%I #14 Dec 21 2016 12:03:40
%S 12,30,902,1360,2450,3730,21475,74945,82208,88282,254677
%N Composite numbers m with sqrt(m) not prime such that T(m) == 1 (mod m), where the central trinomial coefficient T(m) is the coefficient of x^m in the expansion of (x^2+x+1)^m.
%C The author proved in arXiv:1610.03384 that T(p) == 1 (mod p^2) and T(p^2) == T(p) (mod p^3) for each prime p > 3, and conjectured that T(n) == 1 (mod n^2) fails for any composite number n.
%C Conjecture: Besides the listed 11 terms, the sequence has no other terms.
%C By our computation, if the 12th term exists, it should be greater than 6*10^6.
%C T(n) = A002426(n). - _Michael Somos_, Dec 19 2016
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1610.03384">Supercongruences involving Lucas sequences</a>, arXiv:1610.03384 [math.NT], 2016.
%H Zhi-Wei Sun, <a href="http://listserv.nodak.edu/cgi-bin/wa.exe?A2=NMBRTHRY;840717bb.1612">Characterizing primes via central trinomial coefficients</a>, a Message to Number Theory List, Dec. 7, 2016.
%e a(1) = 12 since T(12) = 73789 = 1 + 12*6149.
%t T[0]=1;
%t T[1]=1;
%t T[n_]:=T[n]=((2n-1)T[n-1]+3(n-1)T[n-2])/n;
%t n=0;Do[If[PrimeQ[m]==False&&PrimeQ[Sqrt[m]]==False&&Mod[T[m]-1,m]==0,n=n+1;Print[n," ",m]],{m,2,300000}]
%Y Cf. A000040, A002426, A002808, A277640.
%K nonn,more
%O 1,1
%A _Zhi-Wei Sun_, Dec 19 2016