# Prime Sierpiński Numbers of the First Kind

## Open Question

The only known prime Sierpiński numbers of the first kind are:

 $\displaystyle 1^1 + 1$ $=$ $\displaystyle 2$ $\displaystyle 2^2 + 1$ $=$ $\displaystyle 5$ $\displaystyle 4^4 + 1$ $=$ $\displaystyle 257$

It is an open question as to whether there are any more.

## Progress

Let $S_n$ be the $n$th Sierpiński number of the first kind:

$S_n = n^n + 1$

From Form of Prime Sierpiński Number of the First Kind, it is known that if $S_n$ is prime, then:

$n = 2^{2^k}$

for some integer $k$.

Thus

 $\displaystyle S_n$ $=$ $\displaystyle \left({2^{2^k} }\right)^{\left({2^{2^k} }\right)} + 1$ $\displaystyle$ $=$ $\displaystyle 2^{2^k \times \left({2^{2^k} }\right)} + 1$ $\displaystyle$ $=$ $\displaystyle 2^{2^{k + 2^k} } + 1$ $\displaystyle$ $=$ $\displaystyle 2^{2^m} + 1$ where $m = k + 2^k$

Thus a prime $S_n$ is a Fermat number $F_m$ where $m = k + 2^k$.

The sequence of $m$ begins:

$1, 3, 6, 11, 20, 37, 70, 135, 264, 521, 1034, 2059, 4108, 8205, \ldots$

Thus it remains to investigate the following Sierpiński numbers of the first kind for primality:

$k$ $m = k + 2^k$ $n = 2^{2^k}$ Status of $S_n = n^n + 1 = 2^{2^m} + 1 = F_m$
$0$ $1$ $2$ Prime ($S_2 = F_1 = 5$)
$1$ $3$ $4$ Prime ($S_4 = F_3 = 257$)
$2$ $6$ $16$ Composite with factor $1071 \times 2^8 + 1$
$3$ $11$ $256$ Composite with factor $39 \times 2^{13} + 1$
$4$ $20$ $65 \, 536$ Composite with no factor known
$5$ $37$ $4 \, 294 \, 967 \, 296$ Composite with factor $1 \, 275 \, 438 \, 465 \times 2^{39} + 1$
$6$ $70$ too large unknown
$7$ $135$ unknown
$8$ $264$ unknown
$9$ $521$ unknown
$10$ $1034$ unknown
$11$ $2059$ Composite with factor $591 \, 909 \times 2^{2063} + 1$
$12$ $4108$ unknown
$13$ $8205$ unknown
$14$ $16 \, 398$ unknown
$15$ $32 \, 783$ unknown
$16$ $65 \, 552$ unknown
$17$ $131 \, 089$ unknown