A346071 a(n) is the smallest number m such that m^3 = x^3 + y^3 + z^3, x > y > z > 0, has at least n different solutions.

6, 18, 54, 87, 108, 174, 174, 324, 324, 324, 492, 492, 492, 984, 984, 1296, 1296, 1296, 1440, 1440, 2592, 2592, 2592, 2592, 3960, 3960, 3960, 3960, 4320, 4320, 4320, 5760, 5940, 5940, 5940, 5940, 5940, 5940, 8640, 9900, 9900, 9900, 11880, 11880, 11880, 11880, 11880

Offset: 1

Author

Sebastian Magee, Jul 30 2021

Keywords

• nonn
• more

Comments

a(n) is the smallest number for which there are at least n sets of positive integers (b_i, c_i, d_i) i=1..n which satisfy the equation a(n)^3 = b_i^3 + c_i^3 + d_i^3.

This sequence is related to Euler's sum of powers conjecture. In particular to the case k=3, a(n) is the smallest number that has at least n different solutions to the equation.

The sequences of numbers whose cubes can be expressed as the sum of 3 positive cubes in at least n ways for n = 1, 2, 3, ... form a family of related sequences. This sequence is the sequence of first terms in that family of sequences.

The first of this family is A023042.

Examples

			a(1) = 6 because 6^3 = 5^3 + 4^3 + 3^3; 6 = a(1) = A023042(1).
a(2) = 18 because 18^3 = 15^3 + 12^3 + 9^3 = 16^3 + 12^3 + 2^3.
a(3) = 54 because 54^3 = 45^3 + 36^3 + 27^3 = 48^3 + 36^3 + 6^3 = 53^3 + 19^3 + 12^3.
		

Links

• Wikipedia, Euler's sum of powers conjecture

Crossrefs

Cf. A023042, A025418, A346137, A316359.

Programs

Python

import numpy as np
def residual(a,b,c,d, exp=3):
    return a**exp-b**exp-c**exp-d**exp
def test(max_n,k=3):
    ans=dict()
    for a in range(max_n):
        #print(a)
        for b in range(int(np.ceil((a**k/3)**(1/k))),a):
            n3=a**k-b**k
            for c in range(int(np.ceil((n3/2)**(1/k))),b):
                m3=n3-c**k
                if m3<0:
                    break;
                l=int(np.ceil((m3)**(1/k)))
                options=[l,l-1]
                for d in options:
                    res=residual(a,b,c,d, exp=k)
                    if res==0:
                        if a in ans.keys():
                            ans[a].append((a,b,c,d))
                        else:
                            ans[a]=[(a,b,c,d)]
                        #print("found:",(a,b,c,d))
                        break
                    else:
                        #print("tested: {0}, residual: {1}".format((a,b,c,d),res))
                        if res>0:
                            break
    return ans
def serie(N):
    result=test(N)
    results_by_number_of_answers=[]
    results_by_number_of_answers.append(result)
    temp=dict()
    for k in result.keys():
        if len(result[k])>=2:
            temp[k]=result[k]
    results_by_number_of_answers.append(temp)
    i=3
    while len(temp)>0:
        temp=dict()
        for k in results_by_number_of_answers[-1].keys():
            if len(results_by_number_of_answers[-1][k])>=i:
                temp[k]=result[k]
        if len(temp)>0:
            results_by_number_of_answers.append(temp)
        i+=1
    return [next(iter(a)) for a in results_by_number_of_answers]
#Get the elements of the serie up until A_n>1000
A=serie(1000)
print(A)

Python

from itertools import combinations
from collections import Counter
from sympy import integer_nthroot
def icbrt(n): return integer_nthroot(n, 3)[0]
def aupto(mmax):
    cbs = [i**3 for i in range(mmax+1)]
    cbsset = set(cbs)
    c = Counter(sum(c) for c in combinations(cbs, 3) if sum(c) in cbsset)
    nmax = max(c.values())
    return [min(icbrt(s) for s in c if c[s] >= n) for n in range(1, nmax+1)]
print(aupto(500)) # Michael S. Branicky, Sep 04 2021

Extensions

a(16)-a(31) from Jinyuan Wang, Aug 02 2021

More terms from David A. Corneth, Sep 04 2021