<s>[INST] Compositeness of number $k\cdot 2^n+1$? Every odd prime number can be expressed in the form $k \cdot 2^n+1$ ,where $k$ is an odd number .
For $n>2$ number $k \cdot 2^n+1$ is composite if :

$1.$ $k\equiv 1 \pmod {30} \land (n\equiv 2 \pmod 4 \lor n \equiv 1 \pmod 2 ) $
$2.$ $k\equiv 3 \pmod {30} \land n\equiv 3 \pmod 4$
$3.$ $k\equiv 5 \pmod {30} \land n\equiv 0 \pmod 2$
$4.$ $k\equiv 7 \pmod {30} \land n\equiv 1 \pmod 2$
$5.$ $k\equiv 9 \pmod {30} \land n\equiv 0 \pmod 4$
$6.$ $k\equiv 11 \pmod {30} \land n\equiv 0 \pmod 2$
$7.$ $k\equiv 13 \pmod {30} \land n\equiv 1 \pmod 2$
$8.$ $k\equiv 17 \pmod {30} \land (n\equiv 1 \pmod 4 \lor n\equiv 0 \pmod 2)$
$9.$ $k\equiv 19 \pmod {30} \land (n\equiv 0 \pmod 4 \lor n\equiv 1 \pmod 2)$
$10.$ $k\equiv 21 \pmod {30} \land n\equiv 2 \pmod 4$
$11.$ $k\equiv 23 \pmod {30} \land (n\equiv 3 \pmod 4 \lor n\equiv 0 \pmod 2)$
$12.$ $k\equiv 25 \pmod {30} \land n\equiv 1 \pmod 2$
$13.$ $k\equiv 27 \pmod {30} \land n\equiv 1 \pmod 4$
$14.$ $k\equiv 29 \pmod {30} \land n\equiv 0 \pmod 2$

Are there some other similar relations between coefficient $k$ and exponent $n$ that ensure compositeness of number $k \cdot 2^n+1$ ?
 [/INST] Your findings have been generalized in this paper. Here, a covering system of integers is constructed and that allows for infinitely many numbers of the form $k\times2^n+1$ to be composite in the sequence of Lucas nubers. 
So, for you, I guess the important part is that there are infinitely many Sierpinski numbers i.e. numbers of the form $k\times2^n+1$ are composite.
http://home.wlu.edu/~finchc/Research/RieselLucasArticle11.pdf
Just in case you are wondering what is a covering system, see this wiki entry 
http://en.wikipedia.org/wiki/Covering_system
And this thing by Carl Pomerance
http://www.math.dartmouth.edu/~carlp/PDF/covertalkunder.pdf
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