A023042 -id:A023042 - cp's OEIS frontend

A003072 Numbers that are the sum of 3 positive cubes.

3, 10, 17, 24, 29, 36, 43, 55, 62, 66, 73, 80, 81, 92, 99, 118, 127, 129, 134, 136, 141, 153, 155, 160, 179, 190, 192, 197, 216, 218, 225, 232, 244, 251, 253, 258, 270, 277, 281, 288, 307, 314, 342, 344, 345, 349, 352, 359, 368, 371, 375, 378, 397, 405, 408, 415, 433, 434

Offset: 1

N. J. A. Sloane, David W. Wilson

A119977 is a subsequence; if m is a term then there exists at least one k>0 such that m-k^3 is a term of A003325. - Reinhard Zumkeller, Jun 03 2006
A025456(a(n)) > 0. - Reinhard Zumkeller, Apr 23 2009
Davenport proved that a(n) << n^(54/47 + e) for every e > 0. - Charles R Greathouse IV, Mar 26 2012

			a(11) = 73 = 1^3 + 2^3 + 4^3, which is sum of three cubes.
a(15) = 99 = 2^3 + 3^3 + 4^3, which is sum of three cubes.

K. D. Bajpai, Table of n, a(n) for n = 1..12955 (first 1000 terms from T. D. Noe)
H. Davenport, Sums of three positive cubes, J. London Math. Soc., 25 (1950), 339-343. Coll. Works III p. 999.
Eric Weisstein's World of Mathematics, Cubic Number
Index entries for sequences related to sums of cubes

Subsequence of A004825.
Cf. A003325, A024981, A057904 (complement), A010057, A000578, A023042 (subsequence of cubes).
Cf. A###### (x, y) = Numbers that are the sum of x nonzero y-th powers:
- squares: A000404 (2, 2), A000408 (3, 2), A000414 (4, 2), A047700 (5, 2);
- cubes: A003325 (2, 3), A003072 (3, 3), A003327 (4, 3), A003328 (5, 3), A003329 (6, 3), A003330 (7, 3), A003331 (8, 3), A003332 (9, 3), A003333 (10, 3), A003334 (11, 3), A003335 (12, 3);
- fourth powers: A003336 (2, 4), A003337 (3, 4), A003338 (4, 4), A003339 (5, 4), A003340 (6, 4), A003341 (7, 4), A003342 (8, 4), A003343 (9, 4), A003344 (10, 4), A003345 (11, 4), A003346 (12, 4);
- fifth powers: A003347 (2, 5), A003348 (3, 5), A003349 (4, 5), A003350 (5, 5), A003351 (6, 5), A003352 (7, 5), A003353 (8, 5), A003354 (9, 5), A003355 (10, 5), A003356 (11, 5), A003357 (12, 5);
- sixth powers: A003358 (2, 6), A003359 (3, 6), A003360 (4, 6), A003361 (5, 6), A003362 (6, 6), A003363 (7, 6), A003364 (8, 6), A003365 (9, 6), A003366 (10, 6), A003367 (11, 6), A003368 (12, 6);
- seventh powers: A003369 (2, 7), A003370 (3, 7), A003371 (4, 7), A003372 (5, 7), A003373 (6, 7), A003374 (7, 7), A003375 (8, 7), A003376 (9, 7), A003377 (10, 7), A003378 (11, 7), A003379 (12, 7);
- eighth powers: A003380 (2, 8), A003381 (3, 8), A003382 (4, 8), A003383 (5, 8), A003384 (6, 8), A003385 (7, 8), A003386 (8, 8), A003387 (9, 8), A003388 (10, 8), A003389 (11, 8), A003390 (12, 8);
- ninth powers: A003391 (2, 9), A003392 (3, 9), A003393 (4, 9), A003394 (5, 9), A003395 (6, 9), A003396 (7, 9), A003397 (8, 9), A003398 (9, 9), A003399 (10, 9), A004800 (11, 9), A004801 (12, 9);
- tenth powers: A004802 (2, 10), A004803 (3, 10), A004804 (4, 10), A004805 (5, 10), A004806 (6, 10), A004807 (7, 10), A004808 (8, 10), A004809 (9, 10), A004810 (10, 10), A004811 (11, 10), A004812 (12, 10);
- eleventh powers: A004813 (2, 11), A004814 (3, 11), A004815 (4, 11), A004816 (5, 11), A004817 (6, 11), A004818 (7, 11), A004819 (8, 11), A004820 (9, 11), A004821 (10, 11), A004822 (11, 11), A004823 (12, 11).

a003072 n = a003072_list !! (n-1)
a003072_list = filter c3 [1..] where
   c3 x = any (== 1) $ map (a010057 . fromInteger) $
                       takeWhile (> 0) $ map (x -) $ a003325_list
-- Reinhard Zumkeller, Mar 24 2012

isA003072 := proc(n)
    local x,y,z;
    for x from 1 do
        if 3*x^3 > n then
            return false;
        end if;
        for y from x do
            if x^3+2*y^3 > n then
                break;
            end if;
            if isA000578(n-x^3-y^3) then
                return true;
            end if;
        end do:
    end do:
end proc:
for n from 1 to 1000 do
    if isA003072(n) then
        printf("%d,",n) ;
    end if;
end do: # R. J. Mathar, Jan 23 2016

Select[Range[435], (p = PowersRepresentations[#, 3, 3]; (Select[p, #[[1]] > 0 && #[[2]] > 0 && #[[3]] > 0 &] != {})) &] (* Jean-François Alcover, Apr 29 2011 *)
With[{upto=500},Select[Union[Total/@Tuples[Range[Floor[Surd[upto-2,3]]]^3,3]],#<=upto&]] (* Harvey P. Dale, Oct 25 2021 *)

sum(n=1,11,x^(n^3),O(x^1400))^3 /* Then [i|i<-[1..#%],polcoef(%,i)] gives the list of powers with nonzero coefficient. - M. F. Hasler, Aug 02 2020 */

list(lim)=my(v=List(),k,t); lim\=1; for(x=1,sqrtnint(lim-2,3), for(y=1, min(sqrtnint(lim-x^3-1,3),x), k=x^3+y^3; for(z=1,min(sqrtnint(lim-k,3), y), listput(v, k+z^3)))); Set(v) \\ Charles R Greathouse IV, Sep 14 2015

{n: A025456(n) >0}. - R. J. Mathar, Jun 15 2018

Incorrect program removed by David A. Corneth, Aug 01 2020

A226903 Shiraishi numbers: a parametrized family of solutions c to the Diophantine equation a^3 + b^3 + c^3 = d^3 with d = c+1.

Original entry on oeis.org

5, 18, 53, 102, 197, 306, 491, 684, 989, 1290, 1745, 2178, 2813, 3402, 4247, 5016, 6101, 7074, 8429, 9630, 11285, 12738, 14723, 16452, 18797, 20826, 23561, 25914, 29069, 31770, 35375, 38448, 42533, 46002, 50597, 54486, 59621, 63954, 69659, 74460, 80765, 86058

Offset: 1

Jonathan Sondow, Jun 22 2013

Shiraishi's solutions to a^3 + b^3 + c^3 = d^3 are a = 3n^2; b = 6n^2 - 3n + 1 or 6n^2 + 3n + 1; c = 9n^3 - 6n^2 + 3n - 1 or  9n^3 + 6n^2 + 3n, respectively, for n > 0; and d = c+1. See Smith and Mikami for a derivation.
Shiraishi's formulas show that the sequence is infinite. Hence the sequences A023042 (solutions to x^3 + y^3 + z^3 = w^3), A225908 (solutions to a^3 + b^3 = c^3 - d^3), A225909 (solutions to a^3 + b^3 = (c+1)^3 - c^3) and A226902 (numbers c in A225909) are also infinite.
Shiraishi's solution b = 6n^2 +/- 3n + 1 is the centered triangular numbers A005448 except 1.

			The first two terms are a(1) = 9 - 6 + 3 - 1 = 5 and a(2) = 9 + 6 + 3 = 18. Then Shiraishi's formulas give 3^3 + 4^3 + 5^3 = 6^3 and 3^3 + 10^3 + 18^3 = 19^3.

Shiraishi Chochu (aka Shiraishi Nagatada), Shamei Sampu (Sacred Mathematics), 1826.

David Eugene Smith and Yoshio Mikami, A History of Japanese Mathematics, Open Court, Chicago, 1914; Dover reprint, 2004; pp. 233-235.
Wikipedia (French), Shiraishi Nagatada
Wikipedia (German), Shiraishi Nagatada
Index entries for sequences related to sums of cubes
Index entries for linear recurrences with constant coefficients, signature (1,3,-3,-3,3,1,-1).

Cf. A003325, A005448, A023042, A181123, A225908, A225909, A226902.

a(2n-1) = 9n^3 - 6n^2 + 3n - 1.
a(2n) = 9n^3 + 6n^2 + 3n.
G.f.: x*(5 + 13*x + 20*x^2 + 10*x^3 + 5*x^4 + x^5) / ((1 + x)^3*(1 - x)^4). [Bruno Berselli, Jun 22 2013]
a(n) = (18*n^3 + 27*n^2 + 27*n + 1 - (3*n^2 + 3*n + 1)*(-1)^n)/16. [Bruno Berselli, Jun 22 2013]
a(n) = a(n-1) + 3*a(n-2) - 3*a(n-3) - 3*a(n-4) + 3*a(n-5) + a(n-6) - a(n-7) for n > 7. - Chai Wah Wu, Aug 05 2025

A114923 Primes p such that there exist three primes q, r and s with p^3=q^3+r^3+s^3.

Original entry on oeis.org

709, 1033, 2767, 2791, 2917, 3727, 3769, 5647, 5657, 5737, 7039, 7321, 8089, 8291, 8387, 9433, 9473, 9851, 12073, 12343, 13417, 14083, 14561, 14723, 14831, 14969, 15313, 18127, 19841, 25033, 28081, 28477, 29153, 29179, 32771, 33161, 33199, 33377, 34337, 36713

Offset: 1

Carlos Rivera and Farideh Firoozbakht, Jan 09 2006

nonn

The sets of three primes corresponding to the first seven terms of the sequence are respectively {193,461,631}, {599,691,823}, {103,2179,2213}, {769,1879,2447}, {31,1951,2591}, {1399,1667,3541} and {11,1783,3631}. - Robert G. Wilson v, Jan 09 2006
The sets of three primes corresponding to the next eight terms of the sequence are respectively {2251, 3121, 5171}, {1487, 2731, 5399}, {839, 3691, 5167}, {2099, 2377, 6883}, {3163, 5443, 5843}, {1621, 6323, 6481}, {2357, 4999, 7559} and {1621, 5297, 7589}. - Robert G. Wilson v, Jan 09 2006
The indices of the primes: 127,174,403,406,422,520,525,742,745,754,905,933,1017,1040,1050, ..., . - Robert G. Wilson v, Jan 09 2006
The sets of three primes corresponding to the terms 12073, 12343, 13417, 14083, 14561, 14723, 14831, 14969, 15313, 18127, 19841 and 25033 are respectively {4007, 4327, 11731}, {373, 9209, 10321}, {5099, 7561, 12277}, {4639, 7129, 13259}, {1997, 8599, 13469}, {3881, 6427, 14207}, {6257, 9439, 12959}, {2239, 5189, 14741}, {2269, 2969, 15259}, {2129, 5227, 17971}, {3931, 15263, 16127} and {4093, 19391, 20269}. The indices of the primes: 127, 174, 403, 406, 422, 520, 525, 742, 745, 754, 905, 933, 1017, 1040, 1050, 1168, 1174, 1215, 1446, 1474, 1591, 1661, 1707, 1723, 1738, 1753, 1789, 2077, 2244, 2765. - Farideh Firoozbakht, Jan 27 2006

			The prime number 3769 is in the sequence because we have 3769^3=11^3+1783^3+3631^3 and three numbers 11, 1783 and 3631 are primes.

Charles R Greathouse IV, Table of n, a(n) for n = 1..776 (first 121 terms from Chai Wah Wu)
G. L. Honaker, Jr. and Chris Caldwell, Prime Pages.
Carlos Rivera, P^3=a^3+b^3+c^3, {P, a, b, c} = primes, Puzzle 48, The Prime Puzzles & Problems Connection.

Subset of A023042.

N:= 20000: # to get all terms <= N
Primes:= select(isprime, [2,seq(i,i=3..N,2)]):
P2:= {seq(seq(Primes[i]^3 + Primes[j]^3, j=1..i),i=1..nops(Primes))}:
Q:= convert(map(t->-t^3,Primes),set):
filter:= p -> P2 intersect map(`+`,Q,p^3) <> {}:
select(filter, Primes); # Robert Israel, Jan 11 2016

t = {}; Do[ If[p = (Prime[q]^3 + Prime[r]^3 + Prime[s]^3)^(1/3); PrimeQ[p], AppendTo[t, p]; Print[{p, Prime[s], Prime[r], Prime[q]}]], {q, 3, 1059}, {r, q-1}, {s, r-1}]; t (* Robert G. Wilson v, Jan 09 2006 *)

is(p)=my(p3=p^3,a3,A,c);if(isprimepower(p3-16)==3, return(1)); forprime(a=sqrtnint(p3\3,3),sqrtnint(p3-54,3), a3=a^3; A=p3-a3; forprime(b=3,min(sqrtnint(A,3),a), if(ispower(A-b^3,3,&c) && isprime(c), return(isprime(p))))) \\ Charles R Greathouse IV, Nov 24 2017

a(8)-a(18) from Robert G. Wilson v, Jan 09 2006
a(19)-a(30) from Farideh Firoozbakht, Jan 27 2006
a(31)-a(40) from Chai Wah Wu, Jan 10 2016

A225908 Numbers that are both a sum and a difference of two positive cubes.

Original entry on oeis.org

91, 152, 189, 217, 513, 728, 1027, 1216, 1512, 1736, 2457, 3087, 4104, 4706, 4921, 4977, 5103, 5256, 5824, 5859, 6832, 7657, 8216, 8587, 9728, 10712, 11375, 12096, 12691, 13851, 13888, 14911, 15093, 15561, 16120, 16263, 19000, 19656, 21014, 23058, 23625, 24696

Offset: 1

Jonathan Sondow, Jun 21 2013

nonn

Solutions x to the equations x = a^3 + b^3 = c^3 - d^3 in positive integers.
The intersection of A003325 and A181123. See those sequences for additional comments, references, links and cross-refs.
Suggested by Shiraishi's solutions to Gokai Ampon's equation u^3 + v^3 + w^3 = n^3 (transpose a term from the left side to the right side). See A023042 and A226903.
An infinite subsequence is (A226904(n)+1)^3 - A226904(n)^3.

			3^3 + 4^3 + 5^3 = 6^3, so 3^3 + 4^3 = 91 and 3^3 + 5^3 = 152 and 4^3 + 5^3 = 189 are members.

Shiraishi Chochu (aka Shiraishi Nagatada), Shamei Sampu (Sacred Mathematics), 1826.

Chai Wah Wu, Table of n, a(n) for n = 1..10000 (terms 1..1000 from T. D. Noe)
David Eugene Smith and Yoshio Mikami, A History of Japanese Mathematics, Open Court, Chicago, 1914; Dover reprint, 2004; pp. 233-235.
Wikipedia (French), Shiraishi Nagatada
Wikipedia (German), Shiraishi Nagatada
Index entries for sequences related to sums of cubes

Cf. A003325, A023042, A181123, A226903, A226904.

nn = 3*10^4; t1 = Union[Flatten[Table[x^3 + y^3, {x, nn^(1/3)}, {y, x, (nn - x^3)^(1/3)}]]]; p = 3; t2 = Union[Reap[Do[n = i^p - j^p; If[n <= nn, Sow[n]], {i, Ceiling[(nn/p)^(1/(p - 1))]}, {j, i}]][[2, 1]]]; Intersection[t1, t2] (* T. D. Noe, Jun 21 2013 *)

A306213 Numbers that are the sum of cubes of three distinct positive integers in arithmetic progression.

Original entry on oeis.org

36, 99, 153, 216, 288, 405, 408, 495, 645, 684, 792, 855, 972, 1071, 1197, 1224, 1407, 1548, 1584, 1701, 1728, 1968, 2079, 2241, 2304, 2403, 2541, 2673, 2736, 3051, 3060, 3240, 3264, 3537, 3540, 3888, 3960, 4059, 4131, 4257, 4500, 4587, 4833, 5049, 5160, 5256, 5472, 5643, 5832, 5940, 6336, 6369, 6669

Offset: 1

Antonio Roldán, Jan 29 2019

nonn

Numbers that can be written as 3*a*(a^2 + 2*b^2) = (a-b)^3 + a^3 + (a+b)^3 where 0 < b < a. - Robert Israel, Dec 15 2022

			153 = 1^3 + 3^3 + 5^3, with 3 - 1 = 5 - 3 = 2;
972 = 3^3 + 6^3 + 9^3, with 6 - 3 = 9 - 6 = 3.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A000578, A023042, A024975, A122723, A306212, A306214.

N:= 10000: # for terms <= N
S:= {}:
for a from 1 while a^3 + (a+1)^3 + (a+2)^3 <= N do
  for d from 1 do
    x:= a^3 + (a+d)^3 + (a+2*d)^3;
    if x > N then break fi;
    S:= S union {x}
od od:
sort(convert(S,list)); # Robert Israel, Dec 14 2022

for(n=3, 7000, k=(n/3)^(1/3); a=2; v=0; while(a<=k&&v==0, b=(n-3*a^3)/(6*a); if(b==truncate(b)&&issquare(b), d=sqrt(b), d=0); if(d>=1&&d<=a-1, v=1; print1(n,", ")); a+=1))

w=List(); for(n=3, 7000, k=(n/3)^(1/3); for(a=2, k, for(c=1, a-1, v=(a-c)^3+a^3+(a+c)^3; if(v==n, listput(w,n))))); print(vecsort(Vec(w),,8))

A297305 Numbers k such that k^4 can be written as a sum of five positive 4th powers.

Original entry on oeis.org

5, 10, 15, 20, 25, 30, 31, 35, 40, 45, 50, 55, 60, 62, 65, 70, 75, 80, 85, 89, 90, 93, 95, 100, 103, 105, 110, 115, 120, 124, 125, 130, 135, 140, 145, 150, 155, 160, 165, 170, 175, 178, 180, 185, 186, 190, 195, 200, 205, 206, 210, 215, 217, 220, 225, 230, 233

Offset: 1

Seiichi Manyama, Mar 16 2018

nonn

If k is in the sequence, then k*m is in the sequence for every positive integer m.

			    5^4 =  2^4 +  2^4 +  3^4 +  4^4 +   4^4 (=       625).
   31^4 = 10^4 + 10^4 + 10^4 + 17^4 +  30^4 (=    923521).
   89^4 = 10^4 + 35^4 + 52^4 + 60^4 +  80^4 (=  62742241).
  103^4 =  4^4 + 15^4 + 50^4 + 50^4 + 100^4 (= 112550881).

Jinyuan Wang, Table of n, a(n) for n = 1..500
Titus Piezas III, Ramanujan and the Quartic Equation 2^4+2^4+3^4+4^4+4^4 = 5^4, 2005.
Jinyuan Wang, All solutions to k^4 = a^4 + b^4 + c^4 + d^4 + e^4 with k < 999.
Eric Weisstein's World of Mathematics, Diophantine Equation 4th Powers.
Index to sequences related to Diophantine equations (4,1,5)

Cf. A000583, A003294, A023042, A301601, A386494.

a(43)-a(57) from Jon E. Schoenfield, Mar 17 2018

A328149 Numbers whose set of divisors contains a quadruple (x, y, z, w) satisfying x^3 + y^3 + z^3 = w^3.

Original entry on oeis.org

60, 72, 120, 144, 180, 216, 240, 288, 300, 360, 420, 432, 480, 504, 540, 576, 600, 648, 660, 720, 780, 792, 840, 864, 900, 936, 960, 1008, 1020, 1080, 1140, 1152, 1200, 1224, 1260, 1296, 1320, 1368, 1380, 1440, 1500, 1512, 1560, 1584, 1620, 1656, 1680, 1710

Offset: 1

Michel Lagneau, Jun 07 2020

nonn

The subsequence of numbers of the form 2^i*3^j is 72, 144, 216, 288, 432, 576, 648, 864, 1152, 1296, ...
The corresponding number of quadruples of the sequence is 1, 1, 2, 2, 2, 2, 3, 3, 2, 6, 2, 4, 4, 2, 3, 4, 4, 3, 2, 10, ... (see the sequence A328204).
The set of divisors of a(n) contains at least one primitive quadruple.
Examples:
  The set of divisors of a(1) = 60 contains only one primitive quadruple: (3, 4, 5, 6).
  The set of divisors of a(10) = 360 contains two primitive quadruples: (1, 6, 8, 9) and (3, 4, 5, 6).
From Robert Israel, Jul 06 2020: (Start)
Every multiple of a member of the sequence is in the sequence.
The first member of the sequence not divisible by 6 is a(68) = 2380, which has the quadruple (7, 14, 17, 20).
The first odd member of the sequence is a(1230) = 43065, which has the quadruple (11, 15, 27, 29). (End)

			120 is in the sequence because the set of divisors {1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120} contains the quadruples {3, 4, 5, 6} and {6, 8, 10, 12}. The first quadruple is primitive.

Y. Perelman, Solutions to x^3 + y^3 + z^3 = u^3, Mathematics can be Fun, pp. 316-9 Mir Moscow 1985.

Robert Israel, Table of n, a(n) for n = 1..10000
Fred Richman, Sums of Three Cubes

Cf. A023042, A027750, A096545, A328204, A330893.

with(numtheory):
for n from 3 to 2000 do :
   d:=divisors(n):n0:=nops(d):it:=0:
    for i from 1 to n0-3 do:
     for j from i+1 to n0-2 do :
      for k from j+1 to n0-1 do:
      for m from k+1 to n0 do:
       if d[i]^3 + d[j]^3 + d[k]^3 = d[m]^3
        then
        it:=it+1:
        else
       fi:
      od:
     od:
    od:
    od:
    if it>0 then
    printf(`%d, `,n):
    else fi:
   od:

nq[n_] := If[ Mod[n, 6]>0, 0, Block[{t, u, v, c = 0, d = Divisors[n], m}, m = Length@ d; Do[ t = d[[i]]^3 + d[[j]]^3; Do[u = t + d[[h]]^3; If[u > n^3, Break[]]; If[ Mod[n^3, u] == 0 && IntegerQ[v = u^(1/3)] && Mod[n, v] == 0, c++], {h, j+1, m - 1}], {i, m-3}, {j, i+1, m - 2}]; c]]; Select[ Range@ 1026, nq[#] > 0 &] (* program from Giovanni Resta adapted for the sequence. See A330893 *)

isok(n) = {my(d=divisors(n), m); if (#d > 3, for (i=1, #d-3, for (j=i+1, #d-2, for (k=j+1, #d-1, if (ispower(d[i]^3+d[j]^3+d[k]^3, 3, &m) && !(n%m), return (1));););););} \\ Michel Marcus, Nov 15 2020

A225909 Numbers that are both a sum of two positive cubes and a difference of two consecutive cubes.

Original entry on oeis.org

91, 217, 1027, 4921, 8587, 14911, 31519, 39331, 106597, 117019, 136747, 185257, 195841, 265519, 281827, 616987, 636181, 684019, 712969, 724717, 736561, 955981, 1200169, 1352737, 1405621, 1771777, 2481571, 2756167, 2937331, 4251871, 4996171, 5262901

Offset: 1

Jonathan Sondow, Jun 21 2013

nonn

Solutions x to the equations x = a^3 + b^3 = (c+1)^3 - c^3 in positive integers. The values of c are A226902.
The intersection of A003325 and A003215.
Subsequence of A225908 = numbers that are both a sum and a difference of two positive cubes.
Shiraishi's solution to Gokai Ampon's equation u^3 + v^3 + w^3 = n^3 (see A023042 and A226903) shows that the sequence is infinite.

			3^3 + 4^3 = 6^3 - 5^3 = 91, so 91 is a member.

Shiraishi Chochu (aka Shiraishi Nagatada), Shamei Sampu (Sacred Mathematics), 1826.

Chai Wah Wu, Table of n, a(n) for n = 1..5000
David Eugene Smith and Yoshio Mikami, A History of Japanese Mathematics, Open Court, Chicago, 1914; Dover reprint, 2004; pp. 233-235.
Wikipedia (French), Shiraishi Nagatada
Wikipedia (German), Shiraishi Nagatada
Index entries for sequences related to sums of cubes

Cf. A003215, A003325, A023042, A181123, A225908, A226902, A226903.

Select[#[[2]]-#[[1]]&/@Partition[Range[5000]^3,2,1],Count[ IntegerPartitions[ #,{2}],?(AllTrue[Surd[#,3],IntegerQ]&)]>0&] (* The program uses the AllTrue function from Mathematica version 10 *) (* _Harvey P. Dale, Sep 07 2018 *)

a(n) = (A226902(n)+1)^3 - A226902(n)^3.

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

A066890 Cubes that are the sum of three distinct positive cubes.

Original entry on oeis.org

216, 729, 1728, 5832, 6859, 8000, 13824, 15625, 19683, 21952, 24389, 27000, 46656, 54872, 64000, 68921, 74088, 85184, 91125, 97336, 110592, 125000, 148877, 157464, 175616, 185193, 195112, 216000, 250047, 287496, 300763, 328509, 343000, 357911, 373248, 421875

Offset: 1

Harvey P. Dale, Jan 22 2002

nonn

			729 is included because it is 9^3 and 1^3 + 6^3 + 8^3 = 729.

David Wells, The Penguin Dictionary of Curious and Interesting Numbers (Rev. ed. 1997), pp. 130, 147.

Donovan Johnson, Table of n, a(n) for n = 1..1000

maxCube = (m = 67)^3; Reap[ Do[ bmax = (maxCube - a^3)^(1/3) // Ceiling; Do[ cmax = (maxCube - b^3)^(1/3) // Ceiling; Do[ n = a^3 + b^3 + c^3; If[n <= maxCube, If[ IntegerQ[n^(1/3)], Sow[n]]], {c, b, cmax}], {b, a, bmax}], {a, 1, (maxCube - 2)^(1/3) // Ceiling}]][[2, 1]] // Flatten // Union (* Jean-François Alcover, Mar 07 2013 *)

a(n) = (A023042(n))^3. - Christian N. K. Anderson, Aug 10 2014

Offset corrected by Arkadiusz Wesolowski, Aug 06 2012

cp's OEIS Frontend

A003072 Numbers that are the sum of 3 positive cubes.

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A226903 Shiraishi numbers: a parametrized family of solutions c to the Diophantine equation a^3 + b^3 + c^3 = d^3 with d = c+1.

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A114923 Primes p such that there exist three primes q, r and s with p^3=q^3+r^3+s^3.

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A225908 Numbers that are both a sum and a difference of two positive cubes.

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A306213 Numbers that are the sum of cubes of three distinct positive integers in arithmetic progression.

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A297305 Numbers k such that k^4 can be written as a sum of five positive 4th powers.

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A328149 Numbers whose set of divisors contains a quadruple (x, y, z, w) satisfying x^3 + y^3 + z^3 = w^3.

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