Theorem (Based on work of Hans Havermann, L. Edson Jeffery, Brad Klee, Don Reble, Bob Selcoe, and N. J. A. Sloane) 

              A269254(110) = -1. 

Proof.
Let s(n) denote the sequence s(0)=1, s(1)=111, and
s(n) = 110*s(n-1) - s(n-2).
The initial terms are
1, 111, 12209, 1342879, 147704481, 16246150031, 1786928798929, 196545921732159, 21618264461738561, 2377812544869509551, 261537761671184312049, 28766775971285404815839, 3164083819079723345430241, 348020453322798282592510671, 38279085781688731361830743569, ...

We claim s(n) is never a prime, and consider 3 cases.
(i) n=3k+1:
By reading the definition of s(n) mod 3, we see that s(n) mod 3 is 1,0,2,1,0,2,... and so s(3k+1) is a multiple of 3.

(ii) n=3k:
The sequence s(3k) has generating function (g.f.)
(1+12209*x)/(x^2-1330670*x+1) ..... (Eq1)
Define sequences b(k) and c(k) by
b(0)=1, b(1)=139, thereafter b(k) = 110*b(k-1)-b(k-2),
c(0)=1, c(1)=9661, c(2)=116876761, thereafter c(k) = 12099*(c(k-1)-c(k-2))+c(k-3).
We claim that s(3*k) = b(k)*c(k) for all k >= 0. ..... (Eq2)
This is easily checked to hold for 0 <= k <= 50.
Since b(k) satisfies a second-order recurrence, and
c(k) satisfies a third-order recurrence, we know that b(k)*c(k)
satisfies a linear recurrence of order at most 6. Using gfun [1],
we find that in fact b(k)*c(k) satisfies a second-order recurrence,
and furthermore has g.f. (Eq1). Since the g.f.'s are equal this proves
(Eq2) and shows that these terms are never primes. 
Note that if a sequence is known to satisfy a linear recurrence of order d,
gfun will find it correctly when given 2d+1 terms. 
Note also that it is easy to see that for k>0.  b(k) and c(k) are greater than 1.

(iii) n=3k+2.
This case is essentially identical to the previous case, except that the initial conditions on b(k) and c(k) are different (the recurrences are the same), and 
there is a shift in the subscripts.  More preciely, the sequence s(3k+2) is:
12209, 16246150031, 21618264461738561, 28766775971285404815839, 38279085781688731361830743569, 50936831077090977385276030140145391, ...

We define b(0)=-1, b(1) = 29, thereafter b(k) = 110*b(k-1)-b(k-2),
c(0)=421, c(1)=5091241, c(2)=61593831181, thereafter c(k) = 12099*(c(k-1)-c(k-2))+c(k-3).
Then we claim that s(3k+2) = b(k+1)*c(k).
We use gfun to show that the sequence s(3k+2) has generating function
(12209+x)/(x^2-1330670*x+1) ..... (Eq1)
and that {b(k+1)*c(k), k >= 0} has the same g.f.

This completes the proof.

[1] B. Salvy and P. Zimmermann, “Gfun: a Maple package for the manipulation of generating and holonomic functions in one variable,” Trans. Math. Software, vol. 20, no. 2, pp. 163–177, 1994.

Written by N. J. A. Sloane, Oct 23 2017