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.
$value = $seq->ith($i)
$i'th Proth number. The first number is 3 at
$bool = $seq->pred($value)
Return true if
$valueis 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,
initial value = 1 # to give 3 on first next() call inc = 2 limit = 4 next() 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,
1Z1 1xZ01 1xxZ001 1xxxZ0001
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.
Copyright 2010, 2011, 2012, 2013, 2014 Kevin Ryde
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