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%I #63 Sep 15 2023 07:53:11
%S 0,1,5,25,65,325,1105,4225,5525,27625,71825,138125,160225,801125,
%T 2082925,4005625,5928325,29641625,77068225,148208125,243061325,
%U 1215306625,3159797225,6076533125,12882250225,53716552825,64411251125
%N Squared radii of the circles around (0,0) that contain record numbers of lattice points.
%C The number of lattice points (i,j) on the circle with i^2 + j^2 = a(n) is given by A071385(n).
%C 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.
%C Also, numbers with a record number of divisors all of whose prime factors are of the form 4k + 1. - _Amiram Eldar_, Sep 12 2019
%C Indices of records of A004018. Apart from the first term, also indices of records of A002654. - _Jianing Song_, May 20 2021
%H Ray Chandler, <a href="/A071383/b071383.txt">Table of n, a(n) for n = 1..425</a> (first 97 terms from Ray Chandler, terms 98-365 from Amiram Eldar)
%H James Buddenhagen, <a href="https://groups.google.com/group/sci.math/msg/f2087083936610d3">Circle with 3 lattice points, thread in sci.math</a> (May 2002)
%H Hugo Pfoertner, <a href="http://www.randomwalk.de/sequences/a071383.pdf">Construction of the sequences A071383, A071384, A071385</a>
%F 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
%F lim n ->infinity Log(a(n))/n = 1. [Conjectured by _Benoit Cloitre_, proved by _J. H. Conway_]
%o (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
%o (Python)
%o from math import prod
%o from sympy import isprime
%o primes_congruent_1_mod_4 = [5]
%o def prime_4k_plus_1(i): # the i-th prime that is congruent to 1 mod 4
%o while i>=len(primes_congruent_1_mod_4): # generate primes on demand
%o n = primes_congruent_1_mod_4[-1]+4
%o while not isprime(n): n += 4
%o primes_congruent_1_mod_4.append(n)
%o return primes_congruent_1_mod_4[i]
%o def generate_A054994():
%o TO_DO = {(1,())}
%o while True:
%o radius, exponents = min(TO_DO)
%o yield radius, exponents
%o TO_DO.remove((radius, exponents))
%o TO_DO.update(successors(radius,exponents))
%o def successors(r,exponents):
%o for i,e in enumerate(exponents):
%o if i==0 or exponents[i-1]>e:
%o yield (r*prime_4k_plus_1(i), exponents[:i]+(e+1,)+exponents[i+1:])
%o if exponents==() or exponents[-1]>0:
%o yield (r*prime_4k_plus_1(len(exponents)), exponents+(1,))
%o n,record,radius=1,1,0
%o print(radius, end="") # or record, for A071385
%o for radius,expo in generate_A054994():
%o num_points = 4*prod((e+1) for e in expo)
%o if num_points>record:
%o record = num_points
%o n += 1
%o print (",", radius, end="") # or record, for A071385
%o if n==27: break
%o print()
%o # _Günter Rote_, Sep 12 2023
%Y 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.
%Y Indices of records of Sum_{d|n} kronecker(m, d): A230655 (m=-3), this sequence (m=-4), A279541 (m=-6).
%K nonn
%O 1,3
%A _Hugo Pfoertner_, May 23 2002
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