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A201207
Half-convolution of sequence A000032 (Lucas) with itself.
2
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
0,1
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.
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.
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Jan 03 2012
STATUS
approved