A079279--a humble admission

[Richard Guy]

No, it was just a wild remark, with no basis
except (faulty) intuition.      R.

On Wed, 26 Mar 2003, Matthew Vandermast wrote:

> 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
> >
>