A071383 Squared radii of the circles around (0,0) that contain record numbers of lattice points.

0, 1, 5, 25, 65, 325, 1105, 4225, 5525, 27625, 71825, 138125, 160225, 801125, 2082925, 4005625, 5928325, 29641625, 77068225, 148208125, 243061325, 1215306625, 3159797225, 6076533125, 12882250225, 53716552825, 64411251125

Offset: 1

Author

Hugo Pfoertner, May 23 2002

Keywords

• nonn

Comments

The number of lattice points (i,j) on the circle with i^2 + j^2 = a(n) is given by A071385(n).

In a sci.math posting on May 05 2002 entitled "Circle with 3 lattice points", James R. Buddenhagen asked: Which circles have the property that they pass through more lattice points than any smaller circle? and he gave the terms 1, 25, 65, 325, 1105, 4225, 5525, with the missing 5 added by Ahmed Fares. In the same thread Gerry Myerson mentioned the factorization into primes of the form 4*k+1.

Also, numbers with a record number of divisors all of whose prime factors are of the form 4k + 1. - Amiram Eldar, Sep 12 2019

Indices of records of A004018. Apart from the first term, also indices of records of A002654. - Jianing Song, May 20 2021

Links

• Ray Chandler, Table of n, a(n) for n = 1..425 (first 97 terms from Ray Chandler, terms 98-365 from Amiram Eldar)
• James Buddenhagen, Circle with 3 lattice points, thread in sci.math (May 2002)
• Hugo Pfoertner, Construction of the sequences A071383, A071384, A071385

Crossrefs

Cf. A000448, A048610, A052199, A071384, A071385, A230655, A300162. Subsequence of A054994 (excluding first term). Where records occur in A004018. See A088959 for circles with integer radius.

Indices of records of Sum_{d|n} kronecker(m, d): A230655 (m=-3), this sequence (m=-4), A279541 (m=-6).

Programs

PARI

my(v=list(10^15), rec=0); print1(0, ", "); for(n=1, #v, if(numdiv(v[n])>rec, rec=numdiv(v[n]); print1(v[n], ", "))) \\ Jianing Song, May 20 2021, see program for A054994

Python

from math import prod
from sympy import isprime
primes_congruent_1_mod_4 = [5]
def prime_4k_plus_1(i): # the i-th prime that is congruent to 1 mod 4
    while i>=len(primes_congruent_1_mod_4): # generate primes on demand
        n = primes_congruent_1_mod_4[-1]+4
        while not isprime(n): n += 4
        primes_congruent_1_mod_4.append(n)
    return primes_congruent_1_mod_4[i]
def generate_A054994():
    TO_DO = {(1,())}
    while True:
        radius, exponents = min(TO_DO)
        yield radius, exponents
        TO_DO.remove((radius, exponents))
        TO_DO.update(successors(radius,exponents))
def successors(r,exponents):
    for i,e in enumerate(exponents):
        if i==0 or exponents[i-1]>e:
            yield (r*prime_4k_plus_1(i), exponents[:i]+(e+1,)+exponents[i+1:])
    if exponents==() or exponents[-1]>0:
        yield (r*prime_4k_plus_1(len(exponents)), exponents+(1,))
n,record,radius=1,1,0
print(radius, end="") # or record, for A071385
for radius,expo in generate_A054994():
    num_points = 4*prod((e+1) for e in expo)
    if num_points>record:
        record = num_points
        n += 1
        print (",", radius, end="") # or record, for A071385
        if n==27: break
print()
# Günter Rote, Sep 12 2023

Formula

For n>1 we have 1 < a(n+1)/a(n) <= 5, since one can multiply the points x+iy for which x^2 + y^2 = N by either 2+i or 2-i to get two new sets of points X+iY for which X^2 + Y^2 = 5N. This strictly increases the number since it is easy to see that the two sets aren't the same. - J. H. Conway, Jun 04 2002

lim n ->infinity Log(a(n))/n = 1. [Conjectured by Benoit Cloitre, proved by J. H. Conway]