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NNTP-Posting-Date: Sat, 11 Mar 2017 01:42:34 -0600
From: Gareth Owen <gwowen@gmail.com>
Newsgroups: comp.lang.c
Subject: Re: Program to find consecutive integers
References: <o9t9jl$bkg$1@gioia.aioe.org> <1bpohpg0sy.fsf@pfeifferfamily.net> <o9ti7j$lnp$1@gioia.aioe.org> <slrn3vfsoc4oes.lda.ike@iceland.freeshell.org> <11e8e73e-fc33-48e7-92a7-6064330813cd@googlegroups.com> <slrn3vfsoc5v88.fq0.ike@iceland.freeshell.org>
Date: Sat, 11 Mar 2017 07:42:33 +0000
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Ike Naar <ike@iceland.freeshell.org> writes:

> On 2017-03-10, jameskuyper@verizon.net <jameskuyper@verizon.net> wrote:
>> Other possibilities include:
>> {-14, -13, -12, -11, -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1,
>> 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}
>>
>> I suspect that this is not what insane_goat meant, but until he
>> specifies the problem to be solved more precisely, this seems to
>> qualify.
>
> I was assuming that all numbers were meant to be positive integers.


Let M = 2^N * (odd prime factors)

For every odd factor (not necessarily prime), f of M there's a sequence
length f centred on M/f and a sequence length f*2^(N+1) centred on
M/(f*2^(N+1))

For M=15, odd factors 1,3,5,15
f=1  {15},               {7,8}
f=3  {4,5,6},            {0,1,2,3,4,5}
f=5  {1,2,3,4,5},        {-3,...6}
f=15 {-6...8}            {-14,....15}

For M=8, odd factors 1
Only solutions
f=1 {8}                   {-7...8}

For M=24, odd factors 1,3
f=1 {24}                  {-6..9}
f=3 {7,8,9}               {-23..24}

etc, etc, etc

So all you need to is factorise ... which may be harder than you think