(n, k)*x^k. Then P(n, P(m, x)) = P(n*m, x) for all n and m in Z. - _Michael Somos_, Apr 10 2020

%F Write w(n,k) for the triangle at A115068.  The triangle at A193862 is then given by w(n,n-k).

%F T(n, k) = binomial(n, k)/2 * 2^k. - _Michael Somos_, Apr 10 2020

%e First six rows:

%e 1

%e 2...2

%e 3...6....4

%e 4...12...16...8

%e 5...20...40...40....16

%e 6...30...80...120...96...32

%e Production matrix begins

%e 2......2

%e -1/2...1...2

%e 1/4....0...1...2

%e -1/8...0...0...1...2

%e 1/16...0...0...0...1...2

%e -1/32..0...0...0...0...1...2

%e 1/64...0...0...0...0...0...1...2

%e -1/128.0...0...0...0...0...0...1...2

%e 1/256..0...0...0...0...0...0...0...1...2

%e - _Philippe Deléham_, Jan 29 2014

%t z = 11;

%t p[0, x_] := 1; p[n_, x_] := x*p[n - 1, x] + 1;

%t q[n_, x_] := (2 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}]]  (* A115068 *)

%t TableForm[Table[h[n], {n, 0, z}]]

%t Flatten[Table[h[n], {n, -1, z}]]   (* A193862 *)

%t T[ n_, k_] := Binomial[n, k]/2 2^k; (* _Michael Somos_, Apr 10 2020 *)

%o (PARI) {T(n, k) = binomial(n, k)/2 * 2^k}; /* _Michael Somos_, Apr 10 2020 */

%Y Cf. A115068.

%K nonn,tabl

%O 0,2

%A _Clark Kimberling_, Aug 07 2011