inconsummate numbers

N. J. A. Sloane njas at research.att.com
Wed Jan 17 16:18:09 CET 2001


The following is a message from John Conway about
inconsummate numbers, together with my reply.



Subject: Least inconsummate numbers.
From: John Conway <conway at Math.Princeton.EDU>
Date: Wed, 17 Jan 2001 09:29:57 -0500 (EST)

   Neil - I finally found an explicit rule for the least
inconsummate number m(B) to any base B, but it took one hell
of a long time.  Over the weekend I went to the trouble of
reconstructing the entire proof, but haven't yet had the time
to type it into the machine (it's not terribly long, but will
be hard to type because there are lots of detailed formulae
to get right).

   Here's the answer.   m(2) = 13?, m(3) = 19?, m(4) = 26?
are what they are - I think the values are listed, but have had
trouble remembering these before.

[This is A052491 - njas ]

                                  Thereafter, m(B) is a
two-digit number  k|s  in base  B,  whose first digit  k  is
the smallest number strictly greater than  B/2, and whose
second digit  s  is

    1) for odd bases, alternatingly 1 and 2
          (same parity as k).

    2) for even bases, the first number in the sequence
        2,3,4,5,...   for which  2s+1  is prime to B-1.

   So for instance, in base  B = 106, the first digit is 54,
   and the second digit 5, since  5,7,9 aren't prime to 105,
   but 11 is.


   This theorem gives the first proof that the sequence of
inconsummate numbers to any base exists, since it proves
first that there is one, and then the infinitude of the
sequence follows since  Bm  is inconsummate iff  m  is.

   However, I suspect there may be a largest inconsummate
number that's NOT a multiple of the base.  It might be very
large indeed, which would lead to very interesting sequences!

   Here's a sequence that's interesting in any case - the
list of numbers that are inconsummate to some base.  I haven't
worked it out, but it's easy to do so, since the abave theorem
gives a bound on the base.  It starts  13,... .

   JHC

My reply:

Here's one of the sequences in question:

%I A052491
%S A052491 13,17,29,16,27,30,42,46,62,68,86,92,114,122,147,154,182,192,222,232,
%T A052491 266,278,314,326,367,380,422,436,482,498,546,562,614,632,688,704,762,
%U A052491 782,842,862,926,948,1014,1036,1107,1130,1202,1226,1302,1328,1406,1432
%N A052491 Smallest "inconsummate number" in base n: smallest number such that in base n, no number is this multiple of the sum of its digits.
%O A052491 2,1
%H A052491 <a href="http://gotmath.com/inconsummate">David Radcliffe, Inconsummate Numbers</a>
%K A052491 nonn,nice,easy,base
%Y A052491 Cf. A003635, A058898-A058907.
%e A052491 a(10) = 62, from A003635.
%A A052491 John Conway (conway at Math.Princeton.EDU), Dec 30 2000
%E A052491 Corrected and extended by David Radcliffe (radcliff at uwm.edu), Jan 08 2001. More terms from David Wilson (wilson at aprisma.com), Jan 10 2001

Others are:

%N A058898 Inconsummate numbers in base 2: no number is this multiple of the sum of its digits (in base 2).
%N A058899 Inconsummate numbers in base 3: no number is this multiple of the sum of its digits (in base 3).
%N A058900 Inconsummate numbers in base 4: no number is this multiple of the sum of its digits (in base 4).
%N A058901 Inconsummate numbers in base 5: no number is this multiple of the sum of its digits (in base 5).
%N A058902 Inconsummate numbers in base 6: no number is this multiple of the sum of its digits (in base 6).
%N A058903 Inconsummate numbers in base 7: no number is this multiple of the sum of its digits (in base 7).
%N A058904 Inconsummate numbers in base 8: no number is this multiple of the sum of its digits (in base 8).
%N A058905 Inconsummate numbers in base 9: no number is this multiple of the sum of its digits (in base 9).
%N A003635 Inconsummate numbers in base 10: no number is this multiple of the sum of its digits (in base 10).
%N A058906 Inconsummate numbers in base 11: no number is this multiple of the sum of its digits (in base 11).
%N A058907 Inconsummate numbers in base 12: no number is this multiple of the sum of its digits (in base 12).

Neil






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