A201207 Half-convolution of sequence A000032 (Lucas) with itself.

4, 2, 7, 11, 27, 41, 84, 137, 270, 435, 826, 1338, 2488, 4024, 7353, 11899, 21461, 34723, 61960, 100255, 177344, 286947, 503892, 815316, 1422892, 2302286, 3996619, 6466667, 11173935, 18079805, 31114236

Offset: 0

Author

Wolfdieter Lang, Jan 03 2012

Keywords

• nonn
• easy

Comments

For the definition of the half-convolution of a sequence with itself see a comment on A201204. There the rule for the o.g.f. is given. Here the o.g.f. is (L(x)^2 + L2(x^2))/2, with the o.g.f. L(x)=(2-x)/(1-x-x^2) of A000032, and L2(x) = (4-7*x-x^2)/((1+x)*(1-3*x+x^2)) the o.g.f. of A001254. This leads to the o.g.f given in the formula section.

For the bisection of this sequence see A203570 and A203574.

Crossrefs

Cf. A000032, A000045, A201204, A203570, A203574.

Formula

a(n) = Sum_{k=0..floor(n/2)} L(k)*L(n-k), n >= 0, with the Lucas numbers L(n)=A000032(n).

O.g.f.: (4-2*x-7*x^2+6*x^3-x^4+3*x^5)/((1-3*x^2+x^4)*(1+x^2)*(1-x-x^2)). See a comment above.

a(n) = (1/4)*(2*(2*n+5+(-1)^n)*F(n+1)-(2*n+3+(-1)^n)*F(n)) +(i^n+(-i)^n)/2, n >= 0, with the Fibonacci numbers F(n)=A000045(n) and the imaginary unit i=sqrt(-1). From the partial fraction decomposition of the o.g.f. and the Fibonacci recurrence.