[EMAIL PROTECTED] wrote:
> This is totally weird, and I think I messed up in getting this, but it is
> true. Could someone with a lot of time and a lot of patience and a big
> calculator check this? Thanks.
>
> 1 = 1/(2�- 1) + 1/(2� - 1) + 1/(2^4 - 1) + 1/(2^5 - 1) + ... + 1/(3� - 1) +
> 1/(3� - 1) + 1/(3^4 - 1) + ... + 1/(5� - 1) + 1/(5� - 1) + 1/(5^4 - 1) + ...
>
> In other words, if set x contains integral powers, (2�, 3�, 2�, etc....) and
> x' is the set of integers minus set x. Then this says:
>
> 1 = sum(n=2 to n=infinity (sum( i=1 to i = infinity, 1/((x'i)^n - 1))))
>
> please confirm!!
>
> this is a side formula I found, while working with primes. Hope it is not to
> off topic.
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The above mentioned question is neither weird nor off topic on account of the
following:
1) The general task (including yours) is to have the unity, i.e. the number "1"
as the sum of reciprocals of natural numbers.
2) In case the number of the summands is finite, the denominators are divisors
of a perfect number, - the LCM of them.- ( See: Daniel Shanks: Solved and
unsolved
problems in Number Theory, Vol.I.Spartan books, Washington DC.1962; page 25.)
Easy to prove:
the existance of such a finite summation is equivalent to the existance of
perfect numbers.
3) So the existance (or non-existance) of producing the unity as the sum
of a finite number of reciprocals of odd numbers is equivalent with the
existance (or non-existance)
of odd perfect numbers.
4) Only a comment: The Egyptians also used the reciprocals in their
arithmetics, but none of
the remained sources contained summands having only reciprocals of odd numbers.
Every one of them has at least one reciprocal of even numbers.
Best regards
Pa'l La'ng
Budapest, Hungary
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