A346071 - OEIS

%I #72 Oct 05 2021 22:43:34

%S 6,18,54,87,108,174,174,324,324,324,492,492,492,984,984,1296,1296,

%T 1296,1440,1440,2592,2592,2592,2592,3960,3960,3960,3960,4320,4320,

%U 4320,5760,5940,5940,5940,5940,5940,5940,8640,9900,9900,9900,11880,11880,11880,11880,11880

%N 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.

%C 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.

%C 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.

%C 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.

%C The first of this family is A023042.

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Euler%27s_sum_of_powers_conjecture">Euler's sum of powers conjecture</a>

%e a(1) = 6 because 6^3 = 5^3 + 4^3 + 3^3; 6 = a(1) = A023042(1).

%e a(2) = 18 because 18^3 = 15^3 + 12^3 + 9^3 = 16^3 + 12^3 + 2^3.

%e a(3) = 54 because 54^3 = 45^3 + 36^3 + 27^3 = 48^3 + 36^3 + 6^3 = 53^3 + 19^3 + 12^3.

%o (Python)

%o import numpy as np

%o def residual(a,b,c,d, exp=3):

%o     return a**exp-b**exp-c**exp-d**exp

%o def test(max_n,k=3):

%o     ans=dict()

%o     for a in range(max_n):

%o         #print(a)

%o         for b in range(int(np.ceil((a**k/3)**(1/k))),a):

%o             n3=a**k-b**k

%o             for c in range(int(np.ceil((n3/2)**(1/k))),b):

%o                 m3=n3-c**k

%o                 if m3<0:

%o                     break;

%o                 l=int(np.ceil((m3)**(1/k)))

%o                 options=[l,l-1]

%o                 for d in options:

%o                     res=residual(a,b,c,d, exp=k)

%o                     if res==0:

%o                         if a in ans.keys():

%o                             ans[a].append((a,b,c,d))

%o                         else:

%o                             ans[a]=[(a,b,c,d)]

%o                         #print("found:",(a,b,c,d))

%o                         break

%o                     else:

%o                         #print("tested: {0}, residual: {1}".format((a,b,c,d),res))

%o                         if res>0:

%o                             break

%o     return ans

%o def serie(N):

%o     result=test(N)

%o     results_by_number_of_answers=[]

%o     results_by_number_of_answers.append(result)

%o     temp=dict()

%o     for k in result.keys():

%o         if len(result[k])>=2:

%o             temp[k]=result[k]

%o     results_by_number_of_answers.append(temp)

%o     i=3

%o     while len(temp)>0:

%o         temp=dict()

%o         for k in results_by_number_of_answers[-1].keys():

%o             if len(results_by_number_of_answers[-1][k])>=i:

%o                 temp[k]=result[k]

%o         if len(temp)>0:

%o             results_by_number_of_answers.append(temp)

%o         i+=1

%o     return [next(iter(a)) for a in results_by_number_of_answers]

%o #Get the elements of the serie up until A_n>1000

%o A=serie(1000)

%o print(A)

%o (Python)

%o from itertools import combinations

%o from collections import Counter

%o from sympy import integer_nthroot

%o def icbrt(n): return integer_nthroot(n, 3)[0]

%o def aupto(mmax):

%o     cbs = [i**3 for i in range(mmax+1)]

%o     cbsset = set(cbs)

%o     c = Counter(sum(c) for c in combinations(cbs, 3) if sum(c) in cbsset)

%o     nmax = max(c.values())

%o     return [min(icbrt(s) for s in c if c[s] >= n) for n in range(1, nmax+1)]

%o print(aupto(500)) # _Michael S. Branicky_, Sep 04 2021

%Y Cf. A023042, A025418, A346137, A316359.

%K nonn,more

%O 1,1

%A _Sebastian Magee_, Jul 30 2021

%E a(16)-a(31) from _Jinyuan Wang_, Aug 02 2021

%E More terms from _David A. Corneth_, Sep 04 2021