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Is it possible to add a number of inverses of odd positive integers and form an even number?(You cannot take 1 in the list)

Form is
(1/a)+(1/b)+...=2n

where a,b,... are odd positive integers.

If not does there exist a proof for it??

EDIT: the odd integers must be distinct.
Last edited by Afif_D on Tue Feb 21, 2012 5:03 am UTC, edited 1 time in total.

Afif_D

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You probably meant to say that a,b,... must be distinct. If not, then consider

(1/3) + (1/3) + (1/3) + (1/5) + (1/5) + (1/5) + (1/5) + (1/5) = 2

EDIT: If a,b,... are required to be to distinct, I don't know the answer, but here's a possibly relevant fact:

(1/3) + (1/5) + (1/7) + (1/9) + ... + (1/111) < 2

(1/3) + (1/5) + (1/7) + (1/9) + ... + (1/111) + (1/113) > 2
skullturf

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I don't have an explicit example, but yes, this can be done. See this wiki page, and take x = 2, y = 1.
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NathanielJ

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I'm stupid. Or at least, I was at 7 in the morning.

I made things way harder for myself by starting with 1/3 rather than 1/1.

Note that

1/1 + 1/3 + 1/5 + 1/7 + 1/9 + 1/11 + 1/13 < 2

1/1 + 1/3 + 1/5 + 1/7 + 1/9 + 1/11 + 1/13 + 1/15 > 2

If I try the greedy strategy mentioned in the wiki page, I seem to get something that starts with

1/1 + 1/3 + 1/5 + 1/7 + 1/9 + 1/11 + 1/13 + 1/23 + 1/721 + 1/979007 + 1/661211444787

and I haven't finished finding something that sums to exactly 2. I may need pretty high precision arithmetic to complete this.

Perhaps a non-greedy approach can find a sum with a smaller number of terms?

This seems like a surprisingly tricky problem! I think I like it!

EDIT: Rather than reinventing the wheel, I managed to find an explicit answer online:

https://oeis.org/search?q=13%2C23%2C721 ... &go=Search

EDIT #2: More examples with smaller denominators can be found here:

http://www.ics.uci.edu/~eppstein/numth/ ... d-one.html
skullturf

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Ya i meant to say they must be distinct.
And you cant include 1. Thanx for the help anyway . I have been trying to work out a proof for this but nothing has come up so far.

And i guess that it is not possible to add them up to 2 exactly.

Afif_D

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I mean, sure, the usual odd greedy expansion for 2 doesn't work under your stipulation, because it has a 1 in it. So just take the fractions up to 111, subtract that from 2, and find the odd greedy expansion for that.
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Mo' Money

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I took some ideas from that other thread, and 2 can be expressed as the sum of 1/n for all n>1 dividing 1310112879075 except 2175, 176715, 876329685, 56961429525, and 436704293025. The entire expansion has 2298 terms in it, so writing it out in full would not be the best idea.
Nitrodon

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The odd greedy expansion does not necessarily work, the expansion could be infinite.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.
tomtom2357

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tomtom2357 wrote:The odd greedy expansion does not necessarily work, the expansion could be infinite.

Its true that no one has proven that it works, but there aren't any examples which don't provide a finite expansion either. Given that, its reasonable to expect it to work out okay with 2 (or 3 if we allow a 1).
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jestingrabbit

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(EDIT: fixed a typo)

I spoke too soon when I said at 5 pm yesterday that I was stupid at 7 am. At 7 am, I was smart enough to notice that the original question said we can't use 1 as a denominator. Then, by 5 pm, I had forgotten that.

Nitrodon wrote:I took some ideas from that other thread, and 2 can be expressed as the sum of 1/n for all n>1 dividing 1310112879075 except ...(some of them)

Interesting. Is 1310112879075 (which factors as 3^5 * 5^2 * 7 * 11 * 13 * 17 * 19 * 23 * 29) the smallest odd number n such that the sum of its proper divisors is at least 2n? (For that n, we have (sum of proper divisors of n) = approx 2.000465430n.)

Some simple-minded searching seems to indicate there are no such odd n below 2 million. But of course, 1310112879075 is a lot bigger.

And when I do a Google search on 1310112879075, I get almost no hits, which is a little surprising to me if that is indeed the smallest answer to my question.

But now, after doing some fooling around, I suspect that it probably is. It's possible to show that any such n must be divisible by each of the primes from 3 to 23. And the integer 3*5*7*11*13*17*19*23 doesn't work, so then you can experiment with multiplying by 29 versus multiplying by 3^3 or 5^2, and things like that. I don't have a proof, but I suspect 1310112879075 is the correct answer to the above question.

EDITED APPROXIMATELY TWO HOURS LATER: In fact, I was wrong. The smallest such odd integer is

1018976683725 = 3^3 * 5^2 * 7^2 * 11 * 13 * 17 * 19 * 23 * 29

All I had to do to prove this was loop through odd multiples of 3*5*7*11*13*17*19*23.
Last edited by skullturf on Tue Feb 21, 2012 6:05 pm UTC, edited 1 time in total.
skullturf

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I would even expect that it is possible to get 1=1/(n+k1) + 1/(n+k2)... with even ki for every odd n and a finite number of terms. In other words: You can get 1=... even if the first n-1 terms are not available for your sum.
If that is true, it is possible to get any natural number with such a sum.
mfb

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If all denominators are odd, and the denominators 1 and 3 are both forbidden, we can still make 1.

1 = 1/5 + 1/7 + 1/9 + 1/11 + 1/13 + 1/15 + 1/21 + 1/27 + 1/33 + 1/35 + 1/39 + 1/45 + 1/55 + 1/63 + 1/65 + 1/77 + 1/91 + 1/99 + 1/105 + 1/117 + 1/135 + 1/143 + 1/273 + 1/2079 + 1/135135

How did I find this? Not by a "naive" greedy strategy. Instead, by following ideas in the thread I linked to, as Nitrodon did.

The above equation, multiplied by 135135, says that the number 135135 is equal to the sum of some its proper divisors *smaller* than 135135/3.

Why did I choose n=135135? Because for that n, the sum of *all* the divisors of n is at least (7/3)n, so the sum of all divisors of n *except* n and n/3 is still at least n.

I used a greedy strategy *among* the divisors of 135135.
skullturf

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skullturf wrote:
1 = 1/5 + 1/7 + 1/9 + 1/11 + 1/13 + 1/15 + 1/21 + 1/27 + 1/33 + 1/35 + 1/39 + 1/45 + 1/55 + 1/63 + 1/65 + 1/77 + 1/91 + 1/99 + 1/105 + 1/117 + 1/135 + 1/143 + 1/273 + 1/2079 + 1/135135

Nice finding. But i still believe nobody can produce a 2.

I think we should al just try to work out a proof

Afif_D

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Afif_D wrote:But i still believe nobody can produce a 2.

Even though Nitrodon provided an explicit example that does exactly this?
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NathanielJ

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Afif_D wrote:Nice finding. But i still believe nobody can produce a 2.

I think we should al just try to work out a proof

Nitrodon has already supplied a solution, so trying to work out a proof that it cannot be done would be pointless.

Edit: Ninja'd!

jaap

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Here's an example much like Nitrodon's (with slightly smaller denominators) described in a little more detail.

Define n = 1018976683725 (which has prime factorization 3^3 5^2 7^2 11 13 17 19 23 29).

Why this n? The sum of its divisors is relatively large. (By the way, it's possible to write down an expression for the sum of divisors of a number in terms of its prime factorization.) In particular, for this n, the sum of all proper divisors of n happens to be more than 2n.

This number n has 2304 divisors including itself, so 2303 proper divisors. So I won't list them all. But the first few are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 33 (note no 31), and the largest proper divisor is n/3 = 339658894575.

As mentioned, the sum of *all* proper divisors of n happens to be *more* than 2n = 2037953367450. By trial and error, eliminating *some* proper divisors from our list, it happens to be possible to find a list of *some* proper divisors of n that add up to *exactly* 2n.

Specifically, if we list all proper divisors of n *except* for the following seven divisors, we get a list of proper divisors of n (2296 of them) that happen to add up to exactly 2n.

"Excluded" divisors: 20795442525, 77388675, 124729, 855, 35, 5, 1.

This means we have an equation of the form

2n = 2037953367450 = (a sum of 2296 divisors of n) = 3 + 7 + 9 + 11 + ... + 339658894575.

Now, divide both sides of this equation by n.

The left side is 2. The right side is a sum of 2296 different fractions. When reduced, those fractions are distinct unit fractions with odd denominators. (The original denominator was n and the numerator was a divisor of n).

As it happens, the smallest denominator on the right side is n/3 = 339658894575. The number of terms on the right side is 2296. I have no idea whether my example is "optimal" with respect to either of those measures, but I suspect that it's probably at least close.
skullturf

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Ya . Good one. Thanks. And ya i am an idiot.

Now tell me. Do there exist only finite number of solutions for this kind of problem?

Afif_D

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Afif_D wrote:Ya . Good one. Thanks. And ya i am an idiot.

Now tell me. Do there exist only finite number of solutions for this kind of problem?
No. For example, suppose r = 1/a1 + 1/a2 + ... + 1/an. Write 1/an = 1/(an+2) + 2/(an(an + 2)), then expand 2/(an(an+2)) as a sum of odd Egyptian fractions using any technique. Of course all these fractions will be smaller than 1/(an+2), so you will end up with a new set of distinct odd Egyptian fractions summing to r.
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Mo' Money

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I found a proof by Ernest Croot that for every rational r and every natural N, there is a set of natural xi with [imath]N<x_i<e^{r+o(1)}N such that r=sum(1/xi).
The xi are allowed to be even in his proof, but I don't see a reason why this should be necessary. The upper bound may change, but that is not important here. Apart from the even numbers, the result is much stronger than any other result (or question) here.

I am quite sure now that you can get every positive rational number as a sum of inverse odd numbers.
mfb

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Thanks a lot for the help.

Afif_D

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