Proof that A271676 is fini and full from Altug Alkan with minor edits by Ray Chandler. If n is in this sequence, then 3*n + 4 = k^2 for k is nonzero integer, that is, n = (k^2 - 4)/3 = (k-2)*(k+2)/3. If n is a prime power, that is n = p^t where p is prime and t >= 0. Since p^t = (k-2)*(k+2)/3, there are two possible factorizations: (k-2) and (k+2)/3 are prime powers or (k-2)/3 and (k+2) are prime powers. The first case: (k-2) and (k+2)/3 are prime powers. In this case p^m = (k-2) and p^n = (k+2)/3 where m + n = t. So k = p^m + 2 = 3*p^n - 2. Thus p^m + 4 = 3*p^n, that is 2*2 = p^n*(3 - p^(m-n)). So obviously either n must be 0 or p must be 2, but if n = 0 then p^n = 1 and p^(m-n)=-1 which is a contradiction. For p=2, 2*2 = 2^n*(3 - 2^(m-n)) and it is easy to see that (n = 1 and m = 1) or (n = 2 and m = 3). The second case: (k-2)/3 and (k+2) are prime powers. In this case p^m = (k-2)/3 and p^n = (k+2) where m + n = t. So, k = 3*p^m + 2 and 3*p^m + 4 = p^n, that is, 2*2 = p^n - 3*p^m = p^m*(p^(n-m)-3). So obviously either m = 0 or p must be 2. If m = 0 then p^n = 7, so p = 7 and n = 1. If p = 2, then equation is 2*2 = 2^m*(2^(n-m)-3). It is easy to see that m = 2 and n = 4 is the unique solution pair. So we see that our solutions p^(n+m); 2^(1+1), 7^(1+0), 2^(2+3), 2^(4+2) are only prime powers because of the equations that we have after factorizations in above.