[seqfan] Re: Puzzle

[hv at crypt.org]

When n == 1 (mod m), m/n solves [ n(n-1)/m, (n-1)/m ].
3/(6n+2) solves [ n(6n+2), 2n ].
3/(6n+5) solves [ (4n+3)(6n+5), (n+1)(6n+4), 2n+1 ].

That's enough to cover all 1/n, 2/n, 3/n.

I suspect that all rationals r < 1 are solvable, and that it should be
provable. But I've had no luck with 7/9 yet, either.

Hugo

Frank Adams-Watters <franktaw at netscape.net> wrote:
:1/(N-1) - 1/(N*(N-1)) = 1/N, so take a = N*(N-1), b=N-1.
:
:Franklin T. Adams-Watters
:
:-----Original Message-----
:From: M. F. Hasler <oeis at hasler.fr>
:To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
:Sent: Tue, Jan 20, 2015 9:38 pm
:Subject: [seqfan] Re: Puzzle
:
:
:Hugo makes a good point :
:the sequence of operations can be coded as a list of integers (m(k), 
:k=1..n),
:which mean "add m(k) times r, then take the reciprocal"
:(the last reciprocal being "useless" if it ends in 1).
:
:e.g., for r=1/sqrt(2), one has that operations [1,4,1]
: yield the values [ 1/(1+r)=2-2r ; 1/(2+2r) = 1-r ; 1 ]
:
:for r=1/sqrt(3), one has that [1,3,1] yields the
: values [ 1/(1+r)=3/2-r*3/2 ; 1/(3/2+r*3/2) = 1-r ; 1 ]
:
:E.g., for r = 1/N, one has that [N(N-1), N-1] yields
: values  [ 1/(1+N-1) = 1/N ; 1/(1/N + (N-1)/N) = 1 ]
:
:(Not any 1/N is of the form (a-b)/ab, though, is it?)
:Maximilian
:
:
:> -----Original Message-----
:> From: hv <hv at crypt.org>
:> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
:> Sent: Tue, Jan 20, 2015 8:24 pm
:> Subject: [seqfan] Re: Puzzle
:>
:>
:> It is never useful to use the reciprocal twice in a row, nor to start 
:or
:> end the sequence with it, so we're talking about a sequence [a, b, c, 
:... ]
:> of additions, with a single reciprocal between each set of additions.
:>
:> When the sequence is length 1 (ie with no reciprocals), the only 
:solution
:> is r = 0; when the sequence is length 2, we have r = 0 or r = (a - b) 
:/ ab,
:> with a and b any positive integers.
:>
:> At length 3, we get an ugly quadratic, which simplifies to the same 
:case
:> when a + c = b: abcr^2 + (ab - bc)r + (a + c - b) = 0, if I have it 
:right.
:> I guess it'll only get uglier for longer sequences.
:>
:> The rationals satisfying (a - b) / ab with a, b in Z+ seem like quite
:> an interesting subset of Q to characterize, though.
:>
:> I'm not sure "no irrational" is correct though, I think for example
:> that 1/sqrt(3) works via the sequence [add, rec, add, add, add, rec, 
:add].
:>
:> Hugo
:
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