[seqfan] Re: Probability of identical sequences
pemd70 at yahoo.com
Tue Jan 14 05:08:33 CET 2020
Thank you for your response. The two sequences are identical up to k terms. Finding terms for n>k is beyond our computing limits for one or both sequences.
Let me put in another way.
Let's say A1 and B1 have 50 identical terms, and A2 and B2 have 100 identical terms. Is the probability of A1=B1 equal to the probability of A2=B2?
On Monday, January 13, 2020, 10:48:15 PM EST, Sean A. Irvine <sairvin at gmail.com> wrote:
I'm not sure the question makes sense. If the two (or more) definitions yield the same sequence, are the definitions really different? Just because you are unable to prove the definitions are equivalent, it does not mean that they aren't. Alternatively, are they are cases where the definitions are provably different, but the sequences they generate the same?
On Tue, 14 Jan 2020 at 16:25, Ali Sada via SeqFan <seqfan at list.seqfan.eu> wrote:
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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