A048930 -id:A048930 - 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

A048927 Numbers that are the sum of 5 positive cubes in exactly 2 ways.

Original entry on oeis.org

157, 220, 227, 246, 253, 260, 267, 279, 283, 286, 305, 316, 323, 342, 344, 361, 368, 377, 379, 384, 403, 410, 435, 440, 442, 468, 475, 487, 494, 501, 523, 530, 531, 549, 562, 568, 586, 592, 594, 595, 599, 602, 621, 625, 640, 647, 657, 658, 683, 703, 710

Offset: 1

Eric W. Weisstein

nonn

It appears that this sequence has 15416 terms, the last of which is 2243453. - Donovan Johnson, Jan 11 2013

From a(1) = 157 we see that c(n) = (number of ways n is the sum of 5 cubes) coincides with A010057 = characteristic function of cubes, up to n = 156. This sequence lists the numbers n for which c(n) = 2. See A003328 for c(n) > 0 and A048926 for c(n) = 1. - M. F. Hasler, Jan 04 2023

Donovan Johnson, Table of n, a(n) for n = 1..15416 (terms < 10^8)
Eric Weisstein's World of Mathematics, Cubic Number.

Cf. A003328 (sums of 5 positive cubes), A025404, A048926 (sum of 5 positive cubes in exactly 1 way), A048930, A294736, A343702, A343705, A344237.

Select[ Range[ 1000], (test = Length[ Select[ PowersRepresentations[#, 5, 3], And @@ (Positive /@ #)& ] ] == 2; If[test, Print[#]]; test)& ](* Jean-François Alcover, Nov 09 2012 *)

(waycount(n,numcubes,imax)={if(numcubes==0, !n, sum(i=1,imax, waycount(n-i^3,numcubes-1,i)))}); isA048927(n)=(waycount(n,5,floor(n^(1/3)))==2); \\ Michael B. Porter, Sep 27 2009

def ways (n, left = 5, last = 1):
  a = last; a3 = a**3; c = 0
  while a3 <= n-left+1:
    if left > 1:
       c += ways(n-a3, left-1, a)
    elif a3 == n:
       c += 1
    a += 1; a3 = a**3
  return c
for n in range (1,1000): # to print this sequence
  if ways(n)==2: print(n,end=", ") # in Python2 use, e.g.: print n,
# Minor edits by M. F. Hasler, Jan 04 2023

More terms from Walter Hofmann (walterh(AT)gmx.de), Jun 01 2000

A048931 Numbers that are the sum of 6 positive cubes in exactly 3 ways.

Original entry on oeis.org

221, 254, 369, 411, 443, 469, 495, 502, 576, 595, 600, 648, 658, 684, 704, 711, 720, 739, 746, 753, 760, 765, 767, 772, 774, 779, 786, 793, 811, 818, 828, 835, 844, 854, 863, 866, 874, 880, 884, 886, 892, 893, 899, 905, 910, 919, 928, 929, 935, 936, 937

Offset: 1

Eric W. Weisstein

nonn

It appears that this sequence has 1141 terms, the last of which is 26132. - Donovan Johnson, Jan 09 2013

			221 is in the sequence since 221 = 216+1+1+1+1+1 = 125+64+8+8+8+8 = 64+64+64+27+1+1

Donovan Johnson, Table of n, a(n) for n = 1..1141 (terms < 10^7)
Eric Weisstein's World of Mathematics, Cubic Number.

Cf. A003329, A048929, A048930, A343705, A345512, A345766, A345775, A345815.

Reap[For[n = 1, n <= 1000, n++, pr = Select[ PowersRepresentations[n, 6, 3], Times @@ # != 0 &]; If[pr != {} && Length[pr] == 3, 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]==3, n++; write("b048931.txt", n " " i))) /* Donovan Johnson, Jan 09 2013 */

Corrected and extended by Larry Reeves (larryr(AT)acm.org), Oct 02 2000

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

A345814 Numbers that are the sum of six fourth powers in exactly two ways.

Original entry on oeis.org

261, 276, 291, 341, 356, 421, 516, 531, 596, 771, 885, 900, 965, 1140, 1361, 1509, 1556, 1571, 1636, 1811, 2180, 2596, 2611, 2661, 2691, 2706, 2721, 2741, 2756, 2771, 2786, 2836, 2931, 2946, 2961, 3011, 3026, 3091, 3186, 3201, 3220, 3266, 3285, 3300, 3315

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345559 at term 25 because 2676 = 1^4 + 1^4 + 2^4 + 4^4 + 7^4 = 1^4 + 1^4 + 1^4 + 3^4 + 6^4 + 6^4 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 7^4.

			276 is a term because 276 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 4^4 = 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4.

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

Cf. A048930, A344237, A345559, A345813, A345815, A345824, A346357.

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 == 2])
    for x in range(len(rets)):
        print(rets[x])

A345511 Numbers that are the sum of six cubes in two or more ways.

Original entry on oeis.org

158, 165, 184, 221, 228, 235, 247, 254, 256, 261, 268, 273, 275, 280, 282, 284, 287, 291, 294, 306, 310, 313, 317, 324, 331, 332, 343, 345, 347, 350, 352, 362, 369, 371, 373, 376, 378, 380, 385, 387, 388, 392, 395, 399, 404, 406, 408, 411, 418, 425, 430, 432

Offset: 1

David Consiglio, Jr., Jun 20 2021

nonn

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

Cf. A003329, A048930, A343702, A344806, A345512, A345520, A345559.

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 >= 2])
    for x in range(len(rets)):
        print(rets[x])

A345774 Numbers that are the sum of seven cubes in exactly two ways.

Original entry on oeis.org

131, 159, 166, 173, 185, 192, 211, 236, 243, 257, 264, 269, 274, 276, 288, 290, 292, 295, 299, 300, 302, 307, 309, 311, 314, 320, 321, 325, 332, 333, 337, 339, 340, 344, 348, 351, 353, 355, 358, 359, 360, 363, 372, 384, 385, 386, 388, 389, 393, 395, 398, 403

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345520 at term 8 because 222 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 6^3 = 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 4^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 5^3.

Likely finite.

			159 is a term because 159 = 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3.

Sean A. Irvine, Table of n, a(n) for n = 1..355

Cf. A048930, A345520, A345773, A345775, A345784, A345824.

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 == 2])
    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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A048927 Numbers that are the sum of 5 positive cubes in exactly 2 ways.

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A048931 Numbers that are the sum of 6 positive cubes in exactly 3 ways.

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

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A345814 Numbers that are the sum of six fourth powers in exactly two ways.

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A345511 Numbers that are the sum of six cubes in two or more ways.

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A345774 Numbers that are the sum of seven cubes in exactly two ways.

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A025459 Number of partitions of n into 6 positive cubes.

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