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%I #25 Aug 08 2017 22:20:50
%S 1,2,1,6,6,1,12,24,14,1,20,60,70,30,1,30,120,210,180,62,1,42,210,490,
%T 630,434,126,1,56,336,980,1680,1736,1008,254,1,72,504,1764,3780,5208,
%U 4536,2286,510,1,90,720,2940,7560,13020,15120,11430,5100,1022,1,110,990,4620,13860,28644,41580,41910,28050,11242,2046,1
%N Triangle of numbers where T(n,k) is the number of k-dimensional faces on the polytope that is the convex hull of all permutations of the list (0,1,...,1,2), where there are n - 1 ones, for n > 0. T(0,0) is 1.
%C It appears that these integers, with sign changes, are also in A138106.
%F T(n,n) = 1, n >= 0.
%F T(n,n-1) = 2^(n+1)-2, n > 0.
%F T(n,0) = n(n+1), n > 0.
%F T(n,k) = (n+1)*T(n-1,k)/(n-k-1), 0 <= k < n-1, n >= 2.
%F E.g.f.: ((2*x+1)*exp(z*(2*x+1)) - 2*(x+1)*exp(z*(x+1)) + x^2*exp(z*x)+exp(z))/x^2
%F Sum_{k=0...n-1} T(n,k)*x^(n-k-1) = x^(n+1) - 2(x+1)^(n+1) + (x+2)^(n+1) (conjectured). - _Kevin J. Gomez_, Jul 25 2017
%e Triangle begins:
%e 1;
%e 2, 1;
%e 6, 6, 1;
%e 12, 24, 14, 1;
%e 20, 60, 70, 30, 1;
%e ...
%e Row 2 describes a regular hexagon.
%e Row 3 describes the cuboctahedron.
%t Join @@ (CoefficientList[#,
%t x] & /@ (Expand[
%t D[((1 + 2 x) Exp[z (1 + 2 x)] - 2 (1 + x) Exp[z (1 + x)] + Exp[z] +
%t x^2 Exp[z x])/x^2, {z, #}] /. z -> 0] & /@ Range[0, 10]))
%K nonn,tabl
%O 0,2
%A _Vincent J. Matsko_, Jun 30 2015
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