0,2

Because (x*c(x))^n + (1/c(x))^n = L(n,-x)= sum(A034807(n,k)*(-x)^k,k=0..floor(n/2)) the sequence starts with the Lucas polynomial L(10,-x) (of degree 5).

O.g.f.: 1/c(x)^10 = P(11, x) - x*P(10, x)*c(x) with the o.g.f. c(x):=(1-sqrt(1-4*x))/(2*x) of A000108 (Catalan numbers) and the polynomials P(n, x) defined in A115139. Here P(11, x)=1-9*x+28*x^2-35*x^3+15*x^4-x^5 and P(10, x)=1-8*x+21*x^2-20*x^3+5*x^4.

a(n)=-C10(n-10), n>=10, with C10(n):= (n+4) (eighth convolution of Catalan numbers). a(0)=1, a(1)=-10, a(2)=35, a(3)=-50, a(4)=25, a(5)=-2, a(6)=a(7)=a(8)=a(9)=0. [1, -10, 35, -50, 25, -2] is row n=10 of signed A034807 (signed Lucas polynomials).

Cf. A115148 (ninth convolution).

Sequence in context: A001890 A010818 A065195 this_sequence A044087 A022702 A044468

Adjacent sequences: A115146 A115147 A115148 this_sequence A115150 A115151 A115152

sign,easy

Wolfdieter Lang (wolfdieter.lang_AT_physik_DOT_uni-karlsruhe_DOT_de), Jan 13 2006