A079279--a humble admission

[Matthew Vandermast]

Dear Professor Guy,

Do you mean "for this sequence," or "for future generalizations of the
sequence"?  If you mean "for this sequence," I believe not, because the
maximum distance between the nth term and the (n + 3)rd term is 10.
Therefore, a number will never be able to join the sequence because of its
being a multiple of 11, or a higher prime; no number that is a candidate for
membership in the sequence will ever share that factor (or higher) in common
with the 3 numbers that will determine its membership or
non-membership--namely, the 3 most recent members of the sequence.  It all
comes down to the presence or absence of the numbers 2, 3, 5, and 7 in the
number's prime factorization; in any case, this is what I have aimed to
prove (although I would have liked to be able to make this more explicit in
my original message; thanks for giving me a chance to do so).

My original conclusion was wrong because of the distance of 10 between the
16 mod 30 terms and (three terms down the line, until the number 203 joins
the sequence) the 26 mod 30 terms.  Because this distance is greater than 7,
the first 16 mod 30 term that is a multiple of 7 (i.e., 196) draws a 23 mod
30 term (i.e., 203) into the sequence (203 is also coprime to 201 and 202,
the other two key terms, which also enables it to join the sequence).  I
should have seen this possibility, of course!  But this can never happen
with primes greater than 10, or at least I think I've proven this.  (I am
far from being a trained mathematician, and it may still be that I have
skipped steps I should have made more explicit, or maybe there is just a gap
in my reasoning somewhere; I've tried to prevent this, though.)

If you mean "for future generalizations of this sequence":   maybe, maybe
not.  I've looked at the sequence that starts 1,2,3,4, and continues by
always including the next integer a)larger than any that has yet appeared
and b)that shares factors with some but not all of the previous 4 terms.
This also appears to depend on the modulus 210, for the same reasons, and
also apparently includes all the numbers that are in this sequence, plus
some others.  (I'll have to do some double-checking before I submit the
sequence, though--if it is not already in by now.)  It could be that all
such sequences depend on the modulus 210.  As for the generalization
starting with 1,2, and depending on the most recent 2 numbers, this quickly
reaches 2,4, and terminates.

Best regards,
Matthew


----- Original Message -----
From: "Richard Guy" <rkg at cpsc.ucalgary.ca>
To: "Matthew Vandermast" <ghodges14 at msn.com>
Cc: <seqfan at ext.jussieu.fr>
Sent: Wednesday, March 26, 2003 11:43 AM
Subject: Re: A079279--a humble admission


Straight off the top of my head, with no logical
justification:  will you later have to consider
mod 2310, then mod 30030, ...  ???   R.

On Wed, 26 Mar 2003, Matthew Vandermast wrote:

> Hello SeqFans,
>
> I conjectured about A079279 that it contains all 3 mod 6 and 4 mod 6
numbers, all 2 mod 6 numbers that are not 20 mod 30, and no other numbers
except 1.  (The sequence starts with 1, 2, and 3, and from then on the next
member is the smallest integer a) greater than any that have yet appeared,
and b) such that it shares factors with some, but not all, of the previous
three members; it begins 1, 2, 3, 4, 8, 9, 10, 14, 15, 16 ..)  Both Jon
Perry and I thought we had proved this.
> Unfortunately, as I've written to Jon already, the original conjecture was
incorrect, and so were our "proofs"!
>
> My conjecture is correct as written until the phrase "and no other numbers
except 1."  Here's where it goes wrong:
>
> 94th member:  196 (4 mod 6)
> 95th member:  201 (3 mod 6)
> 96th member:  202 (4 mod 6)
> 97TH MEMBER: 203 (5 mod 6; shares factors with 196 (7), but not 201 or
202)
> 98TH MEMBER: 204 (0 mod 6; shares factors with 201 and 202, but not 203)
>
> Then the sequence continues according to the original conjecture until 413
and 414, and in general for all numbers 203 or 204 mod 210 (which are all
members).
>
> I believe I now have a proof (submitted for you to check, if you like)
that the sequence contains all 3 mod 6 and 4 mod 6 numbers, all 2 mod 6
numbers that are not 20 mod 30, all 203 mod 210 and 204 mod 210 numbers, and
no other numbers except 1--in other words, that the residues mod 210 that
always indicate a member of the sequence are:
>
> 2,3,4,8,9,10,14,15,16,21,22,26,27,28,
> 32,33,34,38,39,40,44,45,46,51,52,56,57,58,
> 62,63,64,68,69,70,74,75,76,81,82,86,87,88,
> 92,93,94,98,99,100,104,105,106,111,112,116,117,118,
> 122,123,124,128,129,130,134,135,136,141,142,146,147,148,
> 152,153,154,158,159,160,164,165,166,171,172,176,177,178,
> 182,183,184,188,189,190,194,195,196,201,202,203,204,206,207,208,
>
> and there are no other members except 1.  (Unfortunately, when I submitted
the sequence I worked it out only through 135 or so; otherwise I would have
caught my mistake.)
>
> Why does the cycle repeat mod 210?  Because, as you can verify, the
maximum distance between a(n ) and a(n +3), through 210, is 10 (between 16
and 26, 46 and 56, etc.)  So it never matters to a number's membership in
the sequence, at least through the number 210, whether or not it is a
multiple of 11--by the time a number might qualify for membership by virtue
of sharing the factor 11 with one of the previous 3 numbers, the vacancy
(the next place in the sequence) has already been filled.  It does matter
whether or not it is a multiple of 2, 3, 5 or 7; and the product of those
numbers is 210.  (Could the cycle repeat in accordance with a smaller
divisor of 210 as a modulus?  No.  The aliquot divisors of 210 are 1, 2, 3,
5, 6, 7, 10, 14, 15, 21, 30, 35, 42, 70, and 105.  All of the divisors
greater than 1, except 30 and 42, have both multiple and non-multiples in
the sequence, so they are too narrow as criteria.  203 is the first 23 mod
30 number in the sequence, and also the first 35 mod 42 number.  So 210 is
necessary.)
>
> So since 210 is 2*3*5*7, the product of all the factors that matter (up to
210), the kind of conclusion that I asserted (wrongly) about mod 30 in my
comment to the database in February, does work for mod 210, with the
appropriate changes--namely, that the next-largest number that shares
factors with some but not all of consecutive 2, 3, and 4 mod 210 numbers is
always the next 8 mod 210 number; the next-largest satisfying this condition
with respect to a 3, 4, and 8 mod 210 number-cluster is always the next 9
mod 210 number; . . . ; the next-largest with this relation to a 196, 201,
and 202 mod 210 number-cluster is always the next 203 mod 210 number; and so
on until the cycle comes around again to the beginning.
>
> There are 100 (instead of 98) different residues of 210 that always
indicate a member of the sequence, and A079279 (100m +n) = 210m + A079279
(n ) for n >1.
>
> I will be grateful to anyone who checks this, and apologize for my earlier
error.
>
> Yours sincerely,
> Matthew Vandermast
> ghodges14 at msn.com
>
> P.S.  My server appears to be having problems right now, so I may not be
able to reply to you right away.  Sorry for any inconvenience.--mjv
>