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Math-NumSeq-55
River stage zero No dependents
/ Math::NumSeq::LucasNumbers

NAME

Math::NumSeq::LucasNumbers -- Lucas numbers

SYNOPSIS

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

DESCRIPTION

The Lucas numbers, L(i) = L(i-1) + L(i-2) starting from 1,3

1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, 843, 1364,...

This is the same recurrence as the Fibonacci numbers (Math::NumSeq::Fibonacci), but a different starting point.

FUNCTIONS

See "FUNCTIONS" in Math::NumSeq for behaviour common to all sequence classes.

$seq = Math::NumSeq::LucasNumbers->new ()

Create and return a new sequence object.

($i, $value) = $seq->next()

Return the next index and value in the sequence.

When $value exceeds the range of a Perl unsigned integer the return is a Math::BigInt to preserve precision.

Random Access

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

Return the $i'th Lucas number.

$bool = $seq->pred($value)

Return true if $value is a Lucas number.

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

Return an estimate of the i corresponding to $value. See "Value to i Estimate" below.

FORMULAS

Ith

Fibonacci F[k] and Lucas L[k] can be calculated together by a powering algorithm with two squares per doubling,

F[2k] = (F[k]+L[k])^2/2 - 3*F[k]^2 - 2*(-1)^k
L[2k] =                   5*F[k]^2 + 2*(-1)^k

F[2k+1] =    ((F[k]+L[k])/2)^2 + F[k]^2
L[2k+1] = 5*(((F[k]+L[k])/2)^2 - F[k]^2) - 4*(-1)^k

At the last step, ie. the lowest bit of i, only L[2k] or L[2k+1] needs to be calculated for the return, not the F[] too.

For any trailing zero bits of i, final doublings L[2k] can also be done with just one square as

L[2k] = L[k]^2 - 2*(-1)^k

The main double/step formulas can be applied until the lowest 1-bit of i is reached, then the L[2k+1] formula for that bit, followed by the single squaring for any trailing 0-bits.

Value to i Estimate

L[i] increases as a power of phi, the golden ratio,

L[i] = phi^i + beta^i    # exactly

So taking a log (natural logarithm) to get i, and ignoring beta^i which quickly becomes small,

log(L[i]) ~= i*log(phi)
i ~= log(L[i]) / log(phi)

Or the same using log base 2 which can be estimated from the highest bit position of a bignum,

log2(L[i]) ~= i*log2(phi)
i ~= log2(L[i]) / log2(phi)

This is very close to the Fibonacci formula (see "Value to i Estimate" in Math::NumSeq::Fibonacci), being bigger by

Lestimate(value) - Festimate(value)
  = log(value) / log(phi) - (log(value) + log(phi-beta)) / log(phi)
  = -log(phi-beta) / log(phi)
  = -1.67

On that basis, it could be close enough to take Lestimate = Festimate-1 (or vice-versa) and share code between the two.

SEE ALSO

Math::NumSeq, Math::NumSeq::Fibonacci

HOME PAGE

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

LICENSE

Copyright 2010, 2011, 2012 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.

Math-NumSeq is distributed in the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License for more details.

You should have received a copy of the GNU General Public License along with Math-NumSeq. If not, see <http://www.gnu.org/licenses/>.