10=m^p-n^q?
Neil Fernandez
primeness at borve.demon.co.uk
Mon Oct 7 15:55:08 CEST 2002
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:
>
>10, 14, 22, 29, 31, 34, 42, 50, 52, 54, 58, 60, 62, 66, 68, 70, 72,
> 76, 78, 82, 84, 86, 88, 90, 92, 94, 96, 98, 102, 108, 110, 111,
> 112, 114, 118, 120, 122, 124, 126, 130, 132, 134, 136, 140, 142,
> 146, 150, 153, 156, 158, 160, 162, 164, 174, 176, 177, 178, 182,
> 186, 188, 190, 192, 193, 194
(13^3)-(3^7) = 2197 - 2187 = 10
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
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