A002376 -id:A002376 - cp's OEIS frontend

A188462 Least number of 5th powers needed to represent n.

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12

Offset: 1

Jean-François Alcover, Apr 01 2011

nonn

Vaughan & Wooley (1995) prove that a(n) <= 17 for large enough n; in fact it is conjectured that a(n) <= 6 for large enough n. The maximum value is a(223) = 37. - Charles R Greathouse IV, Jul 05 2017

			33 = 2^5 + 1^5 (least decomposition) hence a(33) = 2.

T. D. Noe, Table of n, a(n) for n = 1..10000
R. C. Vaughan and T. D. Wooley, Waring's problem: a survey, Number theory for the millennium, III (Urbana, IL, 2000), 301-340, A K Peters, Natick, MA, 2002.
Robert C. Vaughan and Trevor D. Wooley, Further improvements in Waring's problem, Acta Mathematica 174:2 (1995), pp. 147-240.

Cf. A002828 (squares), A002376 (cubes), A002377 (4th powers), A374012 (6th powers).

Cnt5[n_] := Module[{k = 1}, While[Length[PowersRepresentations[n, k, 5]] == 0, k++]; k]; Array[Cnt5, 105] (* T. D. Noe, Apr 01 2011 *)

from itertools import count
from sympy.solvers.diophantine.diophantine import power_representation
def A188462(n):
    if n == 1: return 1
    for k in count(1):
        try:
            next(power_representation(n,5,k))
        except:
            continue
        return k # Chai Wah Wu, Jun 25 2024

A274459 Least number of perfect powers that add up to n.

Original entry on oeis.org

1, 2, 3, 1, 2, 3, 4, 1, 1, 2, 3, 2, 2, 3, 4, 1, 2, 2, 3, 2, 3, 3, 4, 2, 1, 2, 1, 2, 2, 3, 2, 1, 2, 2, 2, 1, 2, 3, 3, 2, 2, 3, 2, 2, 2, 3, 3, 2, 1, 2, 3, 2, 2, 2, 3, 3, 2, 2, 2, 3, 2, 3, 2, 1, 2, 3, 3, 2, 3, 3, 3, 2, 2, 2, 3, 2, 3, 3, 3, 2, 1, 2, 3, 3, 2

Offset: 1

Sergio Pimentel, Jun 23 2016

nonn

Least number of perfect powers (A001597) needed to add up to n.
This sequence is close to but not exactly equal to A063274.
a(n) is at most 4 since any number can be written as a sum of 4 squares (Lagrange's theorem), but it is possible that for a sufficiently large n, a(n) < 4.
a(n) <= a(i) + a(n-i) for 1 <= i <= n-1. (for computational ease, the maximum value for i can be chosen as floor(n/2)). a(1991) = 4. for 1992 <= k <= 20000, there is no k such that a(k) = 4. - David A. Corneth, Jun 24 2016 [Next such k is 25887, see A113505. - Vaclav Kotesovec, Jun 25 2016]

			a(31) = 2 since 31 can be written as the sum of two (31 = 3^3 + 2^2 = 27 + 4) but no fewer than two perfect powers.

David A. Corneth, Table of n, a(n) for n = 1..10000

Cf. A001597, A002376, A002377, A002828, A002804, A079611, A063274, A188462, A225926.

Cf. A063275 (indices for which a(n)=3), A113505 (indices for which a(n)=4).

nn = 72; t = Select[Range@ nn, # == 1 || GCD @@ FactorInteger[#][[All, 2]] > 1 &]; Table[Min@ Map[Length, Select[IntegerPartitions@ n, AllTrue[#, MemberQ[t, #] &] &]], {n, nn}] (* Michael De Vlieger, Jun 23 2016, after Ant King at A001597 *)

lista(n) = {my(v = vector(n)); for(i = 2,sqrtint(n), for(j = 2, logint(n, i), v[i^j] = 1)); v[1]=1; v[2]=2; for(i=3, #v, if(v[i]==0, v[i] = vecmin(vector( i\2, k,v[k] + v[i-k]))));v} \\ David A. Corneth, Jun 24 2016; corrected by Peter Schorn, Jun 09 2022

More terms from Michael De Vlieger, Jun 23 2016

Terms from a(74) from David A. Corneth, Jun 24 2016

A374012 Least number of 6th powers needed to represent n.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16

Offset: 1

Seiichi Manyama, Jun 25 2024

a(703) = 73.

Pillai, S. S. (1940) On Waring’s problem g(6) = 73. Proc. Indian Acad. Sci. 12A: 30-40

Seiichi Manyama, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Waring's Problem.

Cf. A002828, A002376, A002377, A188462.

Cf. A002804, A018886.

a_vector(n, k=6) = my(v=vector(n), cnt=0, d=0, p=1, s=sum(j=1, sqrtnint(n, k), x^j^k)+x*O(x^n)); while(cnt

from itertools import count
from sympy.solvers.diophantine.diophantine import power_representation
def A374012(n):
    if n == 1: return 1
    for k in count(1):
        try:
            next(power_representation(n,6,k))
        except:
            continue
        return k # Chai Wah Wu, Jun 25 2024

a(n) <= 73.

A084355 Least number of positive cubes needed to represent n!.

Original entry on oeis.org

1, 1, 2, 6, 3, 5, 5, 4, 4, 3, 3, 3, 3, 4, 3, 3, 3, 3, 3, 3, 3

Offset: 0

Hugo Pfoertner, Jun 22 2003

nonn

			a(4)=3 because 4!=24=2^3+2^3+2^3.
a(0)=1 because 0!=1=1^3.
a(1)=1 because 1!=1=1^3.
a(2)=2 because 2!=2=1^3+1^3.
a(3)=6 because 3!=6=1^3+1^3+1^3+1^3+1^3+1^3.
a(4)=3 because 4!=24=2^3+2^3+2^3.
a(5)=5 because 5!=120=1^3+3^3+3^3+4^3+1^3.
a(6)=5 because 6!=720=4^3+6^3+6^3+6^3+2^3.
a(7)=4 because 7!=5040=1^3+5^3+17^3+1^3.
a(8)=4 because 8!=40320=2^3+10^3+34^3+2^3.
a(9)=3 because 9!=362880=52^3+56^3+36^3.
a(10)=3 because 10!=3628800=96^3+140^3+4^3.
a(11)=3 because 11!=39916800=222^3+303^3+105^3.
a(12)=3 because 12!=479001600=214^3+777^3+47^3.
a(13)=4 because 13!=6227020800=106^3+255^3+1838^3+33^3.
a(14)=3 because 14!=87178291200=1344^3+4392^3+312^3.
a(15)=3 because 15!=1307674368000=2040^3+10908^3+1092^3.
a(16)=3 because 16!=20922789888000=8400^3+27040^3+8240^3.
a(17)=3 because 17!=355687428096000=22848^3+69984^3+9984^3.
a(18)=3 because 18!=6402373705728000=54060^3+184080^3+18900^3.
From _Donovan Johnson_, May 17 2010: (Start)
a(19)=3 because 19!=121645100408832000=131040^3+331200^3+436320^3.
a(20)=3 because 20!=2432902008176640000=87490^3+1034430^3+1098440^3.
(End)

Cf. A000142, A002376, A002325, A003072, A003327, A003328, A003329.

a(n,up,dw,k)=local(i,m);if(k==1,if(n==round(sqrtn(n,3))^3,return(1),return(-1)),forstep(i=up,dw,-1,m=n-i^3;if(a(m,min(i,floor(sqrtn(m,3))),ceil(sqrtn(m/(k-1),3)),k-1)==1,return(1)))) for(n=0,18,for(k=1,9,if(a(n!,floor(sqrtn(n!,3)),ceil(sqrtn(n!/k,3)),k)==1,print1(k", ");break))) \\ Herman Jamke (hermanjamke(AT)fastmail.fm), Apr 01 2007

a(n)=A002376(n!).

More terms from David W. Wilson, Jun 23 2003
More terms from Herman Jamke (hermanjamke(AT)fastmail.fm), Apr 01 2007
a(19)-a(20) from Donovan Johnson, May 17 2010

A100854 Least number of positive cubes that sum to n^2.

Original entry on oeis.org

1, 4, 2, 2, 4, 3, 7, 1, 3, 4, 6, 4, 5, 6, 3, 4, 4, 5, 5, 6, 4, 4, 4, 2, 5, 5, 1, 4, 5, 4, 4, 2, 4, 5, 4, 4, 4, 6, 4, 4, 5, 4, 6, 5, 4, 6, 6, 3, 4, 5, 5, 6, 3, 4, 4, 5, 5, 5, 3, 4, 5, 4, 4, 1, 4, 5, 5, 4, 4, 6, 3, 3, 5, 6, 5, 4, 4, 3, 5, 4, 2, 5, 5, 3, 5, 5, 3, 6, 5, 3, 4, 6, 5, 5, 4, 3, 5, 2, 4, 3

Offset: 1

Giovanni Teofilatto, Jan 08 2005

			a(2)=4 because 4=1+1+1+1;
a(3)=2 because 9=1+8;
a(4)=2 because 16=8+8.

Cf. A002376 (least number of positive cubes needed to represent n).

nn=100^2; cnt=Table[10, {nn}]; maxN=Floor[nn^(1/3)]; Do[v={a, b, c, d, e, f, g, h, i}; n=Plus@@(v^3); If[n>0 && n<=nn, cnt[[n]]=Min[cnt[[n]], 9-Count[v, 0]]], {a, 0, maxN}, {b, a, maxN}, {c, b, maxN}, {d, c, maxN}, {e, d, maxN}, {f, e, maxN}, {g, f, maxN}, {h, f, maxN}, {i, h, maxN}]; Table[cnt[[n^2]], {n, 100}] (T. D. Noe)

a(n) = A002376(n^2). - R. J. Mathar, May 06 2016

Corrected and extended by T. D. Noe, Jan 10 2005

A338497 Least number of odd cubes needed to represent n.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 3

Offset: 1

Ilya Gutkovskiy, Oct 31 2020

nonn

Index entries for sequences related to sums of cubes

Cf. A002376, A016755, A101412.

A354761 Least number of squares and cubes that add up to n.

Original entry on oeis.org

1, 2, 3, 1, 2, 3, 4, 1, 1, 2, 3, 2, 2, 3, 4, 1, 2, 2, 3, 2, 3, 3, 4, 2, 1, 2, 1, 2, 2, 3, 2, 2, 2, 2, 2, 1, 2, 3, 3, 2, 2, 3, 2, 2, 2, 3, 3, 3, 1, 2, 3, 2, 2, 2, 3, 3, 2, 2, 3, 3, 2, 3, 2, 1, 2, 3, 3, 2, 3, 3, 3, 2, 2, 2, 3, 2, 3, 3, 3, 2, 1, 2, 3, 3, 2, 3

Offset: 1

Peter Schorn, Jun 06 2022

nonn

a(n) <= 4 since any number can be written as a sum of 4 squares (Lagrange's theorem).

Sequence first differs from A063274, A225926 and A274459 at n = 32 since 32 is a powerful number, a prime power and a perfect power but neither a square nor a cube.

			a(1) = 1, a(4) = 1 (4 = 2^2), a(7) = 4 (7 = 2^2 + 1^2 + 1^2 + 1^2), a(8) = 1 (8 = 2^3), a(12) = 2 (12 = 2^3 + 2^2), a(17) = 2 (17 = 4^2 + 1^2), a(32) = 2 (32 = 4^2 + 4^2).

Cf. A002760, A063274, A225926, A274459, A002376, A002828, A000290, A000578.

lista(n) = {my(v = vector(n)); for(j = 2, 3, for(i = 2, sqrtnint(n, j), v[i^j] = 1)); v[1]=1; v[2]=2; for(i=3, #v, if(v[i]==0, v[i] = vecmin(vector(i\2, k, v[k] + v[i-k])))); v}

cp's OEIS Frontend

A188462 Least number of 5th powers needed to represent n.

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A274459 Least number of perfect powers that add up to n.

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A374012 Least number of 6th powers needed to represent n.

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A084355 Least number of positive cubes needed to represent n!.

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A100854 Least number of positive cubes that sum to n^2.

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A338497 Least number of odd cubes needed to represent n.

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A354761 Least number of squares and cubes that add up to n.

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