A048931 -id:A048931 - cp's OEIS frontend

A003329 Numbers that are the sum of 6 positive cubes.

6, 13, 20, 27, 32, 34, 39, 41, 46, 48, 53, 58, 60, 65, 67, 69, 72, 76, 79, 83, 84, 86, 90, 91, 95, 97, 98, 102, 104, 105, 109, 110, 116, 117, 121, 123, 124, 128, 130, 132, 135, 136, 137, 139, 142, 143, 144, 146, 147, 151, 153, 154, 156, 158, 160, 161, 162, 163, 165, 170

Offset: 1

N. J. A. Sloane

As the order of addition doesn't matter we can assume terms are in increasing order. - David A. Corneth, Aug 01 2020

			From _David A. Corneth_, Aug 01 2020: (Start)
1647 is in the sequence as 1647 = 3^3 + 3^3 + 5^3 + 5^3 +  7^3 + 10^3.
3319 is in the sequence as 3319 = 5^3 + 5^3 + 5^3 + 6^3 + 10^3 + 12^3.
4038 is in the sequence as 4038 = 3^3 + 3^3 + 6^3 + 8^3 +  8^3 + 14^3. (End)

David A. Corneth, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Cubic Number.

Cf. A048929, A048930, A048931.
Cf. A057907 (Complement)
Cf. A###### (x, y) = Numbers that are the sum of x nonzero y-th powers:
  A000404 (2, 2), A000408 (3, 2), A000414 (4, 2), A047700 (5, 2),
  A003325 (2, 3), A003072 (3, 3), A003327 .. A003335 (4 .. 12, 3),
  A003336 .. A003346 (2 .. 12, 4), A003347 .. A003357 (2 .. 12, 5),
  A003358 .. A003368 (2 .. 12, 6), A003369 .. A003379 (2 .. 12, 7),
  A003380 .. A003390 (2 .. 12, 8), A003391 .. A004801 (2 .. 12, 9),
  A004802 .. A004812 (2 .. 12, 10), A004813 .. A004823 (2 .. 12, 11).

max = 200; cmax = Ceiling[(max - 5)^(1/3)]; cc = Array[c, 6]; iter = Sequence @@ Transpose[ {cc, Join[{1}, Most[cc]], Table[cmax, {6}]}]; Union[ Reap[ Do[ a = Total[cc^3]; If[a <= max, Sow[a]], Evaluate[iter]]][[2, 1]]] (* Jean-François Alcover, Oct 23 2012 *)

(A003329_upto(N,k=6,m=3)=[i|i<-[1..#N=sum(n=1,sqrtnint(N,m), 'x^n^m, O('x^N))^k], polcoef(N,i)])(200) \\ M. F. Hasler, Aug 02 2020

from collections import Counter
from itertools import combinations_with_replacement as multi_combs
def aupto(lim):
  c = filter(lambda x: x<=lim, (i**3 for i in range(1, int(lim**(1/3))+2)))
  s = filter(lambda x: x<=lim, (sum(mc) for mc in multi_combs(c, 6)))
  counts = Counter(s)
  return sorted(k for k in counts)
print(aupto(170)) # Michael S. Branicky, Jun 13 2021

More terms from Eric W. Weisstein

A048930 Numbers that are the sum of 6 positive cubes in exactly 2 ways.

Original entry on oeis.org

158, 165, 184, 228, 235, 247, 256, 261, 268, 273, 275, 280, 282, 284, 287, 291, 294, 306, 310, 313, 317, 324, 331, 332, 343, 345, 347, 350, 352, 362, 371, 373, 376, 378, 380, 385, 387, 388, 392, 395, 399, 404, 406, 408, 418, 425, 430, 432, 436, 437, 441

Offset: 1

Eric W. Weisstein

nonn

It appears that this sequence has 1094 terms, the last of which is 21722. - Donovan Johnson, Jan 09 2013

			158 is in the sequence since 158 = 64+64+27+1+1+1 = 125+8+8+8+8+1

Donovan Johnson, Table of n, a(n) for n = 1..1094 (terms < 10^7)
Eric Weisstein's World of Mathematics, Cubic Number.

Cf. A003329, A048927, A048929, A048931, A345511, A345774, A345814.

Reap[For[n = 1, n <= 500, n++, pr = Select[ PowersRepresentations[n, 6, 3], Times @@ # != 0 &]; If[pr != {} && Length[pr] == 2, Print[n, pr]; Sow[n]]]][[2, 1]] (* Jean-François Alcover, Jul 31 2013 *)

mx=10^6; ct=vector(mx); cb=vector(99); for(i=1, 99, cb[i]=i^3); for(i1=1, 99, s1=cb[i1]; for(i2=i1, 99, s2=s1+cb[i2]; if(s2+4*cb[i2]>mx, next(2)); for(i3=i2, 99, s3=s2+cb[i3]; if(s3+3*cb[i3]>mx, next(2)); for(i4=i3, 99, s4=s3+cb[i4]; if(s4+2*cb[i4]>mx, next(2)); for(i5=i4, 99, s5=s4+cb[i5]; if(s5+cb[i5]>mx, next(2)); for(i6=i5, 99, s6=s5+cb[i6]; if(s6>mx, next(2)); ct[s6]++)))))); n=0; for(i=6, mx, if(ct[i]==2, n++; write("b048930.txt", n " " i))) /* Donovan Johnson, Jan 09 2013 */

Terms corrected by Donovan Johnson, Jan 09 2013

A048929 Numbers that are the sum of 6 positive cubes in exactly 1 way.

Original entry on oeis.org

6, 13, 20, 27, 32, 34, 39, 41, 46, 48, 53, 58, 60, 65, 67, 69, 72, 76, 79, 83, 84, 86, 90, 91, 95, 97, 98, 102, 104, 105, 109, 110, 116, 117, 121, 123, 124, 128, 130, 132, 135, 136, 137, 139, 142, 143, 144, 146, 147, 151, 153, 154, 156, 160, 161, 162, 163, 170

Offset: 1

Eric W. Weisstein

It appears that this sequence has 841 terms, the last of which is 19417. This means that all numbers greater than 19417 can be written as the sum of six positive cubes in at least two ways. - T. D. Noe, Dec 13 2006

T. D. Noe, Table of n, a(n) for n=1..841
Eric Weisstein's World of Mathematics, Cubic Number.

Cf. A003329, A048930, A048931.

Cf. A057907 (numbers not the sum of six positive cubes)

Select[ Range[200], Length[ Select[ PowersRepresentations[#, 6, 3], And @@ (Positive /@ #) &]] == 1 &] (* Jean-François Alcover, Oct 25 2012 *)

from collections import Counter
from itertools import combinations_with_replacement as multi_combs
def aupto(lim):
  c = filter(lambda x: x<=lim, (i**3 for i in range(1, int(lim**(1/3))+2)))
  s = filter(lambda x: x<=lim, (sum(mc) for mc in multi_combs(c, 6)))
  counts = Counter(s)
  return sorted(k for k in counts if counts[k]==1)
print(aupto(20000)) # Michael S. Branicky, Jun 13 2021

A345512 Numbers that are the sum of six cubes in three or more ways.

Original entry on oeis.org

221, 254, 369, 411, 443, 469, 495, 502, 576, 595, 600, 626, 648, 658, 684, 704, 711, 720, 739, 746, 753, 760, 765, 767, 772, 774, 779, 786, 793, 811, 818, 828, 830, 835, 837, 844, 854, 856, 863, 866, 873, 874, 880, 884, 886, 891, 892, 893, 899, 905, 910, 919

Offset: 1

David Consiglio, Jr., Jun 20 2021

nonn

			254 is a term because 254 = 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 4^3 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 5^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3.

Sean A. Irvine, Table of n, a(n) for n = 1..10000

Cf. A048931, A343704, A344807, A345511, A345513, A345521, A345560.

from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**3 for x in range(1, 1000)]
for pos in cwr(power_terms, 6):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v >= 3])
    for x in range(len(rets)):
        print(rets[x])

A345815 Numbers that are the sum of six fourth powers in exactly three ways.

Original entry on oeis.org

2676, 2851, 2916, 4131, 4226, 4241, 4306, 4371, 4481, 4850, 5346, 5411, 5521, 5586, 5651, 6561, 6611, 6756, 6771, 6801, 6821, 6836, 6851, 6931, 7106, 7235, 7475, 7491, 7666, 7841, 7906, 7971, 8146, 8211, 8321, 8386, 8451, 8531, 8706, 9011, 9156, 9171, 9186

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345560 at term 18 because 6626 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 9^4 = 2^4 + 2^4 + 2^4 + 3^4 + 7^4 + 8^4 = 2^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4.

			2851 is a term because 2851 = 1^4 + 1^4 + 1^4 + 4^4 + 6^4 + 6^4 = 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4.

Sean A. Irvine, Table of n, a(n) for n = 1..10000

Cf. A048931, A344244, A345560, A345814, A345816, A345825, A346358.

from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**4 for x in range(1, 1000)]
for pos in cwr(power_terms, 6):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v == 3])
    for x in range(len(rets)):
        print(rets[x])

A345766 Numbers that are the sum of six cubes in exactly four ways.

Original entry on oeis.org

626, 830, 837, 856, 873, 891, 947, 954, 982, 1008, 1026, 1052, 1053, 1071, 1094, 1097, 1106, 1109, 1134, 1143, 1150, 1153, 1172, 1195, 1208, 1227, 1234, 1253, 1267, 1278, 1279, 1283, 1286, 1290, 1297, 1316, 1323, 1324, 1358, 1361, 1368, 1369, 1376, 1395, 1403

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345513 at term 12 because 1045 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 10^3 = 1^3 + 1^3 + 4^3 + 5^3 + 5^3 + 9^3 = 1^3 + 2^3 + 3^3 + 4^3 + 6^3 + 9^3 = 3^3 + 3^3 + 6^3 + 6^3 + 6^3 + 7^3 = 4^3 + 4^3 + 4^3 + 5^3 + 6^3 + 8^3.

			830 is a term because 830 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 8^3 = 1^3 + 3^3 + 3^3 + 5^3 + 5^3 + 6^3 = 1^3 + 3^3 + 3^3 + 3^3 + 4^3 + 7^3 = 2^3 + 2^3 + 3^3 + 3^3 + 6^3 + 6^3.

Sean A. Irvine, Table of n, a(n) for n = 1..1211

Cf. A048931, A344035, A345513, A345767, A345776, A345816.

from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**3 for x in range(1, 1000)]
for pos in cwr(power_terms, 6):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v == 4])
    for x in range(len(rets)):
        print(rets[x])

A345775 Numbers that are the sum of seven cubes in exactly three ways.

Original entry on oeis.org

222, 229, 248, 255, 262, 281, 283, 285, 318, 346, 370, 374, 377, 379, 381, 396, 400, 407, 412, 419, 426, 433, 437, 438, 444, 451, 463, 472, 475, 477, 489, 494, 501, 505, 507, 510, 522, 529, 533, 536, 559, 564, 566, 568, 570, 577, 578, 584, 585, 592, 594, 596

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345521 at term 28 because 470 = 1^3 + 1^3 + 1^3 + 1^3 + 5^3 + 5^3 + 6^3 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 6^3 + 6^3 = 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 + 5^3 = 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 6^3.

Likely finite.

			229 is a term because 229 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 5^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3.

Sean A. Irvine, Table of n, a(n) for n = 1..390

Cf. A048931, A345521, A345774, A345776, A345785, A345825.

from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**3 for x in range(1, 1000)]
for pos in cwr(power_terms, 7):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v == 3])
    for x in range(len(rets)):
        print(rets[x])

A025459 Number of partitions of n into 6 positive cubes.

Original entry on oeis.org

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

Offset: 0

David W. Wilson

nonn

Robert Israel, Table of n, a(n) for n = 0..10000
Index entries for sequences related to sums of cubes

Cf. A003329, A048929, A048930, A048931.

A025459 := proc(n)
    local a,x,y,z,u,v,wcu ;
    a := 0 ;
    for x from 1 do
        if 6*x^3 > n then
            return a;
        end if;
        for y from x do
            if x^3+5*y^3 > n then
                break;
            end if;
            for z from y do
                if x^3+y^3+4*z^3 > n then
                    break;
                end if;
                for u from z do
                    if x^3+y^3+z^3+3*u^3 > n then
                        break;
                    end if;
                    for v from u do
                        if x^3+y^3+z^3+u^3+2*v^3 > n then
                            break;
                        end if;
                        wcu := n-x^3-y^3-z^3-u^3-v^3 ;
                        if isA000578(wcu) then
                            a := a+1 ;
                        end if;
                    end do:
                end do:
            end do:
        end do:
    end do:
end proc: # R. J. Mathar, Sep 15 2015
# Alternative:
N:= 200:
G:= mul(1/(1-y*x^(k^3)),k=1..floor(N^(1/3))):
C6:= coeff(series(G,y,7),y,6):
S:= series(C6,x,N+1):
seq(coeff(S,x,i),i=0..N); # Robert Israel, May 10 2020

a[n_] := Count[PowersRepresentations[n, 6, 3], pr_List /; FreeQ[pr, 0]];
a /@ Range[0, 200] (* Jean-François Alcover, Jun 22 2020 *)
Table[Count[IntegerPartitions[n,{6}],?(AllTrue[Surd[#,3],IntegerQ]&)],{n,0,110}] (* Requires Mathematica version 10 or later *) (* _Harvey P. Dale, Jun 06 2021 *)

a(n) = [x^n y^6] Product_{k>=1} 1/(1 - y*x^(k^3)). - Ilya Gutkovskiy, Apr 23 2019

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A003329 Numbers that are the sum of 6 positive cubes.

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A048930 Numbers that are the sum of 6 positive cubes in exactly 2 ways.

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A048929 Numbers that are the sum of 6 positive cubes in exactly 1 way.

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A345512 Numbers that are the sum of six cubes in three or more ways.

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A345815 Numbers that are the sum of six fourth powers in exactly three ways.

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A345766 Numbers that are the sum of six cubes in exactly four ways.

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A345775 Numbers that are the sum of seven cubes in exactly three ways.

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A025459 Number of partitions of n into 6 positive cubes.

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