In doing unrelated calculations I was a bit flabbergasted when I found that the formula
$f(n)=8 + 3(5^n)$
gives prime numbers for $n=0,1,2,3$; so I checked further and found:
Out of the first 401 natural numbers (that is, $n=0,1,...,400$) 19 of these have $f(n)$ prime. They are: 0, 1, 2, 3, 7, 9, 14
19,
21,
24,
38 ,
48,
49,
59,
69,
86,
131,
174,
399. (Easily checked on Wolfram-Alpha)
I do not at all know what a reasonable number of primes would be out of this, but I expect fewer than this would be expected, given how large the numbers become. My very lax reasoning is below:
The expected density of primes given the number of primes less than $8+3(5^{400})$ (using $Li(n)=\int_2^n\frac{1}{log(x)}dx$ to estimate $\pi(x)$) is about $Li(3\cdot5^{400})\approx 0.00155$. For 200, $Li(3\cdot 200)\approx0.00357$ is the approximate density. So we should expect somewhere on the order of magitude of a density of $0.002$ for primes in the range $3\cdot 5^{174}$ to $3\cdot 5^{400}$.
However, since by the end of the sequence there is about 1 prime for every 200 attempts, or about $0.005$, which is more than twice the expectation (granted 399 is luckily just scraping under 400).
My question: Is there any way to see if this sequence is more likely to generate primes than other sequences? Further, is this a pattern that is already known in some fashion?
Interesting side note: the same pattern holds for small numbers (I haven't checked very far) for the similar sequence $g(n)=4+3(5^n)$.
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