Finding prime numbers with mod function with respect to given odd number $'a'$ between $2^n$ and $2^{n+1}$

Question

Here are few steps which made sense when analysing the prime numbers

Step 1: For any odd number "$a \in Z^+ $" e.g. 17

Step 2 : $a$ is $2^{n} < a < 2^{n+1}$

Step 3: now get the list starting $ b_i $ where $b_i \equiv 2^{n+1} \pmod a $ as follows

Step 4: $b_1 = 15 $ here, $(32\equiv15 \pmod {17 }$)

Step 5: Then for each step double the result till we again reach the $2^{n}$ as follows

$ b_{i +1}= 2\times b_i(30\equiv13 \pmod {17 }$), in this case

$ 26\equiv9 \pmod {17 }$

$ 18\equiv1 \pmod {17} $

and so on till $b_i = 2^n$

When we make a list it will be $[15, 13, 9, 1,2,4,8,16]$ and the list count $l$ is $8$

$$m = \frac{a-1}{mod\_list\_length} $$

Now $m = \frac{(a-1)}{l}= 2 $

When we run a program in python obviously we get $m = \frac{(a-1)}{l} $.

$a$ is almost always a prime number for any odd when $m \in Z^+$. Python program below to check primality of odd +ve integer. Perhaps we get exceptions!

import math
import time

def find_power_of_two(a):
    if a <= 0 or a % 2 == 0:
        raise ValueError("Input must be a positive odd integer.")
    return int(math.log2(a))

begin = time.time()
# Example usage:
a = int(input("Enter a positive odd integer: "))
n = find_power_of_two(a)
print(f"The value of n such that 2^n < {a} < 2^(n+1) is: {n}")
M = 1 << (n + 1)
b = M % a
l = []
sum = 0
while b != (1 << n):
    while b < (1 << (n - 1)):
        l.append(b)
        sum += b
        b = b << 1
    l.append(b)
    sum += b
    b = (b << 1) % a
l.append(1 << n)

print(l)
sum += 1 << n
m = (a - 1) / len(l)
print('The ratio to odd integer less 1 by number of moduli is ', m)
print('sum of all moduli are ', sum)
print('The ratio between sum of moduli and a is', sum / a)
if (a - 1) % len(l) == 0:
    print(f'{a} is a Prime number')
else:
    print(f'{a} is not a prime number')
end = time.time()
print('Time taken to test is', end - begin)

Following python code gives the results by method below and exceptions also displayed. For example, for odd numbers upto 10000, the exceptions (that are not divisible by 3 or 5) are [341, 1387, 1729, 2047, 2701, 2821, 3277, 4033, 4369, 4681, 5461, 6601, 7957, 8321, 8911] These exceptions are falsely detected as Prime numbers by mod function.

from decimal import*
from fractions import Fraction
import time
getcontext().prec = 1000
n = int(input("Enter a number till which you need Prime numbers by moduli method \n:"))
i = 1
l = []
for i in range(9,n):#(2 is only even Prime number, 3 leads to infinite loop. Therefore I have not started from 3)#
    if i%2 !=0 and i % 5 != 0 and i % 3 != 0:
        l = l + [i]
#print(l)# List of Odd numbers out of divisibility by other than 2, 3, 5
begin = time.time()
nl = []
for j in l:
    i = 0
    for i in range(j):
        if 2 ** i < j < 2 ** (i + 1):
            break
    m = 2 ** (i - 1)
    #print("The numerator for n is = ",m)
    M = 2 ** (i + 1)
    Max = 2 ** (i + 2)
    lst = []

    #print(j)
    while M != m:
        M = M % j
        lst = lst + [M]
        M = M * 2
    lst = lst + [m]
    lst = lst + [m * 2]
    #print('List of remainders in the list for given odd number is', lst)
    rl = lst

    #print('Total number of moduli are', len(rl))
    ratio = (j - 1) / len(lst)
    #print("The ratio between (n-1) and Nom is   ", Fraction((j - 1), len(lst)))
    # returns Fraction
    # print((n-1)//len(lst),",", Fraction((n-1)%len(lst)),"/",len(lst))
    if (j - 1) % len(lst) == 0:
        nl = nl + [j]
print("list of Prime numbers out of Odd numbers is\n",nl)
print(len(nl))
end = time.time()
print("Time taken to calculate list of Prime numbers is", end - begin)
ls = []
a = 0
for num in range(9, n):
    if num > 2:
        for i in range(2, num):
            if num % i == 0:
                break
        else:
            a += 1
            ls = ls + [num]
            # print(a, '|',num )
# print(ls)
print(len(ls))

# using sieve method
# Uncommon elements in List
res_list = []
for i in nl:
    if i not in ls:
        res_list.append(i)
for i in ls:
    if i not in nl:
        res_list.append(i)

    # printing the uncommon
print("The uncommon of two lists is : " + str(res_list))
print(len(res_list))

Even though I thought of relationship with Fermat's little theorem, the nonprime numbers are not exactly the Carmichael numbers.)

One of the important observation I made is due to its association with mod function with $2^n$, any prime of $2^n -1$ will have high $m $ which turns out very fast in running to give the output as prime number or not. Next will be prime numbers of $2^n +1$ forms.

My question is "Is there any way to identify the $m$ as predictable integers?