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<DIV><FONT face=Arial size=2>Dear seqfans</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>I found, but i am not probably the first, that
starting fron any pair of real OR complex numbers a(1) and b(1)</FONT><FONT
face=Arial size=2>, the two associated sequences defined by the following
genarating functions :</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>(1) a(n) =
sqrt[a(n-1)]+sqrt[b(n-1]</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>(2) b(n) =
sqrt[(a(n-1)+b(n-1)]</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>have always the same limits A and B
where</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>A=3,394547695. . . .= B*(B-1)</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>B=2,409069851 , the greater of the two real roots
of y^6-4*y^5+4*y^4-4*y+4=0</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>To remain in the domain of integers , let us start
with a pair of integer numbers c(1) and d(1)<=c(1) and replace the genarating
functions by</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>(3) c(n) =
floor{sqrt[a(n-1)]+sqrt[b(n-1)]}</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>(4) d(n)
=floor{sqrt[a(n-1)]+sqrt[b(n-1)]}</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>We find that for any starting pair, the two
associated sequences have always the same limits C=3 and D=2</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>Starting in this way with big enough values of c(1)
and d(1)<= c(1), and reversing the sequences we get two finite sequences, as
long as we want , beginning with the pair e(1)=3 and f(1)=2 and
whose terms satisfy </FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>the </FONT><FONT face=Arial
size=2>functions and inegality</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>(5) e(n-1)=floor{sqrt[e(n)] +
sqrt[f(n)]}</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>(6) f(n-1)=floor{sqrt[e(n)+f(n)]}
</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>(7) e(n) >= f(n)</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>but these functions don't define univocally e(n)
and f(n) as successors of e(n-1) and f(n-1)</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>(for example the pair (3 3) has only one
successor pair = (8 1) but 25 different </FONT><FONT face=Arial size=2>pairs,
from (5 4) to (14 1), have for ancestors the pair (4 3) </FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>I wonder if it's possible to define generating
functions enabling to build a pair of infinite sequence of this type
(different of the sequences of constant terms 3,3,3 . . . and,2,2,2 . . . ), i.e
starting with the pair (3 2), </FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>and satisfying the conditions (5), (6)
ans (7) (maybe by </FONT><FONT face=Arial size=2>adding a rule fixing
the </FONT><FONT face=Arial size=2>choice among all the pairs e(n) and f(n)
satisfying these relations)</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>Is somebody interested by this
problem?</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>Best regards to each of you.<BR><BR>ph
LALLOUET</FONT></DIV></BODY></HTML>