A383176 - OEIS

A383176

If p = A002313(n) is a prime such that p = x^2 + y^2, then a(n) is the largest integer k that satisfies x^2 + y^2 - k*x*y > 0.

1, 2, 2, 4, 2, 6, 2, 3, 2, 3, 2, 2, 10, 3, 2, 3, 2, 2, 6, 2, 2, 14, 7, 2, 4, 16, 2, 2, 3, 8, 2, 2, 2, 3, 2, 2, 2, 3, 20, 6, 2, 2, 3, 5, 2, 4, 2, 2, 2, 2, 24, 3, 5, 2, 2, 6, 2, 4, 2, 26, 5, 2, 13, 3, 2, 2, 2, 2, 5, 2, 3, 2, 7, 5, 2, 2, 2, 3, 2, 7, 5, 2, 2, 3

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OFFSET

1,2

COMMENTS

Fermat's Christmas theorem states that if p = 2 or if p is congruent to 1 modulo 4 (A002313), then p is written as a sum of 2 squares uniquely. Thus, if A002313(n) = x^2 + y^2, for certain integers x and y, then a(n) is the largest integer k such that x^2 + y^2 - k*x*y > 0.

a(n) >= 2, for n > 1. If p > 2 and p = x^2 + y^2, since x != y, then it is satisfied that 0 < (x - y)^2 = x^2 + y^2 - 2x*y < x^2 + y^2 - x*y. The equality a(n) = 2 is given when |x - y| < phi*min{x, y}.

LINKS

Table of n, a(n) for n=1..84.

EXAMPLE

Since A002313(8) = 53 and 53 = 2^2 + 7^2, we have that 53 - 3*2*7 > 0 and 53 - 4*2*7 < 0, then a(8) = 3.

PROG

(Python)

from itertools import count, islice

from sympy import isprime

from sympy.solvers.diophantine.diophantine import cornacchia

def A383176_gen(): # generator of terms

yield 1

for p in count(5, 4):