Sequence Inspired By Sylvester's Sequence

Paul D. Hanna pauldhanna at juno.com
Sun Feb 6 05:25:21 CET 2005


Leroy,
         Just a few observations regarding your first sequence, 
which begins:
2,3,7,23,187,4643,898567,4192705463,3768229999719547,
15799096080675502645110083,59534627833852486970959777827956950667527,
940593305274293766389990332525291876515204489479064901503485974103,
 
The next term a(13) is 107 digits. 
 
Number of digits of first 25 terms is:
1,1,1,2,3,4,6,10,16,26,41,66,107,173,280,453,732,1184,1916,
3100,5015,8115,13130,21244,34373,
  
Asymptotics:
------------- 
Limit_{n->inf} log(a(n+1))/log(a(n)) = (sqrt(5)+1)/2 = 1.6180339887...
  
Limit_{n->inf} log(a(n))/((sqrt(5)+1)/2)^n = 0.471775871...

 
I think it is interesting that the ratio of the terms have fractional
parts that converge: 
 
Limit_{n->inf} fraction( a(2n+2)/a(2n+1) )  =  
0.9963258537736462872363456738414777019251922501451462595067
51265907424466948546566097677517246961832243212658328089363
10607326862416219726610814679929105111872945248837938615836
766055065685346323870537...
 
Limit_{n->inf} fraction( a(2n+1)/a(2n) )  = 
0.5781727423883458291832179527899399497969420087427440931499
03158291493544435908826920146242378149405816497712693028483
95401574346069439667634103549206044884112611658093899939284
555238191148056017194947...
  
Paul
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On Sat, 5 Feb 05 15:43:49 -0700 Leroy Quet <qq-quet at mindspring.com>
writes:
> I was inspired by Sylvester's sequence (sequence A000058: 
> a(n)=a(n-1)*(a(n-1)-1)+1) to come up with this recursive sequence 
> where 
> every term is coprime with every other:
>  
> a(1) and a(2) = integers where GCD(a(1),a(2)) = 1.
> 
> a(n+2) = a(n) *(a(n) +a(n+1)) - a(n+1).
> 
> 
> We can rewrite the recursion as
> 
> a(n+2) = - a(n+1) + (a(1)+a(2)) * product{k=1 to n} a(k) 
> 
> From this last recursion we can see that each term is coprime to 
> every 
> other.
> 
> 
> With a(1) = 2 and a(2) = 3 the sequence begins
> 2, 3, 7, 23, 187, 4643,...
> 
> (Not in EIS. I will submit the terms above myself.)
> 
> Is there a direct way of calculating the terms?
..





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