so each digit will have its corresponding number that increases by itself every digit, so the first digit is 1 the second digit is 2 the third digit is 4 etc etc. calling these numbers "BN"

so kinda like this:

0 0 0 0 0 0 (binary)

32 16 8 4 2 1 (BNs)

so say you want to get the decimal number of the binary number 011010. you get the BNs corresponding with the places where there is a 1 and add them all together. like this:

0 1 1 0 1 0

16+8+4 = 26

im not really sure if you can do this with numbers higher than 63 (aka more than 6 binary digits)

this is right! this is how binary to decimal conversion works. the BNs are the the powers of two: each digit is the previous increased by itself => each digit is the previous multiplied by 2 => nth digit is 2 to the nth power.

positional base for a given B works the same way if you change "add all the powers of two where there is a one" to "add all the powers of K multiplied by the corresponding digit" (these are equivalent if K is 2; multiplying something by 0 is 0)

say you want to get the decimal number of the octal number 5756. let's get the powers that we need:

8^{3} = 512

8^{2} = 64

8^{1} = 8

8^{0} = 1

and 5*512 + 7*64 + 5*8 + 6*1 = 3054. this will work for every octal(/binary) number!

yes, in the same way that in decimal, each digit from the right corresponds to the next highest power of ten, in binary each digit from the right corresponds to the next highest power of two. this is true of any base.