cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-7 of 7 results.

A003328 Numbers that are the sum of 5 positive cubes.

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, 157, 162, 164, 169, 171, 172, 176, 181, 183, 188, 190, 192, 194, 195, 199
Offset: 1

Views

Author

N. J. A. Sloane

Keywords

Comments

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

Examples

			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)

Links

Crossrefs

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).

Programs

  • PARI
    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
  • Python
    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

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

Views

Author

Eric W. Weisstein

Keywords

Comments

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

Links

Crossrefs

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

Programs

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

Extensions

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

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

Views

Author

David Consiglio, Jr., Apr 26 2021

Keywords

Comments

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).

Examples

			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.

References

  • 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.

Links

Crossrefs

Equivalent sequences for 1 way: A048926; 2 ways: A048927; 1 or more ways: A003328; 3 or more ways: A343704.
Cf. A003327.

Programs

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

A344190 Numbers that are the sum of five fourth powers in exactly one way.

Original entry on oeis.org

5, 20, 35, 50, 65, 80, 85, 100, 115, 130, 145, 165, 180, 195, 210, 245, 290, 305, 320, 325, 355, 370, 385, 405, 420, 435, 450, 500, 530, 545, 560, 580, 595, 610, 625, 629, 644, 659, 674, 675, 689, 690, 709, 724, 739, 754, 755, 770, 785, 789, 800, 804, 819, 850, 865, 869, 899, 914, 929, 930, 949, 964, 979, 994, 1025, 1040
Offset: 1

Views

Author

David Consiglio, Jr., May 11 2021

Keywords

Comments

Differs from A003339 at term 17 because 260 = 1^4 + 1^4 + 1^4 + 1^4 + 4^4 = 1^4 + 2^4 + 3^4 + 3^4 + 3^4

Examples

			35 is a member of this sequence because 35 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4

Links

Crossrefs

Cf. A003339, A048926, A344189, A344237, A344643, A345813.

Programs

  • Python
    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 == 1])
for x in range(len(rets)):
    print(rets[x])

A025403 Numbers that are the sum of 4 positive cubes in exactly 1 way.

Original entry on oeis.org

4, 11, 18, 25, 30, 32, 37, 44, 51, 56, 63, 67, 70, 74, 81, 82, 88, 89, 93, 100, 107, 108, 119, 126, 128, 130, 135, 137, 142, 144, 145, 149, 154, 156, 161, 163, 168, 180, 182, 187, 191, 193, 198, 200, 205, 206, 217, 224, 226, 233, 240, 243, 245, 254, 256, 261, 266, 271, 280
Offset: 1

Views

Author

David W. Wilson

Keywords

Links

Crossrefs

Cf. A003327, A025357, A025395, A025399, A025404, A048926, A344189.

Programs

  • Mathematica
    Reap[For[n = 1, n <= 300, n++, pr = Select[ PowersRepresentations[n, 4, 3], Times @@ # != 0 &]; If[pr != {} && Length[pr] == 1, Print[n, pr]; Sow[n]]]][[2, 1]] (* Jean-François Alcover, Jul 31 2013 *)

Formula

{n: A025457(n) = 1}. - R. J. Mathar, Jun 15 2018

A057906 Positive integers that are not the sum of exactly five positive cubes.

Original entry on oeis.org

1, 2, 3, 4, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 18, 20, 21, 22, 23, 24, 25, 27, 28, 29, 30, 32, 34, 35, 36, 37, 39, 41, 42, 43, 44, 46, 47, 48, 49, 50, 51, 53, 54, 55, 56, 58, 60, 61, 62, 63, 65, 66, 67, 69, 70, 72, 73, 74, 76, 77, 79, 80, 81, 84, 85, 86, 87, 88, 91, 92
Offset: 1

Views

Author

Eric W. Weisstein

Keywords

Comments

It appears that this sequence has 5439 terms, the last of which is 1290740. - T. D. Noe, Dec 13 2006

Links

Crossrefs

Cf. A048926 (numbers that are the sum of five positive cubes in exactly 1 way)
Cf. A003328 (Complement)

Programs

  • Mathematica
    max = 100; Complement[Range[max], Select[Range[max], Count[ PowersRepresentations[#, 5, 3], r_ /; FreeQ[r, 0]] == 1 &]] (* Jean-François Alcover, Oct 23 2012 *)

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

Views

Author

David W. Wilson

Keywords

Comments

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

Links

Crossrefs

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).

Programs

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

Formula

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

Extensions

Second offset from Michel Marcus, Apr 25 2019
Showing 1-7 of 7 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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