A025418 -id:A025418 - cp's OEIS frontend

A025456 Number of partitions of n into 3 positive cubes.

0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 0

David W. Wilson

nonn

If A025455(n) > 0 then a(n + k^3) > 0 for k>0; a(A119977(n))>0; a(A003072(n))>0. - Reinhard Zumkeller, Jun 03 2006
a(A057904(n))=0; a(A003072(n))>0; a(A025395(n))=1; a(A008917(n))>1; a(A025396(n))=2. - Reinhard Zumkeller, Apr 23 2009
The first term > 1 is a(251) = 2. - Michel Marcus, Apr 23 2019

Charles R Greathouse IV, Table of n, a(n) for n = 0..10000
Index entries for sequences related to sums of cubes

Least inverses are A025418.

Cf. A025455, A003108, A003072 (1 or more ways), A008917 (two or more ways), A025395-A025398.

A025456 := proc(n)
    local a,x,y,zcu ;
    a := 0 ;
    for x from 1 do
        if 3*x^3 > n then
            return a;
        end if;
        for y from x do
            if x^3+2*y^3 > n then
                break;
            end if;
            zcu := n-x^3-y^3 ;
            if isA000578(zcu) then
                a := a+1 ;
            end if;
        end do:
    end do:
end proc: # R. J. Mathar, Sep 15 2015

a[n_] := Count[ PowersRepresentations[n, 3, 3], pr_List /; FreeQ[pr, 0]]; Table[a[n], {n, 0, 107}] (* Jean-François Alcover, Oct 31 2012 *)

a(n)=sum(a=sqrtnint(n\3,3),sqrtnint(n,3),sum(b=1,a,my(C=n-a^3-b^3,c);ispower(C,3,&c)&&0Charles R Greathouse IV, Jun 26 2013

a(n) = [x^n y^3] Product_{k>=1} 1/(1 - y*x^(k^3)). - Ilya Gutkovskiy, Apr 23 2019

Second offset from Michel Marcus, Apr 23 2019

A025419 Least sum of 3 distinct positive cubes in exactly n ways.

Original entry on oeis.org

36, 1009, 5104, 13896, 161568, 1296378, 2016496, 2562624, 14926248, 34012224, 69190848, 150547032, 119095488, 1204376256, 952763904, 1592865000, 3974344704, 2176782336, 10077696000, 2985984000, 36330467328, 30723115968, 23887872000, 17414258688, 72825163776, 75686967000, 204141384000, 62099136000, 139314069504, 245784927744, 80621568000, 191102976000, 272097792000, 373248000000

Offset: 1

David W. Wilson

nonn

a(36) = 496793088000. a(37) = 980922617856. a(38) = 209584584000. a(39) = 644972544000. a(42) = 970299000000. - Donovan Johnson, Nov 13 2010

Cf. A025418.

{min k: A025469(k) = n}. - R. J. Mathar, Jun 15 2018

a(6)-a(34) from Donovan Johnson, Nov 13 2010

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

A255018 Smallest number that is the sum of 3 nonnegative cubes in exactly n ways.

Original entry on oeis.org

4, 0, 216, 5104, 13896, 161568, 1259712, 2016496, 2562624, 14926248, 58995000, 34012224, 150547032, 471960000, 119095488, 1259712000, 952763904, 5159780352, 3974344704, 2176782336, 10077696000, 2985984000, 36330467328, 30723115968, 23887872000, 17414258688, 72825163776, 75686967000

Offset: 0

Alex Ratushnyak, Feb 25 2015

nonn

			a(0) = 4 because the smallest number that cannot be represented as a sum of 3 nonnegative cubes is 4.
a(1) = 0 is the sum of three 0's.
a(2) = 216 = 3^3 + 4^3 + 5^3 = 6^3 + 0 + 0.
a(3) = 5104 = 1 + 12^3 + 15^3 = 2^3 + 10^3 + 16^3 = 9^3 + 10^3 + 15^3.

Cf. A000578, A000446, A011541, A025418, A025419.

TOP = 6000000
a = [0]*TOP
for b in range(TOP):
  b3 = b**3
  if b3*3>=TOP: break
  for c in range(b,TOP):
    c3 = b3 + c**3
    if c3>=TOP: break
    for d in range(c,TOP):
      res = c3 + d**3
      if res>=TOP: break
      a[res] += 1
m = max(a)
r = [-1] * (m+1)
for i in range(TOP):
    if r[a[i]]==-1:  r[a[i]]=i
print(r)

More terms from Rémy Sigrist, Jul 14 2020

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A025456 Number of partitions of n into 3 positive cubes.

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A025419 Least sum of 3 distinct positive cubes in exactly n ways.

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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.

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A255018 Smallest number that is the sum of 3 nonnegative cubes in exactly n ways.

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