[seqfan] Re: Is the definition of this sequence correct?
M. F. Hasler
oeis at hasler.fr
Tue Jul 5 04:10:56 CEST 2022
On Mon, Jul 4, 2022, 21:02 Ali Sada wrote:
> I have one question. The ratio of even to odd numbers is 5:2 when n is a
> multiple of 7, up to and including 56. The terms are multiples of 10. But
> that changes at n = 70 (a(n) =100). Why?
We do not expect a ratio of 5/2 = 2.5 but rather of 1+√2 ≈ 2.4:
The n-th odd number is at index (2+sqrt(2))n, and the n-th even number is
at index sqrt(2)n, approximately.
So, up to a given (large enough) index, there should be 1/sqrt(2) ≈ 70%
even numbers and 1/(2+√2) ≈ 30% odd integers, leading to the above
mentioned ratio.
Since this ratio is slightly less than 5/2, it is normal that from time to
time the multiples of 10 are spaced out by 1 more odd term compared to the
most common configuration of 2 odd + 4 even = 6 intermediate terms.
- Maximilian
On Monday, July 4, 2022 at 03:55:48 PM GMT+1, M. F. Hasler wrote:
>
> Oh sorry, the " > " in my final version is wrong again, I mixed up
> reasoning "in the other direction"
> (minimum distance remains the smaller of the two values).
> I think it should be OK with ">=".
> As a redemption, I provide a PARI program:
>
> {A=U=[1]; for(n=1, 99, for(k=U[1]+1,oo,
> (setsearch(U,k) || setsearch(Set(A[max(-n,2-k)..-1]),k-1) ||
> setsearch(Set(A[max(-n,1-k)..-1]), k+1)) && next;
> if(k>U[1]+1, U=setunion(U,[k]), U[1]+=1; while(#U>1&& U[1]+1==U[2],
> U=U[^1])); A=concat(A,k); break));A}
>
> => 1, 2, 4, 6, 8, 3, 10,
12, 5, 14, 16, 7, 18, 20,
22, 9, 24, 26, 11, 28, 30,
32, 13, 34, 36, 15, 38, 40,
42, 17, 44, 46, 19, 48, 50,
21, 52, 54, 56, 23, 58, 60,
25, 62, 64, 66, 27, 68, 70,
29, 72, 74, 31, 76, 78, 80,
33, 82, 84, 35, 86, 88, 90,
37, 92, 94, 39, 96, 98, 41, 100,
102, 104, 43, 106, 108, 45, 110,
112, 114, 47, 116, 118, 49, 120,
122, 124, 51, 126, 128, 53, 130,
132, 55, 134, 136, 138, 57, 140, 142, ...
>
> At that point, all numbers up to 57, and all odd numbers up to 141, are
> used.
>
> - Maximilian
>
>
> On Mon, Jul 4, 2022 at 7:31 AM M. F. Hasler <oeis at hasler.fr> wrote:
>
> > Dear Ali et al.,
> > Now the example is clear and also "the other direction" had been
> clarified
> > in the second proposal of name.
> > This clarification confirms that you can/must actually require a minimum
> > distance of a(n)+1 (because of the "other direction" you clarified using
> > the index m2).
> >
> > Then I would suggest:
> >
> > a(n) is the least positive integer not already in the sequence such that,
> > if a(m) = a(n)+1, then |m - n| > a(n).
> >
> > ( a(1)=1 follows from the definition.
> > If we feel an urge to specify it, I'd suggest to put this to the end so
> > that the main idea comes as early as possible in the "name", considering
> > esp. the truncation of names in the "pop-up titles".)
> >
> > Or maybe better: (all numbers are "already in the sequence"...)
> >
> > Lexicographically first permutation of the positive integers such that,
> if
> > a(m) = a(n)+1, then |m - n| > a(n), for all indices m and n.
> >
> > (We can use "permutation" as shortcut for "...not occurring earlier"
> > because any integer will indeed occur, as soon a possible: the condition
> on
> > the distance can only "delay" its occurrence a little bit.)
> >
> > - Maximilian
> >
> > On Sun, Jul 3, 2022, 21:43 Ali Sada via SeqFan <seqfan at list.seqfan.eu>
> > wrote:
> >
> >>
> >> Thank you, Tom. I really appreciate your response. It will take me some
> >> time to write a VBA program to find the terms. Manually, I will make
> >> mistakes!
> >>
> >> Would any of these two versions work?
> >>
> >> a(1)=1; a(n) is the least positive integer not already in the sequence
> >> such that the absolute distance between a(n) and a(n)+1 is >= a(n).
> >>
> >> or
> >>
> >> a(1)=1; a(n) is the least positive integer not already in the sequence
> >> such that |n-m1|>= a(n) and |n-m2|>= a(n)-1, where m1 and m2 are the
> >> indices of a(n)+1 and a(n)-1 respectively.
> >>
> >> Best,
> >>
> >> Ali
> >>
> >>
> >>
> >>
> >> On Sunday, July 3, 2022 at 10:45:47 PM GMT+1, Tom Duff <
> >> eigenvectors at gmail.com> wrote:
> >>
> >> No, sorry, my definition is bogus. The sequence is more complicated
> than
> >> I
> >> made it out to be.
> >>
> >> On Sun, Jul 3, 2022 at 15:01 Tom Duff <eigenvectors at gmail.com> wrote:
> >>
> >> > I think this should read:
> >> > a(1)=1; a(n+1) is the smallest positive integer, distinct from all
> a(m),
> >> > m<=n, with |a(n+1)-a(n)|>=a(n).
> >> >
> >> > Sequences, not their entries, are “lexicographically earliest”. The
> way
> >> a
> >> > sequence gets to be lexicographically earliest is by picking the
> >> smallest
> >> > eligible entry at each step. “Distance … in both directions” is best
> >> > expressed by explicitly saying that it’s the absolute difference.
> >> > All that said, I’m surprised that this sequence is not already in the
> >> > OEIS. Compute a bunch of terms (it’s easy, you shouldn’t need help)
> and
> >> > search for it. If it’s not there, add it.
> >> >
> >> > On Sun, Jul 3, 2022 at 04:18 Ali Sada via SeqFan <
> seqfan at list.seqfan.eu
> >> >
> >> > wrote:
> >> >
> >> >> Hi everyone,
> >> >>
> >> >> Please check this definition
> >> >>
> >> >> a(1) =1; a(n) is the lexicographically earliest positive integer such
> >> >> that the distance between a(n) and a(n)+1 is >= a(n) in both
> >> directions.
> >> >> (The distance between a(n) and a(m) is |n-m|)
> >> >>
> >> >> a(1) = 1
> >> >> a(2) = 2
> >> >> Now, a(3) cannot be 3, so a(3) = 4.
> >> >> a(4) cannot be 3 nor 5, so a(4) = 6.
> >> >> a(5) cannot be 3 nor 5 nor 7, so a(5) = 8.
> >> >> Now, we can use 3 for a(6) (the distance with 4 is 3).
> >> >> And so on.
> >> >>
> >> >> I would appreciate your help with the correct definition and terms.
> >> >>
> >> >> Best,
> >> >>
> >> >> Ali
> >> >>
> >> >> --
> >> >> Seqfan Mailing list - http://list.seqfan.eu/
> >> >>
> >> >
> >>
> >> --
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> >>
> >>
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> >>
> >
>
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