hv at crypt.org hv at crypt.org
Wed Mar 3 13:31:23 CET 2004

Ralf Stephan <ralf at ark.in-berlin.de> wrote:
:Can you prove or disprove that A073216 = A023745 + 1?
:%S A023745 0,1,2,4,5,8,13,14,17,26,40,41,44,53,80,121,122,125,134,161,242,364,
:%N A023745 Plaindromes: digits in base 3 are in ascending order.
:%S A073216 1,2,3,5,6,9,14,15,18,27,41,42,45,54,81,122,123,126,135,162,243,365,
:%N A073216 The terms of A055235 (sums of two powers of 3) divided by 2.
:I must have myopia but I will write the database entry for you if
:you show the beef.

Yes. Each entry in A023745 by definition is, in base 3, a series of
m '1' digits followed by n '2' digits, with m, n >= 0. Let's write
that as 1{m}2{n}.

Add one to that, and you get 1{m-1}20{n} (if m > 0) or 10{n} (if m = 0).
Double these, and you get: 10{m-1}10{n} (if m > 0) or 20{n} (if m = 0).
These two forms exactly describe the forms of sums of two powers of 3,
the two powers being 3^n and 3^(m+n).

Hugo van der Sanden

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