Least prime of the form $38^n+31$ I search the least n such that 
$$38^n+31$$ 
is prime. 
I checked the $n$ upto $3000$ and found none, so the least prime of that form must have more than $4000$ digits. I am content with a probable prime, it need not be a proven prime.

In general, is there any prime of the form $\frac{ab^n+c}{\gcd(a+c,b-1)}$ with integer $n\ge1$, for fixed integers $a\ge1$, $b\ge2$, $c\ne0$, $\gcd(a,c)=1$, $\gcd(b,c)=1$? It is unknown whether this problem is solvable, e.g. let $a=4$, $b=72$, $c=-1$, the smallest prime of the form $\frac{4\cdot72^n-1}{\gcd(4+1,72-1)}$ with $n\ge1$ is at $n=1119849$
If we want to solve this problem, we must for every such $(a,b,c)$ integer triple, either find a prime of this form or prove that there is no prime of this form, e.g.:
$(a,b,c) = (78557,2,1)$, in which all numbers are divisible by at least one of $3, 5, 7, 13, 19, 37, 73$
$(a,b,c) = (11047,3,1)$, in which all numbers are divisible by at least one of $2, 5, 7, 13, 73$
$(a,b,c) = (47,8,1)$, in which all numbers are divisible by at least one of $3, 5, 13$
$(a,b,c) = (334,10,-1)$, in which all numbers are divisible by at least one of $3, 7, 13, 37$
$(a,b,c) = (9,4,-1)$, in which all numbers factored as difference of squares
$(a,b,c) = (1,8,1)$, in which all numbers factored as sum of cubes
$(a,b,c) = (25,12,-1)$, in which even $n$ factored as difference of squares and odd $n$ is divisible by $13$
$(a,b,c) = (64,936,-1)$, in which even $n$ factored as difference of squares and $n$ divisible by $3$ factored as difference of cubes and other $n$ divisible by either $37$ or $109$, etc.
