Sebastian Magee - cp's OEIS frontend

User: Sebastian Magee

Sebastian Magee has authored 2 sequences.

A346137 Numbers k such that k^3 = x^3 + y^3 + z^3, x > y > z >= 0, has at least 2 distinct solutions.

18, 36, 41, 46, 54, 58, 60, 72, 75, 76, 81, 82, 84, 87, 88, 90, 92, 96, 100, 108, 114, 116, 120, 123, 126, 132, 134, 138, 140, 142, 144, 145, 150, 152, 156, 159, 160, 162, 164, 168, 170, 171, 174, 176, 178, 180, 184, 185, 186, 189, 190, 192, 198, 200, 201, 202, 203

Offset: 1

Sebastian Magee, Jul 30 2021

nonn

This sequence is based on a generalization of Fermat's last theorem with n=3, in which three terms are added. Fermat's Theorem states that there are no solution with only two terms, this sequence shows there are many integers for which there are multiple solutions if three terms are allowed. The sequence is also related to the Taxicab numbers.

			41 is in the sequence because 41^3 = 33^3 + 32^3 + 6^3 = 40^3 + 17^3 + 2^3.

Subsequence of A023042.

Cf. A001235, A346071.

q[k_] := Count[IntegerPartitions[k^3, {3}, Range[0, k-1]^3], ?(UnsameQ @@ # &)] > 1; Select[Range[200], q] (* _Amiram Eldar, Sep 03 2021 *)

from itertools import combinations
from collections import Counter
from sympy import integer_nthroot
def icuberoot(n): return integer_nthroot(n, 3)[0]
def aupto(kmax):
    cubes = [i**3 for i in range(kmax+1)]
    cands, cubesset = (sum(c) for c in combinations(cubes, 3)), set(cubes)
    c = Counter(s for s in cands if s in cubesset)
    return sorted(icuberoot(s) for s in c if c[s] >= 2)
print(aupto(203)) # Michael S. Branicky, Sep 04 2021

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.

Original entry on oeis.org

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

Sebastian Magee, Jul 30 2021

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.

			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.

Wikipedia, Euler's sum of powers conjecture

Cf. A023042, A025418, A346137, A316359.

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)

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

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

More terms from David A. Corneth, Sep 04 2021

cp's OEIS Frontend

User: Sebastian Magee

A346137 Numbers k such that k^3 = x^3 + y^3 + z^3, x > y > z >= 0, has at least 2 distinct solutions.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Python

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.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Python

Python

Extensions