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<DIV><FONT face=Arial color=#0000ff size=2><SPAN
class=804573615-19022003>><FONT face="Times New Roman" color=#000000 size=3>I
believe a(n)-->1/2, which can lead to conjectures related to primes.
</FONT></SPAN></FONT></DIV>
<DIV><FONT face="Times New Roman" color=#000000 size=3><SPAN
class=804573615-19022003></SPAN></FONT> </DIV>
<DIV><FONT face=Arial color=#0000ff size=2><SPAN class=804573615-19022003>This
is possible, but is not suggested by the graph of (n, a(n)) from n = 1
to 10^5. (Of course, the </SPAN></FONT></DIV>
<DIV><FONT face=Arial color=#0000ff size=2><SPAN
class=804573615-19022003>convergence to 1/2 could be clear from looking at
larger n, so the graphical evidence isn't the last word either.)
</SPAN></FONT></DIV>
<DIV><FONT face=Arial color=#0000ff size=2><SPAN class=804573615-19022003>It
appears that there's an infinite subsequence </SPAN></FONT><FONT face=Arial
color=#0000ff size=2><SPAN class=804573615-19022003>a(n_k) each of whose terms
have distance > e for some </SPAN></FONT></DIV>
<DIV><FONT face=Arial color=#0000ff size=2><SPAN class=804573615-19022003>small
constant e > 0--which, if true, would imply divergence.</SPAN></FONT></DIV>
<DIV><FONT face=Arial color=#0000ff size=2><SPAN
class=804573615-19022003></SPAN></FONT> </DIV>
<DIV><FONT face=Arial color=#0000ff size=2><SPAN
class=804573615-19022003>Mathematica code for the graph:</SPAN></FONT></DIV>
<DIV><FONT face=Arial color=#0000ff size=2><SPAN
class=804573615-19022003></SPAN></FONT> </DIV>
<DIV><FONT face=Arial color=#0000ff size=2><SPAN class=804573615-19022003>t =
{1};<BR>gt = 1;<BR>For[i = 2, i <= 10^5, i++,<BR> gt = 1 -
(Prime[i - 1]/Prime[i]) gt;<BR> t = Append[t,
gt]];<BR>ListPlot[t]</SPAN></FONT></DIV>
<DIV><FONT face=Arial color=#0000ff size=2><SPAN
class=804573615-19022003></SPAN></FONT> </DIV>
<DIV><FONT face=Arial color=#0000ff size=2><SPAN
class=804573615-19022003></SPAN></FONT> </DIV>
<DIV><FONT face=Arial color=#0000ff size=2><SPAN class=804573615-19022003>J. L.
Pe</SPAN></FONT></DIV>
<BLOCKQUOTE>
<DIV class=OutlookMessageHeader dir=ltr align=left><FONT face=Tahoma
size=2>-----Original Message-----<BR><B>From:</B> benoit
[mailto:abcloitre@wanadoo.fr]<BR><B>Sent:</B> Tuesday, February 18, 2003 4:05
PM<BR><B>To:</B> Pe Joseph-AJP070<BR><B>Cc:</B>
seqfan@ext.jussieu.fr<BR><B>Subject:</B> Re: A Generalization of the Sequence
Convergence Problem<BR><BR></FONT></DIV>
<P><FONT face=Geneva>Regarding your generalisation. If s(n), s(n+1)-s(n) are
increasing sequences, that seems working. </FONT></P><BR>
<P><FONT face=Geneva>Coming back to the original problem
a(n)=1-(p(n-1)/p(n))*a(n-1), I believe a(n)-->1/2, which can lead to
conjectures related to primes. </FONT></P><BR>
<P><FONT face=Geneva>It's easy to show that limit n-->oo (a(2n)+a(2n+1)) =
1 but that doesn't mean limit n-->oo a(2n)= limit n-->oo a(2n+1)= 1/2 at
all. </FONT></P><BR>
<P><FONT face=Geneva>Since : a(2n)=1/p(2n)*sum(k=1,n,p(2*k)-p(2*k-1)) ;
a(2n+1)=1/p(2n+1)*sum(k=1,n,p(2*k+1)-p(2*k)) </FONT></P><BR>
<P><FONT face=Geneva>a(n)-->1/2 would imply sum(k=1,n,p(2*k)-p(2*k-1)) to
be asymptotic to (2n)*log(2n)/2 or n*log(n) </FONT></P><BR>
<P><FONT face=Geneva>but as one can see
(http://mathworld.wolfram.com/PrimeGaps.html), the behaviour of p(k+1)-p(k) is
quite unknown. </FONT></P><BR>
<P><FONT face=Geneva>More precisely is : </FONT></P><BR>
<P><FONT face=Geneva>sqrt(n)*(1/p(2*n+1)*sum(k=1,n,p(2*k+1)-p(2*k)) -
1/p(2*n)*sum(k=1,n,p(2*k)-p(2*k-1))) bounded ? </FONT></P><BR><BR>
<P><FONT face=Geneva>with (PARI)
delta(n)=sqrt(n)*(1/prime(2*n+1)*sum(k=1,n,prime(2*k+1)-prime(2*k)) -
1/prime(2*n)*sum(k=1,n,prime(2*k)-prime(2*k-1))) </FONT></P><BR><BR>
<P><FONT face=Geneva>Does limit n-->infinity delta(n) exist?
delta(20000)=-0.5846342931598270275... </FONT></P><BR>
<P><FONT face=Geneva>Is 1/prime(2*n+1)*sum(k=1,n,prime(2*k+1)-prime(2*k)) <
1/prime(2*n)*sum(k=1,n,prime(2*k)-prime(2*k-1)) for all n large enough?
</FONT></P><BR>
<P><FONT face=Geneva>If so the sequence of n such that
1/prime(2*n+1)*sum(k=1,n,prime(2*k+1)-prime(2*k)) >=
1/prime(2*n)*sum(k=1,n,prime(2*k)-prime(2*k-1)) </FONT></P><BR>
<P><FONT face=Geneva>could be finite : </FONT></P><BR>
<P><FONT
face=Geneva>1,2,3,4,5,7,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100,101,102,103,104,105,106,107,108,109,110,125,126,127,128,129,239,240,241,242,243,244,261,262,263,264,267,269,274,275,276,277,278,281,282,283,284,285,286,287,288,289,290,291,292,293,294,295,296,297,298,299,300,301,302,303,304,305,306,307,308,309,310,311,313,314
</FONT></P><BR>
<P><FONT face=Geneva>Is 314 the last term? </FONT></P><BR><BR><BR>
<P><FONT face=Geneva>Benoit Cloitre </FONT></P>
<P><FONT face=Geneva>abcloitre@wanadoo.fr
</FONT></P><BR><BR><BR><BR><BR><BR><BR><BR><BR>
<P>Le mardi, 18 fév 2003, à 17:12 Europe/Paris, Pe Joseph-AJP070 a écrit :
</P><BR>
<P>Here is a generalization of the sequence convergence problem I posted </P>
<P>yesterday. (For convenience, I append </P>
<P>the original problem at the end of this message.) </P><BR>
<P>============================================================================
</P><BR>
<P>Let s(n) be a sequence that converges to a real number K different from -1.
</P>
<P>Define the "oscillator sequence" a(n) of s(n) by the rules: </P><BR>
<P>a(1) = 1; </P>
<P>a(n) = 1 - (s(n-1)/s(n)) a(n-1) if n > 1. </P><BR>
<P>Note that the original problem below concerns the convergence of the </P>
<P>oscillator sequence of s(n) = p(n). </P><BR>
<P>Examples of s(n) are s(n) = n, s(n) = n^2 and in general, any polynomial
</P>
<P>function in n that is not the zero polyonomial function. </P>
<P>The oscillator sequence of s(n) may converge or diverge depending on s(n).
</P>
<P>Clearly, if s(n) = 1 is the constant sequence mapping </P>
<P>each positive integer to 1, then a(n) diverges--in fact, oscillates between
</P>
<P>0 and 1 (hence the name "oscillator sequence"). </P>
<P>If s(n) = n, it is not hard to prove that a(n) converges to 1/2. </P><BR>
<P>Can you find conditions on s(n) that will ensure the convergence of its
</P>
<P>oscillator sequence? Of course, if s(n) converges, then </P>
<P>lim s(n) = 1/(K + 1). </P><BR>
<P>============================================================================
</P><BR>
<P>J. L. Pe </P><BR>
<P>============================================================================
</P><BR>
<P>Original Problem: </P><BR>
<P>Define the sequence a(n) by: a(1) = 1; a(n) = 1-(p(n-1)/p(n))*a(n-1) if n
> </P>
<P>1, where p(n) denotes the n-th prime. </P>
<P>It's easy to show (an exercise!) that if L = lim a(n) exists, then L = 1/2.
</P>
<P>Can you prove the convergence of a(n) or the divergence of a(n)?
</P><BR><BR></BLOCKQUOTE></BODY></HTML>