[seqfan] Re: Minimal k > n such that (4k+3n)(4n+3k) is square

[Peter Luschny]

DA> Here is a brief proof sketch (slightly more detailed than the
comment I've added to the OEIS entry for A083752)

While studying the proof I made the following observation:

Substitute in a(n) = (4k+3n)(4n+3k) 'k' by 'I' and
n by A001075(n) then sqrt(Re(a(n)) = A001352(n) for n >= 1.
This gives a transformation A001075 -> A001352.

A001075: 1, 2,  7, 26,  97,  362, 1351,  5042, 18817,  70226, ..
A001352: 1, 6, 24, 90, 336, 1254, 4680, 17466, 65184, 243270, ..

Conversely, if Re(a(n)) is a square we can go back via
the imaginary part: Im(a(n))/25 = A001075(n).

In other words: A001352(n)^2 + 25*A001075(n)*I starts

1+25*I, 36+50*I, 576+175*I, 8100+650*I, 112896+2425*I, ...

which are the same complex numbers as those a(n) for
which Re(a(n)) is a square, for n>= 1.

Everything conjectural.

Peter