Kevin Ryde > Math-NumSeq > Math::NumSeq::Catalan

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Module Version: 72

NAME

Math::NumSeq::Catalan -- Catalan numbers (2n)! / (n!*(n+1)!)

SYNOPSIS

``` use Math::NumSeq::Catalan;
my \$seq = Math::NumSeq::Catalan->new;
my (\$i, \$value) = \$seq->next;```

DESCRIPTION

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```

Odd

Option `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).

FUNCTIONS

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.

Iterating

`\$seq->seek_to_i(\$i)`

Move the current sequence position to `\$i`. The next call to `next()` will return `\$i` and its corresponding value.

Random Access

`\$value = \$seq->ith(\$i)`

Return the `\$i`'th value.

`\$i = \$seq->value_to_i_estimate(\$value)`

Return an estimate of the i corresponding to `\$value`.

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").

http://user42.tuxfamily.org/math-numseq/index.html