A002804 -id:A002804 - cp's OEIS frontend

A185187 Smallest number for which the greedy algorithm fails to find the sum of n-th powers with at most A002804 terms.

23, 50, 160, 466, 1432, 4362, 12960, 39138, 117416, 353274, 1059824, 3183570, 9550712, 28668522, 86038336, 258246082, 774607176, 2324083674, 6973299600, 20918850226, 62758647832, 188280137802, 564857190624, 1694571571874, 5083681161192, 15251177701306

Offset: 2

J. Lowell, Feb 19 2011

nonn

For n > 2, a(n) = 3^n + (floor(3^n//2^n) - 1)*2^n + (2^n - 1), with A002804(n)+1 terms in the greedy representation. - Michael S. Branicky, Dec 15 2021

			23 qualifies for a(2) because 23 as a sum of squares with the greedy algorithm is 16+4+1+1+1 (5 terms,) but A002804(2) = 4.
50 qualifies for a(3) because 50 as a sum of cubes with the greedy algorithm is 27+8+8+1+1+1+1+1+1+1 (10 terms,) but A002804(3) = 9.

Michael S. Branicky, Table of n, a(n) for n = 2..2095

Cf. A002804.

# exhaustive search
from sympy import integer_nthroot
def g(n): twon = (1 << n); return twon + 3**n//twon - 2
def greedy(k, n):
    if k < (1 << n): return k
    bigpow = integer_nthroot(k, n)[0]**n
    m, r = divmod(k, bigpow)
    return  m + greedy(r, n)
def a(n):
    k, gn = 2**n, g(n)
    while greedy(k, n) <= gn: k += 1
    return k
print([a(n) for n in range(2, 12)]) # Michael S. Branicky, Dec 15 2021

# direct computation based on formula
def a(n): return 23 if n == 2 else 3**n + (3**n//2**n-1)*2**n + (2**n-1)
print([a(n) for n in range(2, 28)]) # Michael S. Branicky, Dec 15 2021

a(6) and beyond from Michael S. Branicky, Dec 15 2021

A002377 Least number of 4th 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, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 5, 1, 2, 3

Offset: 1

N. J. A. Sloane

No terms are greater than 19, see A002804. - Charles R Greathouse IV, Aug 01 2013

Seven values of n need the maximum of 19 fourth powers. These form the arithmetic progression {79, 159, 239, 319, 399, 479, 559} each term being congruent to 79 mod 80. For n < 625 the available fourth powers are congruent to 1 or 16 mod 80, requiring 4*16 + 15*1 to sum to 79. However, 625 = 5^4 is congruent to 65 and 1*65 + 14*1 = 79. So for n > 625 and congruent to 79, only 15 fourth powers are needed to satisfy the mod 80 arithmetic. - Peter Munn, Apr 12 2017

D. H. Lehmer, Guide to Tables in the Theory of Numbers. Bulletin No. 105, National Research Council, Washington, DC, 1941, p. 82.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 1..14000
C. A. Bretschneider, Zerlegung der Zahlen bis 4100 in Biquadrate, J. Reine Angew. Math., 46 (1853), 1-28.
Eric Weisstein's World of Mathematics, Biquadratic Number

Cf. A002828, A002376, A188462, A374012.

Cf. A046049, A046050, A099591.

Cnt4[n_] := Module[{k = 1}, While[Length[PowersRepresentations[n, k, 4]] == 0, k++]; k]; Array[Cnt4, 100] (* T. D. Noe, Apr 01 2011 *)
seq[n_] := Module[{v = Table[0, {n}], s, p}, s = Sum[x^(k^4), {k, 1, n^(1/4)}] + O[x]^(n+1); p=1; For[k=1, k <= 19, k++, p *= s; For[i=1, i <= n, i++, If[v[[i]]==0 && Coefficient[p, x, i] != 0, v[[i]] = k]]]; v];
seq[100] (* Jean-François Alcover, Sep 28 2019, after Andrew Howroyd *)

seq(n)={my(v=vector(n), s=sum(k=1, sqrtint(sqrtint(n)), x^(k^4)) + O(x*x^n), p=1); for(k=1, 19, p*=s; for(i=1, n, if(!v[i] && polcoeff(p,i), v[i]=k))); v} \\ Andrew Howroyd, Jul 06 2018

from itertools import count
from sympy.solvers.diophantine.diophantine import power_representation
def A002377(n):
    if n == 1: return 1
    for k in count(1):
        try:
            next(power_representation(n,4,k))
        except:
            continue
        return k # Chai Wah Wu, Jun 25 2024

More terms from Arlin Anderson (starship1(AT)gmail.com)

A002376 Least number of positive cubes needed to sum to n.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4, 5, 6, 7, 8, 2, 3, 4, 5, 6, 7, 8, 9, 3, 4, 5, 1, 2, 3, 4, 5, 4, 5, 6, 2, 3, 4, 5, 6, 5, 6, 7, 3, 4, 5, 6, 7, 6, 7, 8, 4, 5, 6, 2, 3, 4, 5, 6, 5, 6, 7, 3, 4, 1, 2, 3, 4, 5, 6, 4, 5, 2, 3, 4, 5, 6, 7, 5, 6, 3, 3, 4, 5, 6, 7, 6, 7, 4, 4, 5, 2, 3, 4, 5, 6, 5, 5, 6, 3, 4, 5, 6, 7, 6, 6

Offset: 1

N. J. A. Sloane

No terms are greater than 9, see A002804. - Charles R Greathouse IV, Aug 01 2013

D. H. Lehmer, Guide to Tables in the Theory of Numbers. Bulletin No. 105, National Research Council, Washington, DC, 1941, p. 81.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
A. R. Zornow, De compositione numerorum e cubis integris positivus, J. Reine Angew. Math., 14 (1835), 276-280.

David A. Corneth, Table of n, a(n) for n = 1..10000 (first 1000 terms from T. D. Noe)
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Eric Weisstein's World of Mathematics, Cubic Number

Cf. A000578, A003325 (numbers requiring 2 cubes), A047702 (numbers requiring 3 cubes), A047703 (numbers requiring 4 cubes), A047704 (numbers requiring 5 cubes), A046040 (numbers requiring 6 cubes), A018890 (numbers requiring 7 cubes), A018888 (numbers requiring 8 or 9 cubes), A055401 (cubes needed by greedy algorithm).

f:= proc(n) option remember;
  min(seq(procname(n - i^3)+1, i=1..floor(n^(1/3))))
end proc:
f(0):= 0:
map(f, [$1..100]); # Robert Israel, Jun 30 2017

CubesCnt[n_] := Module[{k = 1}, While[Length[PowersRepresentations[n, k, 3]] == 0, k++]; k]; Array[CubesCnt, 100] (* T. D. Noe, Apr 01 2011 *)

from itertools import count
from sympy.solvers.diophantine.diophantine import power_representation
def A002376(n):
    if n == 1: return 1
    for k in count(1):
        try:
            next(power_representation(n,3,k))
        except:
            continue
        return k # Chai Wah Wu, Jun 25 2024

The g.f. conjectured by Simon Plouffe in his 1992 dissertation,

-(-1-z-z^2-z^3-z^4-z^5-z^6+6*z^7)/(z+1)/(z^2+1)/(z^4+1)/(z-1)^2, is incorrect: the first wrong coefficient is that of z^26. - Robert Israel, Jun 30 2017

More terms from Arlin Anderson (starship1(AT)gmail.com)

A079611 Waring's problem: conjectured values for G(n), the smallest number m such that every sufficiently large number is the sum of at most m n-th powers of positive integers.

Original entry on oeis.org

1, 4, 4, 16, 6, 9, 8, 32, 13, 12, 12, 16, 14, 15, 16, 64, 18, 27, 20, 25

Offset: 1

N. J. A. Sloane, Jan 28 2003

The only certain values are G(1) = 1, G(2) = 4 and G(4) = 16.

See A002804 for the simpler problem of Waring's original conjecture, which does not restrict the bound to "sufficiently large" numbers. - M. F. Hasler, Jun 29 2014

			It is known that every sufficiently large number is the sum of 16 fourth powers, and 16 is the smallest number with this property, so a(4) = G(4) = 16. (The numbers 16^k*31 are not the sum of fewer than 16 fourth powers.)

G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, 5th ed., Oxford Univ. Press, 1979, th. 395 (shows G(4) >= 16).
R. C. Vaughan and T. D. Wooley, Waring's problem: a survey, pp. 285-324 of Surveys in Number Theory (Urbana, May 21, 2000), ed. M. A. Bennett et al., Peters, 2003.

H. Davenport, On Waring's problem for fourth powers, Annals of Mathematics, 40 (1939), 731-747. (Shows that G(4) <= 16.)
Wikipedia, Waring's Problem.
Trevor D. Wooley, On Waring's problem for intermediate powers, arXiv:1602.03221 [math.NT], 2016.

Cf. A002376, A002377, A002804, A174406.

Entry revised Jun 29 2014

A271099 Number of ordered ways to write n as u^3 + v^3 + 2x^3 + 2y^3 + 3*z^3, where u, v, x, y and z are nonnegative integers with u <= v and x <= y.

Original entry on oeis.org

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

Offset: 0

Zhi-Wei Sun, Mar 30 2016

nonn

Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 1, 10, 14, 15, 17, 22, 38, 39, 45, 47, 50, 52, 76, 102, 103, 188, 295, 366, 534.
(ii) Any natural number n can be written as s^4 + t^4 + 2*u^4 + 2*v^4 + 3*x^4 + 3*y^4 + 7*z^4, where s, t, u, v, x, y and z are nonnegative integers. Also, each natural number n can be written as r^5 + s^5 +  t^5 + u^5 + 2*v^5 + 4*w^5 + 6*x^5 + 9*y^5 +12*z^5, where r, s, t, u, v, w, x, y and z are nonnegative integers.
(iii) In general, for any integer k > 2, there are 2*k-1 positive integers c(1), c(2), ..., c(2k-1) such that {c(1)*x(1)^k + c(2)*x(2)^k + ... + c(2k-1)*x(2k-1)^k: x(1),x(2),...,x(k) = 0,1,2,...} = {0,1,2,3,...} and that c(1)+c(2)+...+c(2k-1) = g(k), where g(k) = 2^k+floor((3/2)^k)-2 as given by A002804.
This conjecture is stronger than the classical Waring problem on sums of k-th powers. Concerning parts (i) and (ii) of the conjecture, we note that 1+1+2+2+3 = 9 = g(3), 1+1+2+2+3+3+7 = 19 = g(4) and 1+1+1+1+2+4+6+9+12 = 37 = g(5).
We have verified that a(n) > 0 for all n = 0..10^6, and that part (ii) of the conjecture holds for n up to 10^5. Concerning part (iii) for k = 6, we conjecture that any natural number can be written as x(1)^6+x(2)^6+x(3)^6+x(4)^6+x(5)^6+3*x(6)^6+5*x(7)^6+6*x(8)^6+10*x(9)^6+18*x(10)^6+26*x(11)^6 with x(1),x(2),...,x(11) nonnegative integers. Note that 1+1+1+1+1+3+5+6+10+18+26 = 73 = g(6). - Zhi-Wei Sun, Mar 31 2016

			a(1) = 1 since 1 = 0^3 + 1^3 + 2*0^3 + 2*0^3 + 3*0^3.
a(10) = 1 since 10 = 0^3 + 2^3 + 2*0^3 + 2*1^3 + 3*0^3.
a(14) = 1 since 14 = 1^3 + 2^3 + 2*0^3 + 2*1^3 + 3*1^3.
a(15) = 1 since 15 = 0^3 + 2^3 + 2*1^3 + 2*1^3 + 3*1^3.
a(17) = 1 since 17 = 0^3 + 1^3 + 2*0^3 + 2*2^3 + 3*0^3.
a(22) = 1 since 22 = 0^3 + 1^3 + 2*1^3 + 2*2^3 + 3*1^3.
a(38) = 1 since 38 = 2^3 + 3^3 + 2*0^3 + 2*0^3 + 3*1^3.
a(39) = 1 since 39 = 2^3 + 3^3 + 2*1^3 + 2*1^3 + 3*0^3.
a(45) = 1 since 45 = 0^3 + 3^3 + 2*1^3 + 2*2^3 + 3*0^3.
a(47) = 1 since 47 = 1^3 + 3^3 + 2*0^3 + 2*2^3 + 3*1^3.
a(50) = 1 since 50 = 0^3 + 2^3 + 2*1^3 + 2*2^3 + 3*2^3.
a(52) = 1 since 52 = 1^3 + 3^3 + 2*0^3 + 2*0^3 + 3*2^3.
a(76) = 1 since 76 = 2^3 + 4^3 + 2*1^3 +2*1^3 + 3*0^3.
a(102) = 1 since 102 = 0^3 + 2^3 + 2*2^3 + 2*3^3 + 3*2^3.
a(103) = 1 since 103 = 1^3 + 2^3 + 2*2^3 + 2*3^3 + 3*2^3.
a(188) = 1 since 188 = 3^3 + 4^3 + 2*0^3 + 2*2^3 + 3*3^3.
a(295) = 1 since 295 = 1^3 + 6^3 + 2*0^3 + 2*3^3 + 3*2^3.
a(366) = 1 since 366 = 2^3 + 3^3 + 2*0^3 + 2*5^3 + 3*3^3.
a(534) = 1 since 534 = 1^3 + 8^3 + 2*1^3 + 2*2^3 + 3*1^3.

M. B. Nathanson, Additive Number Theory: The Classical Bases, Grad. Texts in Math., Vol 164, Springer, 1996, Chapters 2 and 3.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000

Cf. A000578, A000583, A000584, A001014, A002804.

CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)]
Do[r=0;Do[If[CQ[n-3z^3-2x^3-2y^3-u^3],r=r+1],{z,0,(n/3)^(1/3)},{x,0,((n-3z^3)/4)^(1/3)},{y,x,((n-3z^3-2x^3)/2)^(1/3)},{u,0,((n-3z^3-2x^3-2y^3)/2)^(1/3)}];Print[n," ",r];Continue,{n,0,70}]

A271237 Number of ordered ways to write n as u^3 + 2v^3 + 3x^3 + 4y^3 + 5z^3, where u, v, x, y and z are nonnegative integers.

Original entry on oeis.org

1, 1, 1, 2, 2, 3, 3, 3, 4, 3, 4, 3, 3, 3, 2, 3, 2, 3, 1, 2, 3, 2, 2, 1, 4, 3, 2, 3, 3, 5, 3, 4, 6, 4, 5, 4, 6, 4, 4, 3, 5, 5, 3, 6, 3, 6, 4, 4, 6, 3, 5, 4, 4, 4, 3, 4, 5, 7, 4, 6, 4, 5, 6, 4, 10, 2, 6, 8, 3, 7, 4, 8, 6, 5, 5, 4, 5, 2, 6, 1, 5, 3, 3, 8, 5, 7, 6, 6, 9, 6, 7, 6, 6, 5, 5, 6, 4, 6, 6, 8, 1

Offset: 0

Zhi-Wei Sun, Apr 02 2016

nonn

Conjecture: We have {u^3+a*v^3+b*x^3+c*y^3+d*z^3: u,v,x,y,z = 0,1,2,...} = {0,1,2,...} whenever (a,b,c,d) is among the following 32 quadruples: (1,2,2,3), (1,2,2,4), (1,2,3,4), (1,2,4,5), (1,2,4,6), (1,2,4,9), (1,2,4,10), (1,2,4,11), (1,2,4,18), (1,3,4,6), (1,3,4,9), (1,3,4,10), (2,2,4,5), (2,2,6,9), (2,3,4,5), (2,3,4,6), (2,3,4,7), (2,3,4,8), (2,3,4,9), (2,3,4,10), (2,3,4,12), (2,3,4,15), (2,3,4,18), (2,3,5,6), (2,3,6,12), (2,3,6,15), (2,4,5,6), (2,4,5,8), (2,4,5,9), (2,4,5,10), (2,4,6,7), (2,4,7,10).
In particular, this implies that a(n) > 0 for all n = 0,1,2,... We guess that a(n) = 1 only for n = 0, 1, 2, 18, 23, 79, 100.
If {m*u^3+a*v^3+b*x^3+c*y^3+d*z^3: u,v,x,y,z = 0,1,2,...} = {0,1,2,...} with 1 <= m <= a <= b <= c <= d,  then m = 1, and we can show that (a,b,c,d) must be among the 32 quadruples listed in the conjecture (cf. Theorem 1.2 of the linked 2017 paper).
Conjecture verified for all the 32 quadruples up to 10^11. - Mauro Fiorentini, Jul 09 2023
It is known that there are exactly 54 quadruples (a,b,c,d) with 1 <= a <= b <= c <= d such that {a*w^2+b*x^2+c*y^2+d*z^2: w,x,y,z = 0,1,2,...} = {0,1,2,...}.
See also A271099 and A271169 for conjectures refining Waring's problem.
We also conjecture that if P(u,v,x,y,z) is one of the four polynomials u^6+v^3+2*x^3+4*y^3+5*z^3 and a*u^6+v^3+2*x^3+3*y^3+4*z^3 (a = 5,8,12) then any natural number can be written as P(u,v,x,y,z) with u,v,x,y,z nonnegative integers. - Zhi-Wei Sun, Apr 06 2016
Conjecture verified for all the 4 polynomials up to 10^11. - Mauro Fiorentini, Jul 09 2023

			a(2) = 1 since 2 = 0^3 + 2*1^3 + 3*0^3 + 4*0^3 + 5*0^3.
a(18) = 1 since 18 = 2^3 + 2*1^3 + 3*1^3 + 4*0^3 + 5*1^3.
a(23) = 1 since 23 = 0^3 + 2*2^3 + 3*1^3 + 4*1^3 + 5*0^3.
a(79) = 1 since 79 = 1^3 + 2*3^3 + 3*2^3 + 4*0^3 + 5*0^3.
a(100) = 1 since 100 = 2^3 + 2*1^3 + 3*3^3 + 4*1^3 + 5*1^3.

S. Ramanujan, On the expression of a number in the form a*x^2 + b*y^2 + c*z^2 + d*w^2, Proc. Cambridge Philos. Soc. 19(1917), 11-21.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
L. E. Dickson, Quaternary quadratic forms representing all integers, Amer. J. Math. 49(1927), 39-56.
Zhi-Wei Sun, A result similar to Lagrange's theorem, J. Number Theory 162(2016), 190-211.
Zhi-Wei Sun, New conjectures on representations of integers (I), Nanjing Univ. J. Math. Biquarterly 34(2017), no. 2, 97-120.

Cf. A000578, A002804, A001014, A271099, A271169.

CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)]
Do[r=0;Do[If[CQ[n-5z^3-4y^3-3x^3-2v^3],r=r+1],{z,0,(n/5)^(1/3)},{y,0,((n-5z^3)/4)^(1/3)},{x,0,((n-5z^3-4y^3)/3)^(1/3)},{v,0,((n-5z^3-4y^3-3x^3)/2)^(1/3)}];Print[n," ",r];Continue,{n,0,100}]

A174406 a(n) = smallest number u such that almost every number is the sum of at most u n-th powers of positive numbers.

Original entry on oeis.org

1, 4, 4, 15

Offset: 1

N. J. A. Sloane, Nov 27 2010

A variant of Waring's problem.
"Almost all" means that the exceptions have zero density.
Only three other values of the sequence are known: a(8) = 32, a(16) = 64, and a(32) = 128. The cited survey by Vaughan and Wooley shows that G_1(8) = 32, G_1(16) = 64, and G_1(32) = 128.  The quantity G_1(5) has not been evaluated, nor have G_1(6) and G_1(7). - David Covert, Jun 29 2016

R. C. Vaughan and T. D. Wooley, Waring’s problem: a survey, Number Theory for the Millennium, III (Urbana, IL, 2000), A K Peters, Natick, MA, 2002, pp. 301-340.

Cf. A002804, A079611.

a(5)-a(7) removed by David Covert, Jun 29 2016

A252486 Smallest k such that n^6 = a_1^6+...+a_k^6 where all the a_i are positive integers less than n.

Original entry on oeis.org

64, 36, 15, 29, 22, 21, 15, 19, 15, 17, 15, 16, 14, 15, 13, 12, 11, 11, 13, 14, 12, 13, 13, 12, 12, 12, 12, 12, 11, 11, 11, 11, 11, 13, 11, 11, 11, 10, 11, 11, 11, 11, 11, 10, 11, 11, 11, 11, 11, 11, 11, 11, 9, 11, 10, 11, 11, 11, 9, 10, 11, 11, 11, 11, 10

Offset: 2

M. F. Hasler, Dec 17 2014

nonn

Inspired by Fermat's Last Theorem: 2 never occurs in this sequence.
No n is known for which a(n)<7, according to the MathWorld page. The values 7, 8, 9, 10 and 11 occur first at indices 1141, 251, 54, 39, 18, cf. sequence A252476.
I conjecture that the sequence is bounded by the initial term. Probably even a(3)=36, a(5)=29, a(6)=22 and some more are followed only by smaller terms.
From results on Waring's problem, it is known that all a(n) <= A002804(6) = 73, and a(n) <= 24 for all sufficiently large n. - Robert Israel, Aug 17 2015

Giovanni Resta, Table of n, a(n) for n = 2..200
Jean-Charles Meyrignac, Computing Minimal Equal Sums Of Like Powers
Manfred Scheucher, Sage Script
Eric W. Weisstein, Diophantine Equation--6th Powers
Eric W. Weisstein, Waring's Problem

Cf. A161882, A161883, A161884, A161885.

M:= 10^8:
R:= Vector(M, 74, datatype=integer[4]):
for p from 1 to floor(M^(1/6)) do
  p6:= p^6;
  if p > 1 then A[p]:= R[p6] fi;
  R[p6]:= 1;
  for j from p6+1 to M do
    R[j]:= min(R[j], 1+R[j - p6]);
  od
od:
F:= proc(n, k, ub)
   local lb, m, bestyet, res;
   if ub <= 0 then return -1 fi;
   if n <= M then
     if n = 0 then return 0
     elif R[n] > ub then return -1
     else return R[n]
     fi
   fi;
   lb:= floor(n/k^6);
   if lb > ub then return -1 fi;
   bestyet:= ub;
   for m from lb to 0 by -1 do
     res:= procname(n-m*k^6, k-1, bestyet-m);
     if res >= 0 then
       bestyet:= res+m;
     fi
   od:
   return bestyet
end proc:
for n from floor(M^(1/6))+1 to 50 do
   A[n]:= F(n^6, n-1, 73)
od:
seq(A[n], n=2..50); # Robert Israel, Aug 17 2015

a[n_] := Module[{k}, For[k = 7, True, k++, If[IntegerPartitions[n^6, {k}, Range[n-1]^6] != {}, Print[n, " ", k]; Return[k]]]];
Table[a[n], {n, 2, 100}] (* Jean-François Alcover, Jul 29 2023 *)

a(n,verbose=0,m=6)={N=n^m;for(k=3,64,forvec(v=vector(k-1,i,[1,n\sqrtn(k+1-i,m)]),ispower(N-sum(i=1,k-1,v[i]^m),m,&K)&&K>0&&!(verbose&&print1("/*"n" "v"*/"))&&return(k),1))}

from itertools import count
from sympy.solvers.diophantine.diophantine import power_representation
def A252486(n):
    m = n**6
    for k in count(2):
        try:
            next(power_representation(m,6,k))
        except:
            continue
        return k # Chai Wah Wu, Jun 25 2024

More terms from Manfred Scheucher, Aug 15 2015

a(53)-a(66) from Giovanni Resta, Aug 17 2015

A271169 Number of ordered ways to write n as s^5 + t^5 + 2u^5 + 3v^5 + 4w^5 + 5x^5 + 7y^5 + 14z^5, where s,t,u,v,w,x,y,z are nonnegative integers with s <= t.

Original entry on oeis.org

1, 1, 2, 2, 3, 4, 4, 6, 5, 7, 6, 7, 7, 6, 8, 6, 8, 6, 7, 7, 6, 8, 6, 8, 6, 7, 7, 6, 7, 5, 6, 4, 5, 4, 3, 4, 3, 4, 3, 4, 4, 4, 5, 4, 5, 4, 5, 5, 4, 5, 4, 5, 4, 5, 5, 4, 5, 4, 5, 4, 4, 4, 3, 3, 4, 3, 3, 3, 4, 5, 3, 6, 4, 7, 5, 5, 7, 4, 8, 4, 7

Offset: 0

Zhi-Wei Sun, Mar 31 2016

nonn

Conjecture: a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 1, 2602.
Note that 1+1+2+3+4+5+7+14 = 37. In 1964 J.-R. Chen proved that any natural number can be written as the sum of 37 fifth powers of nonnegative integers.
For k = 2,3,4,... define s(k) as the smallest positive integer s such that {a(1)*x(1)^k+...+a(s)*x(s)^k: x(1),...,x(s) = 0,1,2,...} = {0,1,2,...} for some positive integers a(1), ..., a(s), and t(k) as the least positive integer t such that {a(1)*x(1)^k+...+a(t)*x(t)^k: x(1),...,x(t) = 0,1,2,...} = {0,1,2,...} for some positive integers a(1), ..., a(t) with a(1)+...+a(t) = g(k), where g(.) is given by A002804. Then s(k) <= t(k) <= g(k). Part (iii) of the conjecture in A271099 implies that t(k) <= 2k-1 for k > 2. It is easy to see that s(2) = t(2) = 4. Our computation suggests that s(3) = t(3) = 5, s(4) = t(4) = 7, s(5) = t(5) = 8 (which is smaller than 2*5-1), and s(6) = t(6) = 10. We conjecture that s(k) = t(k) for any integer k > 1, and that each natural number can be written as x(1)^6+x(2)^6+x(3)^6+2*x(4)^6+3*x(5)^6+5*x(6)^6+6*x(7)^6+10*x(8)^6+18*x(9)^6+26*x(10)^6, where x(1),x(2),...,x(10) are nonnegative integers. Note that 1+1+1+2+3+5+6+10+18+26 = 73 = g(6).
We also conjecture that any natural number can be written as s^5+t^5+2*u^5+3*v^5+4*w^5+6*x^5+8*y^5+12*z^5, with s,t,u,v,w,x,y,z nonnegative integers. Note that 1+1+2+3+4+6+8+12 = 37 = g(5). - Zhi-Wei Sun, Apr 04 2016

			a(1) = 1 since 1 = 0^5 + 1^5 + 2*0^5 + 3*0^5 + 4*0^5 + 5*0^5 + 7*0^5 + 14*0^5.
a(2602) = 1 since 2602 = 0^5 + 1^5 + 2*4^5 + 3*2^5 + 4*1^5 + 5*1^5 + 7*0^5 + 14*2^5.

J.-R. Chen, Waring's Problem for g(5)=37, Sci. Sinica 13(1964), 1547-1568.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, New conjectures on representations of integers (I), Nanjing Univ. J. Math. Biquarterly 34(2017), no. 2, 97-120.

Cf. A000578, A000583, A000584, A001014, A002804, A271099.

FQ[n_]:=FQ[n]=IntegerQ[n^(1/5)]
Do[r=0;Do[If[FQ[n-14z^5-7y^5-5x^5-4w^5-3v^5-2u^5-s^5],r=r+1],{z,0,(n/14)^(1/5)},{y,0,((n-14z^5)/7)^(1/5)},{x,0,((n-14z^5-7y^5)/5)^(1/5)},{w,0,((n-14z^5-7y^5-5x^5)/4)^(1/5)},{v,0,((n-14z^5-7y^5-5x^5-4w^5)/3)^(1/5)},{u,0,((n-14z^5-7y^5-5x^5-4w^5-3v^5)/2)^(1/5)}, {s,0,((n-14z^5-7y^5-5x^5-4w^5-3v^5-2u^5)/2)^(1/5)}];Print[n," ",r];Label[aa];Continue,{n,0,80}]

A379396 Numbers that can be written in exactly one way as a sum of at most nine positive third powers.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 31, 38, 39, 45, 46, 47, 50, 52, 53

Offset: 1

Patrick De Geest, Dec 22 2024

The 'nine' is not arbitrary. Waring stated that every natural number can be expressed as a sum of at most nine cubes. (Cf. A002804)

			7 is in the sequence since there is only 1^3+1^3+1^3+1^3+1^3+1^3+1^3.
53 is in the sequence since there is only 1^3+1^3+2^3+2^3+2^3+3^3;

Eric Weisstein's World of Mathematics, Waring's Problem
Wikipedia, Waring's Problem
Index entries for sequences related to sums of cubes

Cf. A002804, A025453, A379397, A379398, A379399, A379400

upto(n) = my(v=vector(n), maxb=sqrtnint(n,3)); forvec(x=vector(9,i,[0,maxb]), s=sum(i=1,9,x[i]^3); if(0x==1,v,1) \\ David A. Corneth, Dec 23 2024

cp's OEIS Frontend

A185187 Smallest number for which the greedy algorithm fails to find the sum of n-th powers with at most A002804 terms.

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A002377 Least number of 4th powers needed to represent n.

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A002376 Least number of positive cubes needed to sum to n.

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A079611 Waring's problem: conjectured values for G(n), the smallest number m such that every sufficiently large number is the sum of at most m n-th powers of positive integers.

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A271099 Number of ordered ways to write n as u^3 + v^3 + 2*x^3 + 2*y^3 + 3*z^3, where u, v, x, y and z are nonnegative integers with u <= v and x <= y.

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A271237 Number of ordered ways to write n as u^3 + 2*v^3 + 3*x^3 + 4*y^3 + 5*z^3, where u, v, x, y and z are nonnegative integers.

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A174406 a(n) = smallest number u such that almost every number is the sum of at most u n-th powers of positive numbers.

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A271099 Number of ordered ways to write n as u^3 + v^3 + 2x^3 + 2y^3 + 3*z^3, where u, v, x, y and z are nonnegative integers with u <= v and x <= y.

A271237 Number of ordered ways to write n as u^3 + 2v^3 + 3x^3 + 4y^3 + 5z^3, where u, v, x, y and z are nonnegative integers.

A271169 Number of ordered ways to write n as s^5 + t^5 + 2u^5 + 3v^5 + 4w^5 + 5x^5 + 7y^5 + 14z^5, where s,t,u,v,w,x,y,z are nonnegative integers with s <= t.