[seqfan] Re: A230655 & A071383 and (3, 1)- and (4, 1)-highly composite numbers.

[israel at math.ubc.ca]

For a number n whose prime factors are == 1 (mod 4), each prime factor p 
can be written uniquely as x^2+y^2 with 0<x<y, corresponding to the 
Gaussian prime factorization p = (x+iy)(x-iy). The number of positive 
integer solutions of n = x^2+y^2 should then be the same as the number of 
divisors of n. Thus (unless n itself is a square) the number of points of 
Z^2 on the circle of radius sqrt(n) is 4 times the number of divisors of n.

Cheers,
Robert

On Sep 11 2019, Ami Eldar wrote:

>Hello,
>
>There are variations of highly composite numbers (numbers with a record
>number of divisors, A002182) in literature in which the divisors are
>restricted to have prime factors only of the form a*k+b (a and b coprime, k
>= 0, 1, 2, ... ). These numbers may be called (a,b)-highly composite
>numbers (the usual highly composite numbers are then (1, 0)-highly
>composites).
>
>I have calculated the sequences of (3,1)- and (4,1)-highly composite
>numbers, and apparently they are already in the OEIS, but with a different
>interpretation: A230655 and A071383 which are the sequences of squared
>radii of the circles around a point of hexagonal (A230655) or square
>(A071383) lattice that contain record numbers of lattice points.
>
>I have compared terms with all available data in these two sequences (17 in
>A230655 and 97 in A071383) and found that they are the same.
>
>In both sequences, divisors are not mentioned, but in A071383 it is said
>that all the terms are products of consecutive primes of the form 4k+1
>starting from 5, with nonincreasing exponents, which is necessary, but not
>sufficient, for the terms to be highly composite.
>
>Can it be proven that the two definitions are equivalent? If yes, I think
>that it worthwhile to mention it.  In addition, in that case I can easily
>extend the data with many more terms.
>
>Best,
>
>Amiram
>
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