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2*A014335 - A203578. Difference of the exponential convolution of A000045 (Fibonacci) with itself and the corresponding exponential half-convolution.
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%I #12 Nov 02 2013 21:35:06

%S 0,0,0,3,8,35,75,371,888,3891,9445,40755,102323,426803,1091167,

%T 4469555,11625960,46805811,123364443,490156851,1306737465,5132989235,

%U 13816838695,53753361203,145912841523,562912506675,1539304050375,5894896300851,16225419029303,61732155503411,170909837010835

%N 2*A014335 - A203578. Difference of the exponential convolution of A000045 (Fibonacci) with itself and the corresponding exponential half-convolution.

%C See A203578 for the exponential (or binomial) half-convolution of A000045 (Fibonacci). The present sequence has to be added to this sequence in order to obtain the (full) exponential convolution 2*A014335.

%F a(n) = sum(binomial(n,k)*F(k)*F(n-k), k=floor(n/2)+1..n), n>=0, with the Fibonacci numbers A000045(n).

%F E.g.f.: exp(x)*(cosh((2*phi-1)*x)-1)/5 - (BesselI(0,2*phi*x) + BesselI(0,2*(phi-1)*x) - 2*BesselI(0,2*I*x))/10. See the e.g.f. of A203578, also for phi and BesselI.

%F Bisection: a(2*k) =((2^(2*k) - binomial(2*k,k))*L(2*k)/2 - (1 - (-1)^k*binomial(2*k,k)))/5 and a(2*k+1) = (2^(2*k)*L(2*k+1) - 1)/5 = A203578(2*k), k>=0, with the Lucas numbers L(n)= A000032(n). Compare with A203578. See A000346(k-1), with A000346(-1)=0, for (2^(2*k) - binomial(2*k,k))/2, k>=0.

%Y Cf. A000045, A014335, A203578.

%K nonn

%O 0,4

%A _Wolfdieter Lang_, Jan 16 2012