A sequence based on consecutive primes
Nick Hobson
nickh at qbyte.org
Mon May 7 16:20:12 CEST 2007
Hi Seqfans,
Suppose p and q are consecutive primes. Let
a = sqrt((p^2 + q^2)/2 - 1). (1)
It is easy to see that if q = p + 2 then a is an integer. Is the converse
true?
If we consider the smallest prime q > p + 2 which makes a an integer, we
generate the following sequence, where 0 is used if there is no such q:
11, 263, 59, 23, 101, 109, 0, 151, 193, 79, 269, 277, 311, 0, 179, 83,
83003, 479, 487, 181, 563, 571, 613, 1201, 157, 141509, 739, 773, 479
So, for example, q = 11 is the smallest prime > 5 such that sqrt((3^2
+ q^2)/2 - 1) is an integer. There is no such q for p = 19 and p = 47
(consider (1) modulo 7), but I can find no similar proof of impossibility
for p = 127, which would generate the next term in the above sequence.
Source: M500 Magazine: http://www.m500.org.uk/
Nick
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