A079279--a humble admission

Wed Mar 26 15:21:23 CET 2003

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  
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