[seqfan] Re: Probability of identical sequences
djsycamore at yahoo.co.uk
Tue Jan 14 20:59:52 CET 2020
Ali’s question is both interesting and familiar, though I agree that a discussion about probabilities of the equivalence of different names is not really helpful since the only interesting probabilities in this case are 0 and 1.
I suppose it’s not unusual for an oeis user to define a new sequence, calculate a few terms, then discover that it’s already in the data base under a different Name. The challenge then is to prove the equivalence of two different names, as has now been done in Ali’s case. I came upon the following example yesterday, before reading his post:
A025480 is a(2*n)=n, a(2*n+1)=a(n), giving:
0,0,1,0,2,1,3,0,4,2,5,1,6,3,7,0,8,4... Zeros occur throughout this sequence (a(2^k-1)=0; k>=0 ), and if these terms are all removed we find that the complimentary subsequence of non zero terms is 1,2,1,3,4,2,5,1,6,3,7,8,4... which is apparently A131987, another fractal sequence. If this is correct then A025480 contains a fractal proper subsequence other than itself, and this might be interesting enough to add as a comment in the entry. But is it really true? Up to ~ 120 terms yes it is, but that’s not the same as proving it. Whilst the definition of A025480 is straightforward, that of A131987 is perhaps less so, although the example included in the entry shows clearly how the sequence is built. Perhaps a clue lies in the positions of zeros in A035480? It would be interesting to see a proof of equivalence.
I have come across other examples of seeing apparently irreconcilable names for the same sequence, A063116 being a case in point. Perhaps there are some cases where the names just don’t ever match, and we have to settle for an uncomfortable coincidence?
Sean’s point about provably different definitions producing identical sequences is intriguing, but I know of no example.
It has been suggested that if there is a new sequence identical to an existing one but with a different name and no proof of equivalence, then the second version should be submitted and retained until a proof is available, then killed off. Is this really a solution the editors would accept?
> On 14 Jan 2020, at 04:34, Robert Dougherty-Bliss <robert.w.bliss at gmail.com> wrote:
> This question is tricky to interpret, and in general I suspect that the
> answer is "undefined." I will say that it probably has no empirically
> satisfying answer.
> Any reasonable statistical test for this question will satisfy two things:
> 1. A sufficiently long string of "matches" will raise the probability
> arbitrarily close to 1.
> 2. A "mismatch" will instantly return probability 0.
> The first point is a problem since you can superficially construct distinct
> sequences that agree for arbitrarily many terms. This problem is shared by
> all statistical tests, but the second point is really killer. The test is
> far too sensitive. Why even have a probability if the answer is basically
> "all or nothing"? What good would it do you?
> On Mon, Jan 13, 2020, 22:25 Ali Sada via SeqFan <seqfan at list.seqfan.eu>
>> Hi Everyone,
>> If we have two sequences, A and B, with different definitions. However,
>> when we calculate k terms for each sequence, all of these terms are
>> identical. If we can’t prove that definition A equals to definition B,
>> what’s the probability that A and B are identical? Is it zero, since we are
>> dealing with infinite terms? Or is it a function of k?
>> This question came to me when I was trying this algorithm: Exchange n and
>> 2n. Each term gets changed only once.
>> a(1)=2 and a(2)=1.
>> a(3)=6 and a(6)=3
>> The sequence I got was
>> 2, 1, 6, 8, 10, 3, 14, 4, 18, 5, 22, 24, 26, 7, 30, 32, 34, 9, 38, 40, 42,
>> 11, 46, 12, 50, 13
>> This sequence seems identical to A073675 (Rearrangement of natural numbers
>> such that a(n) is the smallest proper divisor of n not included earlier but
>> if no such divisor exists then a(n) is the smallest proper multiple of n
>> not included earlier, subject always to the condition that a(n) is not
>> equal to n.)
>> This example might not reflect my question above exactly. I am just trying
>> to show why I asked the question.
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