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I ran into the following sequence:
$$f_n=n\cdot2^n-1.$$
Apparently, for $n>1$, $f_n$ will yield a prime number. It will not list all of them, however.
My question is: Is this true for all $n$?
I currently have a program running that has thus far checked that it is true for all $n\leq49$. What do you guys think?
Here is a list of numbers $1\leq n\leq 100$ for which $n2^n-1$ is not prime:
$$1,4, 5, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24,$$ $$ 25, 26, 27, 28, 29, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45,$$ $$46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65,$$ $$66, 67, 68, 69, 70, 71, 72, 73, 74, 76, 77, 78, 79, 80, 82, 83, 84, 85, 86, 87,$$ $$88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100.$$
These are called Woodall numbers. See https://oeis.org/A002234 and references given there. It is conjectured that infinitely many Woodall numbers are prime, but this is still very much an open question.
Your program apparently needs work. Let $n=2^{2k}$. Then $f_n=2^{2k}\times2^{2^{2k}}-1=2^{2k+2^{2k}}-1=(2^{k+2^{2k-1}}+1)(2^{k+2^{2k-1}}-1)$.
So $f_n$ is composite for all positive integer powers of 4.
$16\times 2^{16}-1$ is divisible by $5$.
This is because $2^4 = 16 = 1 \mod 5$.
In general, for every odd prime $p$, for $n = (p-1)^2, \ n2^n - 1$ is divisible by $p$.
$n = 4$ is the first value for which $f_n$ isn't prime.
$4 \times 2^4 - 1 = 63 = 3^2 \times 7$
Notice that if $n = 2^{2a}$ for some positive integer $a,$ then $n2^{n}- 1$ is never prime, because $n 2^{n}-1 = 2^{ 2a+2^{2a}} -1,$ which is divisible by $3$ as $2^{2m}-1$ is divisible by $3$ for any positive integer $m.$
