puzzle
Meeussen Wouter (bkarnd)
wouter.meeussen at vandemoortele.com
Wed Mar 3 13:40:00 CET 2004
the integer digits base 3, of the first are
{{0}, {1}, {2}, {1, 1}, {1, 2}, {2, 2}, {1, 1, 1}, {1, 1, 2}, {1, 2, 2},
{2, 2, 2}, {1, 1, 1, 1}, {1, 1, 1, 2}, {1, 1, 2, 2}, {1, 2, 2, 2}, {2,
2, 2, 2}, ...
while twice the second gives:
{{2}, {1, 1}, {2, 0}, {1, 0, 1}, {1, 1, 0}, {2, 0, 0}, {1, 0, 0, 1}, {1,
0, 1, 0}, {1, 1, 0, 0},
{2, 0, 0, 0}, {1, 0, 0, 0, 1}, {1, 0, 0, 1, 0}, {1, 0, 1, 0, 0}, {1, 1,
0, 0, 0}, {2, 0, 0, 0, 0}
so the first is 1 everywhere but with a step-up to 2 along the way.
Now read it left to right :
(-1 + 3^(p))/(3 - 1) + (-1 + 3^(q))/(3 - 1)
The second is (3^u + 3^v)/2 - 1,
or zero everywhere with 'spikes' at position u and v.
FullSimplify[(3^(u) + 3^(v))/2 -
1 == (-1 + 3^(p))/(3 - 1) + (-1 + 3^(q))/(3 - 1)]
gives
3^p + 3^q == 3^u + 3^v
so p=u and q=v is a solution.
What have I dropped along the way?
Wouter Meeussen
Senior Scientist
R&D dept.
Lipids & Dough div.
Vandemoortele NV
+32 (0) 51 33 21 24 tel
+32 (0) 51 33 21 75 fax
-----Original Message-----
From: Ralf Stephan [mailto:ralf at ark.in-berlin.de]
Sent: woensdag 3 maart 2004 12:13
To: seqfan at ext.jussieu.fr
Subject: puzzle
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.
ralf
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