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A176668 Triangle T(n,k) read by rows: coefficient [x^k] of the polynomial sum_{k=0..infinity} (2*k+1)^n*binomial(x,k) / 2^x.

%I A176668 #16 Nov 02 2013 21:38:33
%S A176668 1,1,1,1,3,1,1,6,6,1,1,8,21,10,1,1,5,45,55,15,1,1,7,30,185,120,21,1,1,
%T A176668 70,-77,245,595,231,28,1,1,72,490,-756,1435,1596,406,36,1,1,-1311,
%U A176668 3762,-546,-2625,6111,3738,666,45,1,1,-1309,-11325,35130,-20895,-1743,20685
%N A176668 Triangle T(n,k) read by rows: coefficient [x^k] of the polynomial sum_{k=0..infinity} (2*k+1)^n*binomial(x,k) / 2^x.
%C A176668 Row sums are A007051(n).
%C A176668 Exponential Riordan array [exp(x), log((exp(2x)+1)/2)]=[exp(x),x+log(cosh(x))]. - _Paul Barry_ Jan 10 2011
%F A176668 From _Peter Bala_, Mar 16 2012. (Start)
%F A176668 The row polynomials of this triangle may be obtained by applying the operator x*d/dx repeatedly to x*(1+x^2)^n = sum {k = 0..n} binomial(n,k)*x^(2*k+1) and evaluating the result at x = 1. The first few results are:
%F A176668 sum {k = 0..n} (2*k+1)*binomial(n,k) = (n+1)*2^n
%F A176668 sum {k = 0..n} (2*k+1)^2*binomial(n,k) = (n^2+3*n^+1)*2^n
%F A176668 sum {k = 0..n} (2*k+1)^3*binomial(n,k) = (n^3+6*n^2+6*n+1)*2^n.
%F A176668 Compare with A209849. (End)
%e A176668 1;
%e A176668 1, 1;
%e A176668 1, 3, 1;
%e A176668 1, 6, 6, 1;
%e A176668 1, 8, 21, 10, 1;
%e A176668 1, 5, 45, 55, 15, 1;
%e A176668 1, 7, 30, 185, 120, 21, 1;
%e A176668 1, 70, -77, 245, 595, 231, 28, 1;
%e A176668 1, 72, 490, -756, 1435, 1596, 406, 36, 1;
%e A176668 1, -1311, 3762, -546, -2625, 6111, 3738, 666, 45, 1;
%e A176668 1, -1309, -11325, 35130, -20895, -1743, 20685, 7890, 1035, 55, 1;
%e A176668 Production matrix begins
%e A176668 1, 1,
%e A176668 0, 2, 1,
%e A176668 0, -1, 3, 1,
%e A176668 0, 1, -3, 4, 1,
%e A176668 0, -1, 4, -6, 5, 1,
%e A176668 0, 1, -5, 10, -10, 6, 1,
%e A176668 0, -1, 6, -15, 20, -15, 7, 1,
%e A176668 0, 1, -7, 21, -35, 35, -21, 8, 1,
%e A176668 0, -1, 8, -28, 56, -70, 56, -28, 9, 1
%e A176668 - _Paul Barry_ Jan 10 2011
%p A176668 A176668 := proc(n,k) sum( (2*l+1)^n*binomial(x,l),l=0..infinity) ; simplify(%/2^x) ; coeftayl(%,x=0,k) ; end proc: # _R. J. Mathar_, Jan 15 2011
%t A176668 p[x_, n_] = Sum[(2*k + 1)^n*Binomial[x, k], {k, 0, Infinity}]/2^x ;
%t A176668 Table[CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x], {n, 0, 10}];
%t A176668 Flatten[%]
%Y A176668 Cf. A007051, A009393.
%K A176668 sign,tabl
%O A176668 0,5
%A A176668 _Roger L. Bagula_, Apr 23 2010