inconsummate numbers
John Conway
conway at Math.Princeton.EDU
Mon Jan 8 21:25:48 CET 2001
Neil's message reminds me that the first message I sent out
on these did not go to seqfan, so I'll briefly sketch the notation
and previous results.
I start from my old theorem that no number is 62 times the sum
of its decimal digits, which is now expressed by saying that "62
is inconsummate in base 10". In fact 62 is the least inconsummate
number in base 10, and this message concerns the problem of
determining the least inconsummate numbers in arbitrary bases.
I write a|b|c|...|y|z for
a.B^n + b.B^(n-1) + ... + y.B + z.
Results so far: every 1-digit number (in a given base B) is
consummate (in that base), and the only possibilities for 2-digit
inconsummates are the numbers of the form a|b with a > B/2
and 1 =< a =< b.
Odd and even bases behave differently, and I've only been
pursuing even bases so far. Also, since bases 2 and 4 behave
specially, I shall suppose B = 2j >= 6. Then it seems that
the least inconsummate number has the form j+1|s , where s
is small compared to the base, and takes the values
3 2 2 2 2 3 2 2 2 2 3 2 2 2 2 4 2 2 2 2 3 for
bases 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 ... .
More precisely, I have the first few of a chain of theorems:
Thm. We have s >= 2, and s = 2 unless B == 1 (mod 5).
Thm. If B == 1 (mod 5), then s >= 3, and s = 3 unless B == 1 (mod 7).
Thm. If B == 1 (modd 5,7), then s >= 4, and s = 4 unless B == 1 (3).
Thm. If B == 1 (modd 3,5,7), then s >= 5, and s = 5 unless B == 1 (11).
Thm. If B == 1 (3,5,7,11), then s >= 6, and s = 6 unless B == 1 (13).
Thm. If B == 1 (3,5,7,11,13), then s >= 8.
(s = 7 doesn't happen, because we already know that B == 1 (15).)
Now I don't know how long this chain continues, but I conjecture that
it is finite - in other words, that s is bounded - and if so, we'll
have a theorem giving s for all even bases (as an ultimately periodic
function of the base). I rather suspect that the odd bases will behave
in a similar way, but want to finish off the even ones before tackling
the odds.
The proofs of the above theorems are quite elegant, and I'll describe
them in my next.
Bye for now,
John Conway
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