Math::NumSeq::ProthNumbers -- Proth number sequence


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


The Proth numbers

    3, 5, 9, 13, 17, 25, 33, 41, 49, 57, 65, 81, 97, 113, ...
    starting i=1

being integers of the form k*2^n+1 for some k and n and where k < 2^n.

The k < 2^n condition means the values in binary have low half 00..01 and high half some value k,

    binary 1xxx0000000...0001

    value    binary        k  n  2^n
      3       11           1  1   2
      5       101          1  2   4
      9      1001          2  2   4
     13      1101          3  2   4
     17      10001         2  3   8
     25      11001         3  3   8
     33     100001         4  3   8
     41     101001         5  3   8
     k part --++-- low part

Taking all k < 2^n duplicates some values, as for example 33 is k=4 n=3 as 4*2^3+1 and also k=2 n=4 as 2*2^4+1. This happens for any even k. Incrementing n on reaching k=2^n-1 makes a regular pattern, per "Ith" below.


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

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

Create and return a new sequence object.

Random Access

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

Return the $i'th Proth number. The first number is 3 at $i==1.

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

Return true if $value is a Proth number, meaning is equal to k*2^n+1 for some k and n with k < 2^n.

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

Return an estimate of the i corresponding to $value. This is roughly sqrt(2*$value).



Successive values can be calculated by keeping track of the 2^n power and incrementing k by adding such a power,

      value = 1   # to give 3 on first next() call
      inc = 2
      limit = 4

      value += inc
      if value >= limit
         inc *= 2       # ready for next time
         limit *= 4
      return value


Taking the values by their length in bits, the values are

       11        1 value  i=1
       101       1 value  i=2
      1x01       2 values i=3,4
      1x001      2 values i=5,6
     1xx001      4 values i=7 to 10
     1xx0001     4 values
    1xxx0001     8 values
    1xxx00001    8 values

For a given 1xxx high part the low zeros, which is the 2^n factor, is the same length and then repeated 1 bigger. That doubling can be controlled by a high bit of the i, so in the following Z is either a zero bit or omitted,


The ith Proth number can be formed from the bits of the index

    i+1 = 1zxx..xx    binary
    k = 1xx..xx
    n = z + 1 + number of x's

The first 1zxxx desired is 10, which is had from i+1 starting from i=1. The z second highest bit makes n bigger, giving the Z above present or omitted.

For example i=9 is bits i+1=1010 binary which as 1zxx is k=0b110=6, n=0+1+2=3, for value 6*2^3+1=49, or binary 110001.

It can be convenient to take the two highest bits of the index i together, so hhxx..xx so hh=2 or 3, then n = hh-1 + number of x's.


Math::NumSeq, Math::NumSeq::CullenNumbers, Math::NumSeq::WoodallNumbers



Copyright 2010, 2011, 2012, 2013, 2014, 2016, 2019, 2020 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 <>.