Math::NumSeq::Catalan -- Catalan numbers (2n)! / (n!*(n+1)!)
use Math::NumSeq::Catalan; my $seq = Math::NumSeq::Catalan->new; my ($i, $value) = $seq->next;
The Catalan numbers
C(n) = binomial(2n,n) / (n+1) = (2n)! / (n!*(n+1)!) 1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, ... starting i=0
From the factorial expression it can be seen the values grow roughly as a power-of-4,
C(i) = C(i-1) * (2i)*(2i-1) / (i*(i+1)) C(i) = C(i-1) * 2*(2i-1)/(i+1) < C(i-1) * 4
values_type => "odd" can give just the odd part of each number, ie. with factors of 2 divided out,
values_type => "odd" 1, 1, 1, 5, 7, 21, 33, 429, 715, 2431, 4199, ... starting i=0
The number of 2s in C(i) is
num2s = (count-1-bits of i+1) - 1
The odd part is always monotonically increasing. When i increments num2s increases by at most 1, ie. a single factor of 2. In the formula above
C(i) = C(i-1) * 2*(2i-1)/(i+1)
it can be seen that C(i) gains at least 1 factor of 2, so after dividing out 2^num2s it's still greater than C(i-1).
See "FUNCTIONS" in Math::NumSeq for behaviour common to all sequence classes.
$seq = Math::NumSeq::Catalan->new ()
$seq = Math::NumSeq::Catalan->new (values_type => $str)
Create and return a new sequence object.
Move the current sequence position to
$i. The next call to
next() will return
$i and its corresponding value.
$value = $seq->ith($i)
$i = $seq->value_to_i_estimate($value)
Return an estimate of the i corresponding to
The current code is based on
C(n) ~= 4^n / (sqrt(pi*n)*(n+1))
but ignoring the denominator there and so simply taking
C(n) ~= 4^n hence i ~= log4(value)
The 4^n term dominates for medium to large
$value (for both plain and "odd").
Copyright 2012, 2013, 2014 Kevin Ryde
Math-NumSeq is free software; you can redistribute it and/or modify it under the terms of the GNU General Public License as published by the Free Software Foundation; either version 3, or (at your option) any later version.
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