A Generalization of the Sequence Convergence Problem

Wed Feb 19 16:54:33 CET 2003

>I believe a(n)-->1/2, which can lead to conjectures related to primes. 

This is possible, but is not suggested by the graph of (n, a(n)) from n = 1
to 10^5. (Of course, the 
convergence to 1/2 could be clear from looking at larger n, so the graphical
evidence isn't the last word either.) 
It appears that there's an infinite subsequence a(n_k) each of whose terms
have distance > e for some 
small constant e > 0--which, if true, would imply divergence.

Mathematica code for the graph:

t = {1};
gt = 1;
For[i = 2, i <= 10^5, i++,
    gt = 1 - (Prime[i - 1]/Prime[i]) gt;
    t = Append[t, gt]];
ListPlot[t]

J. L. Pe

-----Original Message-----
From: benoit [mailto:abcloitre at wanadoo.fr]
Sent: Tuesday, February 18, 2003 4:05 PM
To: Pe Joseph-AJP070
Cc: seqfan at ext.jussieu.fr
Subject: Re: A Generalization of the Sequence Convergence Problem

Regarding your generalisation. If s(n), s(n+1)-s(n) are increasing
sequences, that seems working. 

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. 

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. 

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)) 

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) 

but as one can see (http://mathworld.wolfram.com/PrimeGaps.html), the
behaviour of p(k+1)-p(k) is quite unknown. 

More precisely is : 

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 ? 

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))) 

Does limit n-->infinity delta(n) exist?
delta(20000)=-0.5846342931598270275... 

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? 

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)) 

could be finite : 

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,
56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,8
1,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100,101,102,103,104,
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 

Is 314 the last term? 

Benoit Cloitre 

abcloitre at wanadoo.fr 

Le mardi, 18 fév 2003, à 17:12 Europe/Paris, Pe Joseph-AJP070 a écrit : 

Here is a generalization of the sequence convergence problem I posted 

yesterday. (For convenience, I append 

the original problem at the end of this message.) 

============================================================================

Let s(n) be a sequence that converges to a real number K different from -1. 

Define the "oscillator sequence" a(n) of s(n) by the rules: 

a(1) = 1; 

a(n) = 1 - (s(n-1)/s(n)) a(n-1) if n > 1. 

Note that the original problem below concerns the convergence of the 

oscillator sequence of s(n) = p(n). 

Examples of s(n) are s(n) = n, s(n) = n^2 and in general, any polynomial 

function in n that is not the zero polyonomial function. 

The oscillator sequence of s(n) may converge or diverge depending on s(n). 

Clearly, if s(n) = 1 is the constant sequence mapping 

each positive integer to 1, then a(n) diverges--in fact, oscillates between 

0 and 1 (hence the name "oscillator sequence"). 

If s(n) = n, it is not hard to prove that a(n) converges to 1/2. 

Can you find conditions on s(n) that will ensure the convergence of its 

oscillator sequence? Of course, if s(n) converges, then 

lim s(n) = 1/(K + 1). 

============================================================================

J. L. Pe 

============================================================================

Original Problem: 

Define the sequence a(n) by: a(1) = 1; a(n) = 1-(p(n-1)/p(n))*a(n-1) if n > 

1, where p(n) denotes the n-th prime. 

It's easy to show (an exercise!) that if L = lim a(n) exists, then L = 1/2. 

Can you prove the convergence of a(n) or the divergence of a(n)? 

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