A204451 2*A014335 - A203578. Difference of the exponential convolution of A000045 (Fibonacci) with itself and the corresponding exponential half-convolution.

0, 0, 0, 3, 8, 35, 75, 371, 888, 3891, 9445, 40755, 102323, 426803, 1091167, 4469555, 11625960, 46805811, 123364443, 490156851, 1306737465, 5132989235, 13816838695, 53753361203, 145912841523, 562912506675, 1539304050375, 5894896300851, 16225419029303, 61732155503411, 170909837010835

Offset: 0

Author

Wolfdieter Lang, Jan 16 2012

Keywords

• nonn

Comments

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.

Crossrefs

Cf. A000045, A014335, A203578.

Formula

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).

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.

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.