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A204450 2*A203579 - A204449. Difference between the exponential convolution of A000032 (Lucas) with itself and the corresponding exponential half-convolution.

%I A204450 #10 Mar 30 2012 18:49:34
%S A204450 0,2,6,17,30,177,417,1857,4302,19457,47731,203777,509769,2134017,
%T A204450 5462701,22347777,58104062,234029057,616919457,2450784257,6533317815,
%U A204450 25664946177,69085604341,268766806017,729558799305,2814562533377
%N A204450 2*A203579  - A204449. Difference between the exponential convolution of A000032 (Lucas) with itself and the corresponding exponential half-convolution.
%C A204450 See A204449 for the exponential half-convolution of A000032 (Lucas). The present sequence gives the numbers to be added to A204449 to obtain the corresponding (full) exponential convolution A203579.
%F A204450 a(n) = sum(binomial(n,k)*L(k)*L(n-k),k=floor(n/2)+1..n), n>=0,
%F A204450 with the Lucas numbers L(n)=A000032(n). For n=0 this is 0.
%F A204450 E.g.f.: exp(x)*(cosh(sqrt(5)*x)+1) - (BesselI(0,2*phi*x) + BesselI(0,2*(phi-1)*x) + 2*BesselI(0,2*I*x))/2. Compare this with the e.g.f. of A204449, where phi and BesselI are explained.
%F A204450 Bisection: a(2*k) = (2^(2*k)-binomial(2*k,k))*L(2*k)/2 + 1 - ((-1)^k)*binomial(2*k,k), a(2*k+1) = 2^(2*k)*L(2*k+1)+ 1 = A204449(2*k+1), k>=0.
%e A204450 With A000032 = {2, 1, 3, 4, 7, 11, ...}
%e A204450 a(4) = 4*4*1 + 1*7*2 = 30.
%e A204450 a(5) = 10*4*3 + 5*7*1 + 1*11*2 = 177.
%Y A204450 Cf. A000032, A203579, A204449.
%K A204450 nonn
%O A204450 0,2
%A A204450 _Wolfdieter Lang_, Jan 16 2012