Hi, was wondering if there's a more mathematical approach to calculating a sequence of numbers where each number multiplies itself by 2. so starting with the number 5, and running 6 levels you would end up with

5, 10, 20, 40, 80, 160

I've got this and it works, but wondering if there's a more efficient way of doing it?

int n = 5;
int level = 6;

for (int i = 1; i < level; i++) {

	n *= 2;
}

return n;

thanks

You're performing a multiple of two, n times, which is the same as one multiplication of (2 ^ (n-1))

I.E

5 * 2 ^ (6 - 1) is 5 * 2 ^ 5

ah thank you.

Just gonna slap this here:

5 * Mathf.Pow(2, level - 1);

awesome, thanks.

Say I wanted to get the inverse of that function. Given these numbers:

5, 10, 20, 40, 80, 160

how could I produce:

1, 2, 3, 4, 5, 6

again, I only know how to do this using loops:

int n = 1;
int value = 80;

while (value > 5) {

	value -= value / 2;
	n++;
}

return n;

You are a wizard. Thank you.

Wish I understood these Math functions better. :|

hello again, I've got another interesting sequence problem, more of a difficult pattern this time (imo)

given this sequence of numbers: 1, 5, 10, 30, 60, 300, 600, 1800, 3600, 18000, 36000

can you find a way (without arrays) to return one of these numbers based on an index (1, 2, 3, 4, 5, 6, 7...)?

ignoring 1. The sequence is separated into pairs of double size numbers:

(5, 10)

(30, 60)

(300, 600)

(1800, 3600)

(18000, 36000)

pairs are separated by a multiplication of the preceding number:

1 * 5 = 5

10 * 3 = 30

60 * 5 = 300

600 * 3 = 1800

3600 * 5 = 18000

I'm almost positive there needs to be some tricky modulus going on, but can't seem to crack it.

The very difficult problem is given only an index "7", and the first number real number "5". How do you get 1800.

Your sequence increases by multiples of the repeating sequence [5, 2, 3, 2], so for each multiple of 4 indices, the sequence increases by a factor 60. The remainder of a multiple by 4 can easily be applied afterwards.

Two hints:

1. (index-1)/4 gives you the number of times to multiply by 60.
2. (index-1)%4 gives you the number of factors to multiply as a fraction of 60; for no remainder you multiply by 1 since all factors are covered by a whole number of 60's, for one remainder you multiply by 1*5=5, for two remainders you multiply by 1*5*2=10, and so on for the last factor which I leave up to you to understand.