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<DIV>Seqfans, <BR> Here is my intuitive reasoning for
the conjecture that </DIV>
<DIV>Limit A140101(n)/A140100(n) = t = tribonacci constant. </DIV>
<DIV> </DIV>
<DIV>Since A140100 and A140101 are complements of each other,</DIV>
<DIV>and since limit A140101(n)/A140100(n) = t seems to exist, </DIV>
<DIV>one is motivated by the Beatty Theorem to write: </DIV>
<DIV> A140100(n) = [n*(1+1/t)]</DIV>
<DIV> A140101(n) = [n*(1+t)] ;</DIV>
<DIV>(though this is not true for any t, it is very nearly so). </DIV>
<DIV> </DIV>
<DIV>Likewise, since A140102 and A140103 are complements of each other,</DIV>
<DIV>
<DIV>and since limit A140103(n)/A140102(n) = s seems to exist, </DIV>
<DIV>one is motivated by the Beatty Theorem to write: </DIV>
<DIV> A140102(n) = [n*(1+1/s)]</DIV>
<DIV> A140103(n) = [n*(1+s)] ;</DIV>
<DIV>(though this is not true for any s, it is very nearly so). </DIV>
<DIV> </DIV>
<DIV>From the definition of these sequences:</DIV>
<DIV> A140102(n) = A140101(n) - A140100(n) ;</DIV>
<DIV>
<DIV> A140103(n) = A140101(n) + A140100(n). </DIV>
<DIV>So, roughly speaking, </DIV></DIV></DIV>
<DIV>
<DIV> [n*(1+1/s)] ~ [n*(1+t)] - [n*(1+1/t)] </DIV>
<DIV>
<DIV> [n*(1+s)] ~ [n*(1+t)] + [n*(1+1/t)]
;</DIV>
<DIV>or,</DIV>
<DIV>
<DIV> 1+1/s = (1+t) - (1+1/t) </DIV>
<DIV>
<DIV> 1+s = (1+t) + (1+1/t) </DIV>
<DIV>so that</DIV>
<DIV>
<DIV> 1/s = (t^2 - t - 1)/t </DIV>
<DIV>
<DIV> s = (t^2 + t + 1)/t
.</DIV></DIV></DIV></DIV></DIV>
<DIV>These equations isolate t to satisfy:</DIV>
<DIV> (t^2 + t + 1)*(t^2 - t - 1) = t^2 </DIV>
<DIV>or</DIV>
<DIV> t^4 - 2*t^2 - 2*t - 1 = (t^3 - t^2 - t - 1)*(1+t)
= 0.</DIV>
<DIV> </DIV>
<DIV>Therefore, I reason that t must be the tribonacci constant, i.e.,</DIV>
<DIV>the real-valued t that satisifies: t^3 = t^2 + t + 1.
</DIV>
<DIV> </DIV>
<DIV>This is not even close to a proof, and abuses the Beatty Theorem,</DIV>
<DIV>but it is my intuitive justification for the conjecture. </DIV>
<DIV> </DIV>
<DIV>An actual proof may be quite involved indeed, but it would be nice.</DIV>
<DIV> Paul </DIV>
<DIV> </DIV></DIV></DIV>
<DIV> </DIV>
<DIV>Least positive integers X(n) and Y(n) > X(n) chosen <BR>such that X(n)
does not appear in <BR> {X(k), 1<=k<n} or {Y(k),
1<=k<n} <BR>and Y(n)-X(n) does not appear in <BR> {Y(k)-X(k),
1<=k<n} or {Y(k)+X(k), 1<=k<n} <BR>for n>1, starting
with X(1)=1 and Y(1)=2. </DIV>
<DIV> </DIV>
<DIV> </DIV>
<DIV>On Sun, 8 Jun 2008 01:52:54 -0400 <A
href="mailto:pauldhanna@juno.com">pauldhanna@juno.com</A> writes:</DIV>
<BLOCKQUOTE dir=ltr
style="PADDING-LEFT: 10px; MARGIN-LEFT: 10px; BORDER-LEFT: #000000 2px solid">
<DIV></DIV>
<DIV>Seqfans, <BR>Consider the complementary sequences A140100
and A140101 <BR>determined by the condition that the term-by-term
differences </DIV>
<DIV>and sums must also form a pair of complementary sequences.
<BR> <BR>Can anyone find a proof or even a heuristic argument </DIV>
<DIV>to support the conjecture below that the given limits <BR>involve the
tribonacci constant? <BR> <BR>Thanks, <BR> Paul
<BR> <BR>DEFINITION of sequences. <BR>Least positive integers X(n)
and Y(n) > X(n) chosen <BR>such that X(n) does not appear in
<BR> {X(k), 1<=k<n} or {Y(k), 1<=k<n}
<BR>and Y(n)-X(n) does not appear in <BR> {Y(k)-X(k),
1<=k<n} or {Y(k)+X(k), 1<=k<n} <BR>for n>1,
starting with X(1)=1 and Y(1)=2. <BR> <BR>EXAMPLE.<BR>The first few terms
of the sequences are as follows. <BR>(x, y) (y-x, x+y)
<BR>------ ----------<BR>(1, 2) (1, 3)<BR>(3,
5) (2, 8)<BR>(4, 8) (4, 12)<BR>(6, 11) (5,
17)<BR>(7, 13) (6, 20)<BR>What are the next terms after
this?<BR>Determine the next x to be the least positive integer <BR>not used
earlier as an x or y (here, x = 9), and <BR>the next y-x value will be the
least positive integer <BR>not appearing earlier as a difference y-x or a sum
x+y <BR>(here, y-x = 7; thus, y = 16 and x+y = 25).<BR>Continuing in this way
determines the sequences. <BR> <BR>It would be nice to know
the exact behavior of these sequences. <BR>An
interesting concept to consider is the value of certain limits.
<BR> <BR>CONJECTURE: <BR>The following limits are given by: <BR>(1)
Limit X(n)/n = 1+1/t <BR>(2) Limit Y(n)/n = 1+t <BR>(3) Limit Y(n)/X(n) = t
<BR>(4) Limit [Y(n) + X(n)]/[Y(n) - X(n)] = t^2 <BR>(5) Limit [Y(n)^2 +
X(n)^2]/[Y(n)^2 - X(n)^2] = t <BR>where <BR> t = tribonacci constant
(A058265) = 1.83928675521416113255...<BR>which solves t^3 = 1 + t + t^2.
<BR> <BR>Clearly, given (1) and (2) is true, then (3) is easily deduced,
<BR>while (4) and (5) are true iff t = tribonacci constant. <BR> <BR>The
challenge is to show that (1) and (2) are true and further <BR>that the
parameter t is indeed the tribonacci constant. </DIV>
<DIV> </DIV>
<DIV>If it is easier to prove (3) over (1) and (2), that
would be fine; </DIV>
<DIV>I am mainly wanting a justification that t = tribonacci
constant. <BR> <BR>The x-values form OEIS entry A140100:
<BR>1,3,4,6,7,9,10,12,14,15,17,18,20,21,23,24,26,27,29,30,32,34,<BR>35,37,38,40,41,43,44,46,47,49,51,52,54,55,57,58,60,61,63,64,<BR>66,67,69,71,72,74,75,77,78,80,82,83,85,86,88,89,91,92,94,95,<BR>97,98,100,102,103,105,106,108,109,111,112,114,115,117,119,...<BR> <BR>The
y-values form OEIS entry A140101:
<BR>2,5,8,11,13,16,19,22,25,28,31,33,36,39,42,45,48,50,53,56,59,<BR>62,65,68,70,73,76,79,81,84,87,90,93,96,99,101,104,107,110,<BR>113,116,118,121,124,127,130,133,136,138,141,144,147,150,153,<BR>156,158,161,164,167,170,173,175,178,181,184,187,190,193,195,...<BR> </DIV>
<DIV>See the OEIS entries for 1001 terms computed by <FONT
face="Courier New">Reinhard Zumkeller. </FONT></DIV>
<DIV><FONT face="Courier New">These terms demonstrate that the tribonacci
constant </FONT><FONT face="Courier New">is </FONT></DIV>
<DIV><FONT face="Courier New">at least near the value of
the limit Y(n)/X(n). </FONT></DIV>
<DIV><FONT face="Courier New"></FONT> </DIV></BLOCKQUOTE></BODY></HTML>
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