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

1,1

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.

Links

Table of n, a(n) for n=1..47.

Wikipedia, Euler's sum of powers conjecture

Example

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.

Prog

(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

Crossrefs

Cf. A023042, A025418, A346137, A316359.

Sequence in context: A318484 A079843 A290582 * A076941 A357198 A006779

Adjacent sequences: A346068 A346069 A346070 * A346072 A346073 A346074

Keyword

nonn,more

Author

Sebastian Magee, Jul 30 2021

Extensions

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

More terms from David A. Corneth, Sep 04 2021

Status

approved