[seqfan] Re: Puzzle
M. F. Hasler
oeis at hasler.fr
Wed Jan 21 03:48:50 CET 2015
Obviously, for r=0 one reaches 1 after the first step.
If r>0, you can reach 1 only via +r, so you must be able to reach 1-r.
For rational r=m/n, you must thus be able to reach (n-m)/n, or equivalently
(n-km)/n for any integer k>0, but negative numbers and zero are never
reached so 1 <= k < n/m so m < n.
I.e., r < 1, which includes Franklin's result that r cannot be integer
(except 0).
But I do not agree with his reasoning that r cannot be irrational. (Because
the integer coefficients in the rational fraction can be zero.)
For example, r = sqrt(2)/2 seems to work:
+ ÷ + + + + ÷ +
should yield 1 (where ÷ is reciprocal).
Maximilian
Le 20 janv. 2015 20:05, "David Wilson" <davidwwilson at comcast.net> a écrit :
> Choose any real r >= 0.
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> Starting with n = 1, on the first step add r, on subsequent steps either
> add
> r or take the reciprocal as you choose.
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> For example, if r = 1/4, we can generate the sequence
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> 1, 5/4, 3/2, 7/4, 2, 1/2, 3/4, 1.
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> For which r is it possible to return to 1 as does this sequence?
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