cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

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

Original entry on oeis.org

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

Views

Author

Hugo Pfoertner, May 23 2002

Keywords

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

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]

A071385 Number of points (i,j) on the circumference of a circle around (0,0) with squared radius A071383(n).

Original entry on oeis.org

1, 4, 8, 12, 16, 24, 32, 36, 48, 64, 72, 80, 96, 128, 144, 160, 192, 256, 288, 320, 384, 512, 576, 640, 768, 864, 1024, 1152, 1280, 1536, 1728, 2048, 2304, 2560, 3072, 3456, 3840, 4096, 4608, 5120, 6144, 6912, 7680, 8192, 9216, 10240, 11520, 12288, 13824, 15360
Offset: 1

Views

Author

Hugo Pfoertner, May 23 2002

Keywords

Examples

			Circles with radius 1 and 2 have 4 lattice points on their circumference, so a(1)=4. A circle with radius sqrt(5) passes through 8 lattice points of the shape (2,1), so a(2)=8. A circle with radius 5 passes through 4 lattice points of shape (5,0) and through 8 points of shape (4,3), so a(3)=4+8=12
A071383(11) = 5^2 * 13^2 * 17^1 = 71825. Therefore A071385(11) = 4*(2+1)*(2+1)*(1+1) = 72.
		

Crossrefs

Programs

  • PARI
    my(v=list(10^15), rec=0); print1(1, ", "); for(n=1, #v, if(numdiv(v[n])>rec, rec=numdiv(v[n]); print1(4*rec, ", "))) \\ Jianing Song, May 20 2021, see program for A054994

Formula

a(n) = 4 * Product_{k=1..klim} (e_k + 1), where klim and e_1 >= e_2 >= ... >= e_klim > 0 are known from A071383(n) = Product_{k=1..klim} p_k^e_k, with p_k = k-th prime of the form 4i+1. (J. H. Conway)
a(n) = 4*A000005(A071383(n)) for n > 1.
a(n) = A004018(A071383(n)).
a(n) = A002654(A071383(n)) for n > 1. - Jianing Song, May 20 2021
a(n) = 4*A344470(n-1) for n > 1. - Hugo Pfoertner, Sep 04 2022

A072324 Terms of A071383 such that A071383(n) = 5 * A071383(n-1).

Original entry on oeis.org

5, 25, 325, 27625, 801125, 29641625, 1215306625
Offset: 1

Views

Author

Hugo Pfoertner, Jul 14 2002

Keywords

Comments

This sequence was proposed by Benoit Cloitre.
The next term of this sequence if it exists is greater than 2^63-1 (~9.2*10^18).
No more terms through A071383(365). - Hugo Pfoertner, Sep 12 2019
No more terms through A071383(425). - Ray Chandler, Dec 18 2019

Crossrefs

Cf. A071383.
Showing 1-3 of 3 results.