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 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. 1
 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 (list; graph; refs; listen; history; text; internal format)
 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 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 A006779 A003208 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

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Last modified December 2 11:59 EST 2021. Contains 349440 sequences. (Running on oeis4.)