[seqfan] Re: Puzzle

[Frank Adams-Watters]

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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