Sierpinski number analogue in another base

Question

I was wondering if Sierpinski numbers would work in other bases, and after a little experimentation: yep!

For instance,

157 \cdot 23^{n} - 2

is composite for all $n$ using a tiny covering set. Or,

5515 \cdot 11^{n} + 2.

If I'm mistaken, please let me know.

Picked bases likely to be neutral relative to the numbers that usually make up these covering sets (3,5,7,13,17,19,37,73,etc), and then quick brute force check for a coefficient which seemed to hit those 0-congruences. — Trevor
– Trevor, Commented Dec 10, 2019 at 8:12
Note that these differ from classic Sierpinski numbers not only in the base not being $2$ but also the added constant isn’t $1$ — J. W. Tanner
– J. W. Tanner, Commented Dec 10, 2019 at 8:15

J. W. Tanner · Accepted Answer · 2019-12-10 07:14:33Z

3

$157 \times 23^{2 k + 1} - 2 \equiv 1 \times 2 - 2 = 0 mod 3$ ,

$157 \times 23^{4 k + 2} - 2 \equiv (- 2) (- 1) - 2 = 0 mod 53$ ,

and $157 \times 23^{4 k} - 2 \equiv 2 \times 1 - 2 = 0 mod 5$ ,

so $157 \times 23^{n} - 2$ is composite for all $n$ .

$5515 \times 11^{3 k + 1} + 2 \equiv 5 \times 11 + 2 \equiv 0 mod 19$ ,

$5515 \times 11^{3 k + 2} + 2 \equiv 6 \times 16 + 2 \equiv 0 mod 7$ ,

$5515 \times 11^{2 k} + 2 \equiv 1 \times 1 + 2 \equiv 0 mod 3,$

and $5515 \times 11^{6 k + 3} + 2 \equiv 2 \times (- 1) + 2 \equiv 0 mod 37$ ,

so $5515 \times 11^{n} + 2$ is composite for all $n$ .

answered Dec 10, 2019 at 6:54

J. W. Tanner

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