Number Riddles

Got the 1st one:

x = (2²ⁿ)(2²ⁿ⁺¹-1)

when n = 4, x = 130,816

Will continue working on the other one.

Amazing.....I can't call your answer incorrect because it faithfully reproduces the terms I gave, but it was not the series I had in mind. The answer to the first is much more straight forward.

So congratulations, you found the first of two possible (I hope) solutions to the first sequence. Can you get the other solution to that first sequence? I promise it's an interesting answer (and easier than the other solution you found).

Hi Roger,

That is interesting! Thanks for putting these on here from time to time - when it's slow, the day goes by a little faster.

I'll work on it some after I get home - leaving work now.

Mike

Did you start with 0 or 1? If so can you explain how either n=0 or or n=1 would result in x=6?

As I see it

if n = 0

• 2²ⁿ = 1
• 2²ⁿ⁺¹-1 = 1
• thus (2²ⁿ)(2²ⁿ⁺¹-1)=1

or if n=1

• 2²ⁿ = 4
• 2²ⁿ⁺¹-1 = 7
• thus (2²ⁿ)(2²ⁿ⁺¹-1) = 28

Am I missing something here? If so, please enlighten me.

"perfect" thanks for the site

To answer my own question, n is a list of prime numbers (not necessarily all prime numbers ) that happen to generate perfect numbers when used as n in the forumla(2^n-1)(2ⁿ-1)

Mikerho's formula yields "perfect" numbers, but it doesn't get all of them.

What do you mean by "all of them"?

I am not up to date on it, but as I recall the 6th or so perfect number is not in this series. I don't know if the question about odd perfect numbers has been resolved, either. (Related topic: Mersenne primes.)

Taking your clue, the fifth number of the first series is much bigger than Mikerho's.

That was sort of the right idea, but wrong in detail (> 40 years since I took a class in number theory, or read Martin Gardner's "Mathematical Games" article on perfect numbers).

Great puzzle, by the way; I'm chewing on the second sequence, and will probably continue to so until I still fail to get it....

OK - so, with the gimme's from Tornado, the next number in the first series is 33550336 - the (ascending from zero) 5th "perfect number". Interesting concept, and one I had not seen before. And to think that this concept has been around since Euclid.

You should have put one more number in the series (not that I would have "gotten it").

A couple of updates:

Yes, the first series is the series of Perfect Numbers

I was incorrect when I said the Mikerho had correctly provided an alternate series. As has been pointed out, the first term 6 isn't produced by Mikerho's series (good effort though!).

The formula 2^p−1(2^p − 1) where p=prime numbers will yield all the even perfect numbers as was pointed out earlier by another poster, an interesting formula proven by Euler.

Mikerho's answer using Tornado's suggestion is correct. The next number in the series is 33,550,336.

So the first series has been solved, but what of the second? I'll give a hint, they are related. Remember, the next one is hard (but can be figured out) and won't be straight forward, but will involve the Perfect Number series.

Good Luck!

Two little addenda to the forumla/definition -- p=1 is omitted from the series, and the definition holds true only when (2^p − 1) is prime. For instance, p = 11 does not yield a perfect number. The series you gave for the challenge comes from p=2, 3, 5, and 7. The next number in the series, 33,550,336 is from p=13.

Great challenge, by the way.

Well Done JBTardis! That is exactly correct.

Thanks for playing everyone.

Thanks Roger....nice puzzle.