# [seqfan] Re: Two "dumb" sequences and a question

Nollaig MacKenzie gnaillo at hushmail.com
Wed Nov 30 20:37:01 CET 2022

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I have always liked James Thomson’s treatment of the reflexive

(The Barber, Russell, Grelling). Imagine a set of villagers, and a
boolean matrix M, where a 1 at Mij means “villager i shaves
villager j”.
Thomson used an obvious point that I liked to call “Thomson’s
Truism”:
no row or column of a boolean matrix can be identical to the negation

of the leading diagonal. Whence: no villager shaves all and only
those
villagers who do not shave themselves.
To distort Ali’s statement slightly, suppose a 1 at Mij means “i
.epsilon. A(j)”,
where A(j) is the OEIS sequence with a-number j. Leaving out the
reference
to knowledge, Ali’s AY would have to have an entry corresponding to
the negation
of the leading diagonal of M. So AY cannot be an OEIS sequence.   p {
line-height: 115%; margin-bottom: 0.25cm; background: transparent }

Cheers, N.

On 2022-11-30 at 5:31 AM, "Ali Sada via SeqFan"  wrote:Hi everyone,

Please consider the two sequences below:

1) Sequence AX contains all OEIS sequences where the A number is a
term in the sequence itself. For example, A000027 since 27 is a
positive integer.

2) Sequence AY contains all OEIS sequences where either:
a) the A number is not a term in the sequence (e.g., A000040, since 40
is not a prime number),
or
b) we don’t know if the A number is a term in the sequence or not
(e.g., A329697).

The question here is: Where should the number Y go? If we put it in
sequence AY, then we know where it belongs and that contradicts the
definition of AY.
Also, it couldn’t be part of AX because Y is not a term of AY.

I’m trying to have some basic understanding of set theory and I

Best,

Ali
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