10=m^p-n^q?
Neil Fernandez
primeness at borve.demon.co.uk
Tue Oct 8 01:10:59 CEST 2002
In message <20021007174044.65829.qmail at web9606.mail.yahoo.com>, murthy
amarnath <amarnath_murthy at yahoo.com> writes
>--- Neil Fernandez <primeness at borve.demon.co.uk>
>wrote:
>> In message
>>
><307E0287E4AFD411975600508B4472BB32B927 at exchange.yosh.ac.il>,
>> ZAKIRS <zfseidov at ycariel.yosh.ac.il> writes
>> >Dear all,
>> >i've just speculated that somel integers can't be
>> represented as
>> >"difference of two full powers >=2"
>> >and the list of such (eligible) numbers <200 is:
[snip]
>> A nice set of sequences could be derived from
>> assigning to each n the
>> numbers a, b, p, q where a^p and b^q are the
>> smallest pair of powers
>> with difference n.
>>
>> E.g.
>>
>> n = a^p - b^q
>> =============
>> 1 = 3^2 - 2^3
>> 2 = 3^3 - 5^2
>> 3 = 2^7 - 5^3
>> 4 - 6^2 - 2^5
>> 5 = 3^2 - 2^2
>> etc.
>>
>> Reading vertically:
>>
>> a sequence: 3, 3, 2, 6, 3, ...
>> p sequence: 2, 3, 7, 2, 2, ...
>> b sequence: 2, 5, 5, 2, 2, ...
>> q sequence: 3, 2, 3, 5, 2, ...
>>
>> If there are any integers which cannot be expressed
>> as a^p - b^q, one
>> could just put a,p,b,q equal to 0.
>>
>> Neil
>> --
>> Neil Fernandez
> see this sequence
>ID Number: A069586
>Sequence:
>8,25,125,4,4,0,9,8,16,2187,16,4,243,0,49,9,8,9,8,0,4,27,4,8,
>
>0,0,0,4,0,6859,0,32,16,0,0,0,27,1331,25,9,8,0,0,81,4,243,81,
>
>16,32,0,0,0,0,27,9,8,64,0,0,4,64,0,961,64,16,0,0,0,0,0,0,9,
> 8,169,0,49,4,0,49,0,0,0,0,0,0
>Name: Smallest prime power p^k (k>=2) such that
>there is a gap of exactly n
> between p^k and the next prime power q^l
>(l >= 2); or 0 if no such q^l
> exists.
So my line for 4 should have read
4 = 2^3 - 2^2
:-)
Neil
--
Neil Fernandez
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