<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] There are zillions of such relations. For example: $2^6+1$ is a multiple of $13$, so $k2^n+1$ is composite if $n\equiv6\pmod{12}$ and $k\equiv1\pmod{13}$.  You can make as many of these as you want. 
EDIT: In general, take any numbers $q$ and $s$, and any prime $p$ dividing $q2^s+1$, and any $r$ such that $2^r\equiv1\pmod p$, then $k2^n+1$ is composite (since a multiple of $p$) when $k\equiv q\pmod p$ and $n\equiv s\pmod r$.  
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