[seqfan] Re: Primitive pairs of powerful numbers

franktaw at netscape.net franktaw at netscape.net
Sun Nov 13 23:19:00 CET 2011

If we take n to be any powerful number, then for any y, n*y^2 is also 
powerful. Now ask when n*y^2 + 1 is a perfect square, say x^2; this 
gives us the Pell equation:

x^2 - n*y^2 = 1.

If n is not a perfect square, this will provide infinitely many 
solutions. It isn't immediately obvious to me how many of these 
solutions will be of the form 4*n*(n+1) for some (other) n in A060355.

Of the 23 numbers in A060355, 18 have n+1 as a square. Only 1 is a 

Franklin T. Adams-Watters

-----Original Message-----
From: Charles Greathouse <charles.greathouse at case.edu>

Sequence A060355 has a simple proof of infinitude: each n in the
sequence generates 4n(n+1) also in the sequence.  This immediately
raises the question: are there infinitely many members not of this
form?  I recently added this as sequence A199801, primitive elements
of A060355.

Any good heuristics, if nothing else?

Charles Greathouse

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