a(1)=1. a(n) = the largest integer such that the finite sequence (a(n-1),a(n-2),...a(n-a(n))) occurs somewhere as a subsequence in the finite sequence (a(1),a(2),...,a(n-1)).
Question: Is this sequence periodic after some point?
I guess, perhaps, that one of the easiest ways to prove the sequence isn't periodic, if it isn't, is to prove that it contains arbitrarily large terms.
Proving it IS periodic, if it is, could entail simply calculating enough terms so as to show that the sequence has a string of terms that repeats.
I calculated the terms above by hand, so, of course, I could have made an error; and the correct version of the sequence could already be in the EIS somewhere else. I hope not.
> a(1)=1. a(n) = the largest integer such that the finite sequence > (a(n-1),a(n-2),...a(n-a(n))) occurs somewhere as a subsequence in the > finite sequence (a(1),a(2),...,a(n-1)).
> Question: Is this sequence periodic after some point?
> I guess, perhaps, that one of the easiest ways to prove the sequence > isn't periodic, if it isn't, is to prove that it contains arbitrarily > large terms.
> Proving it IS periodic, if it is, could entail simply calculating > enough terms so as to show that the sequence has a string of terms > that repeats.
> I calculated the terms above by hand, so, of course, I could have made > an error; and the correct version of the sequence could already be in > the EIS somewhere else. > I hope not.
> Thanks, > Leroy Quet
Clearly, a(n+1) <= a(n)+2 as an occurance of a(n), a(n-1), ..., a(n-a(n +1)) within a(1), ..., a(n) implies an occurance of a(n-1), ..., a(n-a(n +1)-1) within a(1), ..., a(n-1).
This implies that you made a mistake. Indee: 1, 1, 2, 1, 3, 1, 3, 3, 2, 1, 2, 3, 5, 1, 1, 2, 2, 2, 3, 2, 3(!), 3, 3, 3, 4, 1, 1, 2, 2, 2, 3, 2, 3, 3, 3, 3, 4, 1, 1, 2, ...
Hey, that looks periodic. Assume we know already that the sequence up to a(n) repeats as above, i.e. 1,1,2,1,3,1,3,3,2,1,2,3,5 followed periodically by 1,1,2,2,2,3,2,3,3,3,3,4. Then it is easily verified that after the last known 4 comes 1 because 4,3 has not occured before; then another 1 because 1,4 has not occured; then a 2 because 1,1,4 has not occured, but 1,1 has; then a 2 because 2,1,1 has not occured, but 2,1 has; and so on until we have added another period.
hagman wrote: > On 20 Okt., 18:38, Leroy Quet <qqq...@mindspring.com> wrote: > > Define the sequence {a(k)} as follows:
> > a(1)=1. a(n) = the largest integer such that the finite sequence > > (a(n-1),a(n-2),...a(n-a(n))) occurs somewhere as a subsequence in the > > finite sequence (a(1),a(2),...,a(n-1)).
> > Question: Is this sequence periodic after some point?
> > I guess, perhaps, that one of the easiest ways to prove the sequence > > isn't periodic, if it isn't, is to prove that it contains arbitrarily > > large terms.
> > Proving it IS periodic, if it is, could entail simply calculating > > enough terms so as to show that the sequence has a string of terms > > that repeats.
> > I calculated the terms above by hand, so, of course, I could have made > > an error; and the correct version of the sequence could already be in > > the EIS somewhere else. > > I hope not.
> > Thanks, > > Leroy Quet
> Clearly, a(n+1) <= a(n)+2 as an occurance of a(n), a(n-1), ..., a(n-a(n > +1)) > within a(1), ..., a(n) implies an occurance of a(n-1), ..., a(n-a(n > +1)-1) > within a(1), ..., a(n-1).
> Hey, that looks periodic. > Assume we know already that the sequence up to a(n) repeats as above, > i.e. > 1,1,2,1,3,1,3,3,2,1,2,3,5 followed periodically by > 1,1,2,2,2,3,2,3,3,3,3,4. > Then it is easily verified that after the last known 4 comes 1 because > 4,3 has not occured before; > then another 1 because 1,4 has not occured; > then a 2 because 1,1,4 has not occured, but 1,1 has; > then a 2 because 2,1,1 has not occured, but 2,1 has; > and so on until we have added another period.
> hagman
Darned gummit. I guess my sequence isn't as interesting as I wished it was. I have sent a correction in to the Encyclopedia of Integer Sequences, giving credit to hagman for the correction.
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