A (Somewhat) Interesting Recursion

Wed Apr 18 18:32:52 CEST 2007

sequence relating to its coordinates is bounded, and is filled in with 
    a(0) ->
  0 1 2 3 4 5
starting values, it seems.
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From: "Andrew Plewe" <aplewe at sbcglobal.net>
To: <seqfan at ext.jussieu.fr>
Subject: RE: Duplicate hunting, cont.
Date: Wed, 18 Apr 2007 13:52:10 -0700
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some point. I'd say the ratio of duplicates I'm reporting to those I'm
skipping is about 1 to 7.
straightforward duplicates:
sqrt(2) [A063448])
	-Andrew Plewe-
From: N. J. A. Sloane [mailto:njas at research.att.com]
Sent: Tuesday, April 17, 2007 7:40 PM
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Subject: Re: Duplicate hunting, cont.
Me:  this is certainly not true!  Even if a million terms agree,
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Subject: Re: Duplicate hunting, cont.
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From: "Andrew Plewe" <aplewe at sbcglobal.net>
Date:  Tue, 17 Apr 2007 17:32:31 -0700
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To: Leroy Quet <qq-quet at mindspring.com>
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Subject: Re: A (Somewhat) Interesting Recursion 
From: hv at crypt.org
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Date: Thu, 19 Apr 2007 13:09:48 +0100
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  perl -we '($x,$y)=@ARGV;++$a[$x];for(1..1000){++$a[$y];print"$x "; \
  ($x,$y)=($y,$a[$x+$y]||0)};print"\n"' 3 3
  0 1 2 3 4 5
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Subject: Re: A (Somewhat) Interesting Recursion 
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sequence which contains arbitrarily large terms.)

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From: Valery Liskovets <liskov at im.bas-net.by>
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Subject: Identification of fractional sequences
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Recently I've encountered a minor problem. I tried to identify
the following sequence of fractions:
  1, 2, 8/3, 25/6, 169/30, 361/45, 3364/315, 36481/2520,...
Superseeker could recognize neither numerators nor denominators.
In this particular case, however, it was easy to guess by
factorizations to augment the latter to n!, resulting in
  1/1!, 4/2!, 16/3!, 100/4!, 676/5!, 5776/6!, 53824/7!, 583696/8!,...
with A111883 as the numerators... Are there any ready program
facilities capable to help in such and more difficult cases?

Valery Liskovets