A048927 -id:A048927 - cp's OEIS frontend

A003328 Numbers that are the sum of 5 positive cubes.

5, 12, 19, 26, 31, 33, 38, 40, 45, 52, 57, 59, 64, 68, 71, 75, 78, 82, 83, 89, 90, 94, 96, 97, 101, 108, 109, 115, 116, 120, 127, 129, 131, 134, 135, 136, 138, 143, 145, 146, 150, 152, 153, 155, 157, 162, 164, 169, 171, 172, 176, 181, 183, 188, 190, 192, 194, 195, 199

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

It seems only a finite number N of positive integers are not in this sequence, and thus a(n) = n - N for all sufficiently large n. Is it true that 2243453, last term of A048927, is sufficiently large in that sense? - M. F. Hasler, Jan 04 2023

			From _David A. Corneth_, Aug 01 2020: (Start)
3084 is in the sequence as 3084 = 5^3 + 5^3 + 5^3 +  8^3 + 13^3.
4385 is in the sequence as 4385 = 4^3 + 4^3 + 9^3 + 11^3 + 13^3.
5426 is in the sequence as 5426 = 8^3 + 9^3 + 9^3 + 12^3 + 12^3. (End)

David A. Corneth, Table of n, a(n) for n = 1..10000 (first 1000 terms from T. D. Noe)
Eric Weisstein's World of Mathematics, Cubic Number.

Cf. A003328, A048926, A048297.
Cf. A057906 (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), A003072 (3, 3), A003325 (3, 2), 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), 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), 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), 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), 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), A003380 (2, 8), A003381 (3, 8), A003382 (4, 8), A003383 (5, 8), A003384 (6, 8), A003385 (7, 8), A003387 (9, 8), A003388 (10, 8), A003389 (11, 8), A003390 (12, 8), 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), 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), 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), A047700 (5, 2).

select( {is_A003328(n,k=5,m=3,L=sqrtnint(abs(n-k+1),m))=if( n>k*L^m || nM. F. Hasler, Aug 02 2020
A003328_upto(N,k=5,m=3)=[i|i<-[1..#N=sum(n=1,sqrtnint(N,m),'x^n^m,O('x^N))^k], polcoef(N,i)] \\ M. F. Hasler, Aug 02 2020

from collections import Counter
from itertools import combinations_with_replacement as combs_w_rep
def aupto(lim):
  s = filter(lambda x: x<=lim, (i**3 for i in range(1, int(lim**(1/3))+2)))
  s2 = filter(lambda x: x<=lim, (sum(c) for c in combs_w_rep(s, 5)))
  s2counts = Counter(s2)
  return sorted(k for k in s2counts)
print(aupto(200)) # Michael S. Branicky, May 12 2021

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

A048926 Numbers that are the sum of 5 positive cubes in exactly 1 way.

Original entry on oeis.org

5, 12, 19, 26, 31, 33, 38, 40, 45, 52, 57, 59, 64, 68, 71, 75, 78, 82, 83, 89, 90, 94, 96, 97, 101, 108, 109, 115, 116, 120, 127, 129, 131, 134, 135, 136, 138, 143, 145, 146, 150, 152, 153, 155, 162, 164, 169, 171, 172, 176, 181, 183, 188, 190, 192, 194, 195

Offset: 1

Eric W. Weisstein

It appears that this sequence has 11062 terms, the last of which is 1685758. This means that all numbers greater than 1685758 can be written as the sum of five positive cubes in at least two ways. - T. D. Noe, Dec 13 2006

T. D. Noe, Table of n, a(n) for n=1..11062
Eric Weisstein's World of Mathematics, Cubic Number.

Cf. A003328, A048927.

Cf. A057906 (numbers not the sum of five positive cubes)

Select[ Range[200], Count[ PowersRepresentations[#, 5, 3], r_ /; FreeQ[r, 0]] == 1 &] (* Jean-François Alcover, Oct 23 2012 *)

from collections import Counter
from itertools import combinations_with_replacement as combs_with_rep
def aupto(lim):
  s = filter(lambda x: x<=lim, (i**3 for i in range(1, int(lim**(1/3))+2)))
  s2 = filter(lambda x: x<=lim, (sum(c) for c in combs_with_rep(s, 5)))
  s2counts = Counter(s2)
  return sorted(k for k in s2counts if s2counts[k] == 1)
print(aupto(196)) # Michael S. Branicky, May 12 2021

A343705 Numbers that are the sum of five positive cubes in exactly three ways.

Original entry on oeis.org

766, 810, 827, 829, 865, 883, 981, 1018, 1025, 1044, 1070, 1105, 1108, 1142, 1145, 1161, 1168, 1226, 1233, 1259, 1289, 1350, 1368, 1424, 1431, 1439, 1441, 1457, 1487, 1492, 1494, 1529, 1531, 1538, 1550, 1555, 1568, 1583, 1587, 1592, 1593, 1594, 1609, 1611, 1613, 1639, 1648, 1665, 1672, 1674, 1688, 1707, 1711

Offset: 1

David Consiglio, Jr., Apr 26 2021

This sequence differs from A343704 at term 20 because 1252 = 1^3 + 1^3 + 5^3 + 5^3 + 10^3 = 1^3 + 2^3 + 3^3 + 6^3 + 10^3 = 3^3 + 3^3 + 7^3 + 7^3 + 8^3 = 3^3 + 4^3 + 6^3 + 6^3 + 9^3. Thus this term is in A343704 but not in this sequence.
Comment from D. S. McNeil, May 13 2021: (Start)
If we weaken positive cubes to nonnegative cubes, Deshouillers, Hennecart, and Landreau (2000) give numerical and heuristic evidence that all numbers past 7373170279850 are representable as the sum of 4 nonnegative cubes.
So if they are right, then eventually we can just take some N and represent each of (N-1^3, N-2^3, N-3^3, N-4^3) as the sum of four cubes and then take 1^3, 2^3, 3^3, or 4^3 as our fifth cube, giving at least four 5-cube representations for N.
So it is very likely that the set of numbers representable by the sum of 5 positive cubes in exactly three ways is finite. (End)
It is conjectured that the number of ways of writing N as a sum of 5 positive cubes grows like C(N)*N^(2/3), where C(N) depends on N but is bounded away from zero by an absolute constant (Vaughan, 1981; Vaughan and Wooley, 2002). So the number will exceed 3 as soon as N is large enough, and so it is very likely that this sequence is finite. However, at present this is an open question. - N. J. A. Sloane, May 15 2021 (based on correspondence with Robert Vaughan and Trevor Wooley).

			827 is a term of this sequence because 827 = 1^3 + 4^3 + 5^3 + 5^3 + 8^3 = 2^3 + 2^3 + 5^3 + 7^3 + 7^3 = 2^3 + 3^3 + 4^3 + 6^3 + 8^3.

R. C. Vaughan, The Hardy-Littlewood Method, Cambridge University Press, 1981.
R. C. Vaughan, Trevor Wooley (2002), Waring's Problem: A Survey. In Michael A. Bennet, Bruce C. Berndt, Nigel Boston, Harold G. Diamond, Adolf J. Hildebrand, Walter Philipp (eds.). Number Theory for the Millennium. III. Natick, MA: A. K. Peters, pp. 301-340.

David Consiglio, Jr. and Sean A. Irvine, Table of n, a(n) for n = 1..18984
Jean-Marc Deshouillers, François Hennecart, and Bernard Landreau, 7373170279850, Math. Comp. 69 (2000), pp. 421-439. Appendix by I. Gusti Putu Purnaba.

Equivalent sequences for 1 way: A048926; 2 ways: A048927; 1 or more ways: A003328; 3 or more ways: A343704.

Cf. A003327.

Select[Range@2000,Length@Select[PowersRepresentations[#,5,3],FreeQ[#,0]&]==3&] (* Giorgos Kalogeropoulos, Apr 26 2021 *)

from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**3 for x in range(1,50)]#n
for pos in cwr(power_terms,5):#m
    tot = sum(pos)
    keep[tot] += 1
rets = sorted([k for k,v in keep.items() if v == 3])#s
for x in range(len(rets)):
    print(rets[x])

A344237 Numbers that are the sum of five fourth powers in exactly two ways.

Original entry on oeis.org

260, 275, 340, 515, 884, 1555, 2595, 2660, 2675, 2690, 2705, 2755, 2770, 2835, 2930, 2945, 3010, 3185, 3299, 3314, 3379, 3554, 3923, 3970, 3985, 4050, 4115, 4145, 4160, 4210, 4290, 4355, 4400, 4465, 4594, 4769, 4834, 5075, 5090, 5155, 5265, 5330, 5395, 5440, 5505, 5570, 5699, 6370, 6545, 6580, 6595, 6660, 6675

Offset: 1

David Consiglio, Jr., May 12 2021

nonn

Differs from A344237 at term 31 because 4225 = 2^4 + 2^4 + 2^4 + 3^4 + 8^4 = 2^4 + 4^4 + 4^4 + 6^4 + 7^4 = 3^4 + 4^4 + 6^4 + 6^4 + 6^4

			340 is a member of this sequence because 340 = 1^4 + 1^4 + 1^4 + 3^4 + 4^4 = 2^4 + 3^4 + 3^4 + 3^4 + 3^4

David Consiglio, Jr., Table of n, a(n) for n = 1..20000

Cf. A048927, A342686, A344190, A344193, A344238, A344244, A345814.

from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**4 for x in range(1,50)]
for pos in cwr(power_terms,5):
    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])

A343702 Numbers that are the sum of five positive cubes in two or more 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, 712, 719, 729, 731, 738, 745, 752, 759, 764, 766, 771, 773, 778, 785

Offset: 1

David Consiglio, Jr., Apr 26 2021

This sequence differs from A048927:
766 = 1^3 + 1^3 + 2^3 + 3^3 + 9^3
    = 1^3 + 4^3 + 4^3 + 5^3 + 8^3
    = 2^3 + 2^3 + 4^3 + 7^3 + 7^3.
So 766 is a term, but not a term of A048927.

			227 = 1^3 + 1^3 + 1^3 + 2^3 + 6^3
    = 2^3 + 3^3 + 4^3 + 4^3 + 4^3
so 227 is a term of this sequence.

David Consiglio, Jr., Table of n, a(n) for n = 1..20000

Cf. A003328, A025406, A048927, A343704, A344238, A344795, A345511.

Select[Range@1000,Length@Select[PowersRepresentations[#,5,3],FreeQ[#,0]&]>1&] (* Giorgos Kalogeropoulos, Apr 26 2021 *)

from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**3 for x in range(1,50)]#n
for pos in cwr(power_terms,5):#m
    tot = sum(pos)
    keep[tot] += 1
rets = sorted([k for k,v in keep.items() if v >= 2])#s
for x in range(len(rets)):
    print(rets[x])

A025404 Numbers that are the sum of 4 positive cubes in exactly 2 ways.

Original entry on oeis.org

219, 252, 259, 278, 315, 376, 467, 522, 594, 702, 758, 763, 765, 802, 809, 819, 856, 864, 945, 980, 1010, 1017, 1036, 1043, 1073, 1078, 1081, 1118, 1134, 1160, 1251, 1352, 1367, 1368, 1374, 1375, 1393, 1397, 1423, 1430, 1458, 1460, 1465, 1467, 1484, 1486

Offset: 1

David W. Wilson

nonn

Eric Weisstein's World of Mathematics, Cubic Number.

Cf. A025358, A025396, A025403, A025405, A025406, A048927, A344193.

{n: A025457(n) = 2}. - R. J. Mathar, Jun 15 2018

A025458 Number of partitions of n into 5 positive cubes.

Original entry on oeis.org

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

Offset: 0

David W. Wilson

nonn

a(n) > 2 at n= 766, 810, 827, 829, 865, 883, 981, 1018, 1025, 1044,... - R. J. Mathar, Sep 15 2015

The first term > 1 is a(157) = 2. - Michel Marcus, Apr 25 2019

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

Column 5 of A320841, which cross-references the equivalent sequences for other numbers of positive cubes.

Positions of values: A057906 (0), A003328 (nonzero), A048926 (1), A048927 (2), A343705 (3), A344035 (4).

A025458 := proc(n)
    local a,x,y,z,u,vcu ;
    a := 0 ;
    for x from 1 do
        if 5*x^3 > n then
            return a;
        end if;
        for y from x do
            if x^3+4*y^3 > n then
                break;
            end if;
            for z from y do
                if x^3+y^3+3*z^3 > n then
                    break;
                end if;
                for u from z do
                    if x^3+y^3+z^3+2*u^3 > n then
                        break;
                    end if;
                    vcu := n-x^3-y^3-z^3-u^3 ;
                    if isA000578(vcu) then
                        a := a+1 ;
                    end if;
                end do:
            end do:
        end do:
    end do:
end proc: # R. J. Mathar, Sep 15 2015

a[n_] := IntegerPartitions[n, {5}, Range[n^(1/3) // Ceiling]^3] // Length;
a /@ Range[0, 157] (* Jean-François Alcover, Jun 20 2020 *)

a(n) = [x^n y^5] Product_{k>=1} 1/(1 - y*x^(k^3)). - Ilya Gutkovskiy, Apr 23 2019

Second offset from Michel Marcus, Apr 25 2019

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A003328 Numbers that are the sum of 5 positive cubes.

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

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A048926 Numbers that are the sum of 5 positive cubes in exactly 1 way.

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A343705 Numbers that are the sum of five positive cubes in exactly three ways.

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A344237 Numbers that are the sum of five fourth powers in exactly two ways.

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A343702 Numbers that are the sum of five positive cubes in two or more ways.

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A025404 Numbers that are the sum of 4 positive cubes in exactly 2 ways.

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A025458 Number of partitions of n into 5 positive cubes.

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