NAME

Math::NumSeq::ReRound -- sequence from repeated rounding up

SYNOPSIS

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

DESCRIPTION

This is the sequence of values formed by repeatedly rounding up to a multiple of i-1,i-2,...,2,1.

1, 2, 4, 6, 10, 12,

For example i=5 is rounded up to a multiple of 4 to give 8, then rounded up to a multiple of 3 to give 9, then rounded up to a multiple of 2 for value 10 at i=5.

When rounding up if a value is already a suitable multiple then it's unchanged. For example i=4 round up to a multiple of 3 to give 6, then 6 round up to a multiple of 2 is 6 unchanged since it's already a multiple of 2.

For i>2 the last step rounds up to a multiple of 2 so the values are all even. They're also always increasing and end up approximately

value ~= i^2 / pi

though there's values both bigger and smaller than this approximation.

Extra Multiples

The extra_multiples option can round up at each step to an extra multiple. For example extra_multiples >= 2 changes i=5 to round up to a multiple of 4 and then step by 2 further multiples of 4, similarly for round up to 3 and 2 further multiples, etc.

FUNCTIONS

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

$seq = Math::NumSeq::ReRound->new (extra_multiples => $integer)

Create and return a new sequence object.

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

Return $i rounded up to multiples of i-1,...,2.

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

Return true if $value is a ReRound value.

FORMULAS

Predicate

The rounding procedure can be reversed to test for a ReRound value.

for i=2,3,4,etc
  remainder = value mod i
  if remainder==i-1 then not a ReRound
  otherwise
  value -= remainder    # round down to multiple of i
stop when value <= i
is a ReRound if value==i (and i is its index)

For example to test 28, it's a multiple of 2, so ok for the final rounding. It's predecessor in the rounding steps was a multiple of 3, so round down to a multiple of 3 which is 27. The predecessor of 27 was a multiple of 4 so round down to 24. But at that point there's a contradiction because if 24 was the value then it's already a multiple of 3 and so wouldn't have gone up to 27. This case where a round-down gives a multiple of both i and i-1 is identified by the remainder value % i == i-1, since the value is already a multiple of i-1 and subtracting an i-1 would leave it still so.

SEE ALSO

Math::NumSeq

HOME PAGE

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

LICENSE

Copyright 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/>.