Finding the balancing point of a subsequence
Without messing with your actual algorithm, right now it's effectively \$O(n^2)\$.


Creating the initial deque is \$O(n)\$
The while loop is essentially \$O(n)\$ (worst case)
Within the while loop you call equi with successively smaller lists \$k=(n, n-1, n-2, \dots, n-5)\$ and equi is \$O(k)\$ (worst case).


So the loop is \$O(n^2)\$ and dominates.

You can get some improvement in running time vs complexity though (maybe 20%) by fixing this line:

is_sequence = equi(list(equi_deque))


to be 

is_sequence = equi(equi_deque)


The conversion to a list is unnecessary, and is an \$O(k)\$ operation.