There was a deleted question that asked;
We know that if $n$ zeros are inserted between every pair of successive terms of the sequence in LFSR (variable changed $x$ to $x^n$) then linear complexity (shortest LFSR which produces the given seq) will be $n$ times the original one.
My question is that: in the new sequence what is the period (assume period of original one is $p$)
The period was easy to answer;
Let $m$ be a sequence output from $L$ bit maximal LFSR. Then we know that the period of the sequence $m$ is $2^L-1$
Let $m'$ be the modified sequence of $m$ such that for every two successive elements $n$ zeros are inserted. An example with $n=3$
m = 1 1 0 1 1 0 1 m' = 1000100000001000100000001000
It is obvious that the period of $m'$ is $(2^L-1)\cdot n$. One can prove this with a counter-argument; if the period is smaller, then remove the added zeroes to find a smaller period for $m$.
What about the linear complexity?
[1,1,0,1,1,0,1,1] has x^2 + x^1 + 1 [1,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0] has x^8 + x^4 + 1
and for the $n=1$ case;
[1,1,0,1,1,0,1,1] has x^2 + x^1 + 1 [1,0,1,0,0,0,1,0,1,0,0,0,1,0] has x^4 + x^2 + 1 [1,1,0,1,1,1,1,1] has x^4 + x^3 [1,0,1,0,0,0,1,0,1,0,1,0,1,0] has x^7 + x^5
Is there a work or a result on the bounds of the Linear complexity of $m'$ related to $L$ and $n$?