2537720637domain: NAppears in sequencesProduct of Lucas numbers and inverted Lucas numbers: a(n)=A000032(n)*A075193(n).at n=22A075269a(n) = Lucas(4*n+1) + 1, or Lucas(2*n)*Lucas(2*n+1).at n=11A081014Partial sums of A005248.at n=22A188378a(n) = L(n)*L(n+1), where L = A000032 (Lucas numbers).at n=22A215602