A001014 -id:A001014 - cp's OEIS frontend

A194769 Sum of distinct nonzero sixth powers.

1, 64, 65, 729, 730, 793, 794, 4096, 4097, 4160, 4161, 4825, 4826, 4889, 4890, 15625, 15626, 15689, 15690, 16354, 16355, 16418, 16419, 19721, 19722, 19785, 19786, 20450, 20451, 20514, 20515, 46656, 46657, 46720, 46721, 47385, 47386, 47449, 47450, 50752, 50753, 50816

Offset: 1

Charles R Greathouse IV, Sep 02 2011

See A001661 for a proof of the formula. - M. F. Hasler, May 15 2020
From Peter Munn, Aug 02 2023: (Start)
11146309947 = A001661(6) is the largest number not in the sequence.
After a(1) = 1, the next term that is in all the analogous sequences for smaller powers is a(86) = 134067 = A364637(6).
If we tightened the sequence requirement so that the sum was of more than one 6th power, we would remove exactly 30 6th powers from the terms: row 6 of A332065 indicates which 6th powers would remain.
(End)

David A. Corneth, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A001014, A001661, A332065, A364637.
A217846 is a subsequence.
Cf. A003997, A003999, A194768 (analogs for 3rd, 4th and 5th powers).

upto(lim)={
    lim\=1;
    my(v=List(),P=prod(n=1,lim^(1/6),1+x^(n^6),1+O(x^(lim+1))));
    for(n=1,lim,if(polcoeff(P,n),listput(v,n)));
    Vec(v)
}

For n > 9108736851, a(n) = n + 2037573096.

More terms from David A. Corneth, Apr 21 2020

Name qualified by Peter Munn, Aug 02 2023

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}]

A343384 Number of ways to write n as [a^3/3] + [b^3/4] + [c^3/5] + [d^6/6] with a,b,c,d positive integers, where [.] is the floor function.

Original entry on oeis.org

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

Offset: 0

Zhi-Wei Sun, Apr 13 2021

nonn

3-4-5-6 Conjecture: a(n) > 0 for all n >= 0.
We have verified a(n) > 0 for all n = 0..10^6.
Conjecture verified up to 2*10^9. - Giovanni Resta, Apr 28 2021

			a(0) = 1 with 0 = [1^3/3] + [1^3/4] + [1^3/5] + [1^6/6].
a(1) = 1 with 1 = [1^3/3] + [1^3/4] + [2^3/5] + [1^6/6].
a(4) = 1 with 4 = [2^3/3] + [2^3/4] + [1^3/5] + [1^6/6].
a(6) = 1 with 6 = [1^3/3] + [3^3/4] + [1^3/5] + [1^6/6].
a(8) = 1 with 8 = [2^3/3] + [3^3/4] + [1^3/5] + [1^6/6].
a(60) = 1 with 60 = [3^3/3] + [4^3/4] + [5^3/5] + [2^6/6].
a(81) = 1 with 81 = [2^3/3] + [6^3/4] + [5^3/5] + [1^6/6].
a(300) = 1 with 300 = [7^3/3] + [5^3/4] + [9^3/5] + [2^6/6].
a(4434) = 1 with 4434 = [11^3/3] + [4^3/4] + [19^3/5] + [5^6/6].

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, Natural numbers represented by floor(x^2/a) + floor(y^2/b) + floor(z^2/c), arXiv:1504.01608 [math.NT], 2015.

Cf. A000578, A001014, A343326, A343368, A343387.

CQ[n_]:=CQ[n]=n>0&&IntegerQ[n^(1/3)];
tab={};Do[r=0;Do[If[CQ[3(n-Floor[x^6/6]-Floor[y^3/5]-Floor[z^3/4])+s],r=r+1],{s,0,2},{x,1,(6n+5)^(1/6)},{y,1,(5(n-Floor[x^6/6])+4)^(1/3)},{z,1,(4(n-Floor[x^6/6]-Floor[y^3/5])+3)^(1/3)}];tab=Append[tab,r],{n,0,100}];Print[tab]

A349957 Number of ways to write n as x^4 + y^2 + (z^2 + 11*16^w)/60, where x,y,z are nonnegative integers, and w is 0 or 1.

Original entry on oeis.org

1, 2, 3, 5, 6, 4, 3, 3, 3, 4, 3, 4, 5, 3, 2, 2, 4, 4, 5, 9, 9, 3, 3, 4, 6, 5, 5, 9, 7, 4, 4, 6, 5, 2, 4, 8, 7, 3, 5, 7, 7, 4, 4, 4, 4, 4, 6, 9, 4, 3, 3, 9, 9, 4, 4, 5, 7, 2, 4, 4, 4, 2, 7, 7, 4, 3, 5, 12, 7, 3, 1, 6, 6, 4, 5, 8, 3, 1, 4, 5, 6, 3, 8, 14, 13, 6, 5, 5, 6, 6, 9, 8, 6, 3, 4, 8, 6, 6, 5, 12

Offset: 1

Zhi-Wei Sun, Dec 06 2021

nonn

Conjecture 1: a(n) > 0 for all n > 0.
This has been verified for n <= 10^6. It seems that a(n) = 1 only for n = 1, 71, 78, 247, 542, 1258, 1907, 5225, 19798.
Conjecture 2: (i) If a is 1 or 3, then each n = 0,1,2,... can be written as a*x^8 + y^2 + (7*z^4 + w^2)/64 with x,y,z,w nonnegative integers.
(ii) If a is among 1,2,5, then each n = 0,1,2,... can be written as a*x^8 + y^2 + (11*z^4 + w^2)/60 with x,y,z,w nonnegative integers.
Conjecture 3: If (a,b, m) is among the triples (1,7,8), (1,11,12), (2,7,8), (3,11,12), (5,7,8), then each n = 0,1,2,... can be written as a*x^4 + y^2 + (b*z^6 + w^2)/m with x,y,z,w nonnegative integers.
Conjecture 4: (i) If F(x,y,z,w) is x^6 + y^2 + (5*z^4 + 3*w^2)/16 or 3x^6 + 2*y^2 + (11*z^4 + w^2)/60, then each n = 0,1,2,... can be written as F(x,y,z,w) with x,y,z,w nonnegative integers.
(ii) If (a,b,m) is among the triples (1,7,64), (1,11,12), (1,11,60), (2,1,25), (2,1,65), (2,11,4), (2,11,20), (2,11,60), (4,2,9), (4,7,64), (5,11,60), (6,1,10), then each n = 0,1,2,... can be written as a*x^6 + y^2 + (b*z^4 + w^2)/m with x,y,z,w nonnegative integers.

			a(1) = 1 with 1 = 0^4 + 0^2 + (7^2 + 11*16^0)/60.
a(16) = 2 with 16 = 0^4 + 0^2 + (28^2 + 11*16)/60 = 1^4 + 2^2 + (22^2 + 11*16)/60.
a(71) = 1 with 71 = 0^4 + 2^2 + (62^2 + 11*16)/60.
a(78) = 1 with 78 = 2^4 + 5^2 + (47^2 + 11*16^0)/60.
a(247) = 1 with 247 = 3^4 + 3^2 + (97^2 + 11*16^0)/60.
a(542) = 1 with 542 = 3^4 + 21^2 + (32^2 + 11*16)/60.
a(1258) = 1 with 1258 = 2^4 + 15^2 + (247^2 + 11*16^0)/60.
a(1907) = 1 with 1907 = 0^4 + 0^2 + (338^2 + 11*16)/60.
a(5225) = 1 with 5225 = 5^4 + 58^2 + (272^2 + 11*16)/60.
a(19798) = 1 with 19798 = 1^4 + 137^2 + (248^2 + 11*16)/60.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167--190.
Zhi-Wei Sun, New Conjectures in Number Theory and Combinatorics (in Chinese), Harbin Institute of Technology Press, 2021.

Cf. A000290, A000583, A001014, A001016, A349942, A349945, A349956.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
tab={};Do[r=0;Do[If[SQ[60(n-x^4-y^2)-11*16^z],r=r+1],{x,0,(n-1)^(1/4)},{y,0,Sqrt[n-1-x^4]},{z,0,1}];tab=Append[tab,r],{n,1,100}];Print[tab]

A369823 S is a "boomerang sequence": replace each digit d of S by its sixth power: the sequence S remains identical to itself if we follow each result with a comma.

Original entry on oeis.org

0, 1, 4096, 0, 531441, 46656, 0, 15625, 729, 1, 4096, 4096, 1, 4096, 46656, 46656, 15625, 46656, 0, 1, 15625, 46656, 64, 15625, 117649, 64, 531441, 1, 4096, 0, 531441, 46656, 4096, 0, 531441, 46656, 1, 4096, 0, 531441, 46656, 4096, 46656, 46656, 15625, 46656, 4096, 46656, 46656, 15625, 46656, 1

Offset: 1

Eric Angelini, Feb 02 2024

S is the lexicographycally earliest sequence of nonnegative integers with this property.

			a(1) = 0, which raised at the 6th power gives 0
a(2) = 1, which raised at the 6th power gives 1
a(3) = 4096
     1st digit is 4, which raised at the 6th power gives 4096
     2nd digit is 0, which raised at the 6th power gives 0
     3rd digit is 9, which raised at the 6th power gives 531441
     4th digit is 6, which raised at the 6th power gives 46656
Etc. We see that the above last column reproduces S.

Eric Angelini and Jean-Marc Falcoz, Boomerang sequences, Personal blog, Feb 1st 2024.

Cf. A369603, A369604, A369798, A369824, A001014.

a[1]=0;a[2]=1;a[3]=4^6;a[n_]:=a[n]=Flatten[IntegerDigits/@Array[a,n-1]][[n]]^6;Array[a,52] (* Giorgos Kalogeropoulos, Feb 04 2024 *)

A369824 S is a "boomerang sequence": replace each digit d of S by its eighth power: the sequence S remains identical to itself if we follow each result with a comma.

Original entry on oeis.org

0, 1, 256, 390625, 1679616, 6561, 43046721, 0, 1679616, 256, 390625, 1, 1679616, 5764801, 43046721, 1679616, 1, 1679616, 1679616, 390625, 1679616, 1, 65536, 6561, 0, 65536, 1679616, 5764801, 256, 1, 0, 1, 1679616, 5764801, 43046721, 1679616, 1, 1679616, 256, 390625, 1679616, 6561, 43046721

Offset: 1

Eric Angelini, Feb 02 2024

S is the lexicographycally earliest sequence of nonnegative integers with this property.

			a(1) = 0, which raised at the 8th power gives 0
a(2) = 1, which raised at the 8th power gives 1
a(3) = 256
     1st digit is 2, which raised at the 8th power gives 256
     2nd digit is 5, which raised at the 8th power gives 390625
     3rd digit is 6, which raised at the 8th power gives 1679616
Etc. We see that the above last column reproduces S.

Éric Angelini and Jean-Marc Falcoz, Boomerang sequences, Personal blog, Feb 1st 2024.

Cf. A369603, A369604, A369798, A369823, A001014, A001016.

a[1]=0;a[2]=1;a[3]=2^8;a[n_]:=a[n]=Flatten[IntegerDigits/@Array[a,n-1]][[n]]^8;Array[a,43] (* Giorgos Kalogeropoulos, Feb 04 2024 *)

A371344 a(n)/144 is the minimum squared volume > 0 of a tetrahedron with integer edge lengths whose largest is n.

Original entry on oeis.org

2, 11, 26, 47, 54, 107, 146, 191, 242, 299, 191, 134, 146, 146, 151, 767, 423, 151, 854, 558, 764, 491, 503, 464, 146, 146, 431, 944, 666, 146, 146, 350, 599, 311, 599, 511, 1719, 2286, 944, 1871, 1679, 990, 2714, 1907, 990, 551, 959, 1199, 1244, 990, 1206, 854, 764

Offset: 1

Hugo Pfoertner, Mar 19 2024

			a(1) = 2 corresponds to the regular tetrahedron with all edges equal to 1. Its volume is sqrt(2/144) = 0.11785113...

Hugo Pfoertner, Table of n, a(n) for n = 1..450
IBM Research, Tetrahedron Volumes, Ponder This Challenge November 2024.
Sascha Kurz, Enumeration of integral tetrahedra, arXiv:0804.1310 [math.CO], 2008.
Hugo Pfoertner, Plot of log_10(n) vs n, using Plot 2.

Cf. A097125, A371072, A371345.
Subset of A371071.
A001014(n)/72 are the corresponding maximum squared volumes.

\\ See A371345. Replace final #Set(Vec(L)) by vecmin(Vec(L))/2
\\ Second version using simple minded loops and triangle inequalities
\\ Not suitable for larger n
a371344(n) = {my (Vmin=oo,w=vector(6)); w[1]=n; for(w2=1,n,w[2]=w2; for(w3=1,n,w[3]=w3; for(w4=1,n,w[4]=w4; for(w5=1,n,w[5]=w5; for(w6=1,n,w[6]=w6;
forperm (w, v, if(v[4]+v[5]0, Vmin=min(Vmin,CM)))))))); Vmin/2}; \\ return value corrected by M. F. Hasler, Dec 02 2024

/* equivalent to the preceding, but simplified */
A371344(n) = {my (Vmin=oo,CM, n2=n^2); forvec(v=vector(5,k,[1,n]), v[4]+v[5]= Vmin || Vmin=CM); Vmin/2} \\ M. F. Hasler, Dec 02 2024

a(33), a(37), a(38), and a(43) corrected by Hugo Pfoertner, Dec 03 2024

A373994 a(n) is the largest digit sum of all n-digit sixth powers.

Original entry on oeis.org

1, 10, 18, 19, 27, 28, 45, 37, 46, 64, 64, 81, 82, 82, 91, 100, 100, 118, 117, 126, 136, 136, 154, 154, 163, 163, 172, 181, 181, 190, 199, 208, 217, 226, 235, 235, 243, 244, 261, 262, 280, 280, 280, 289, 298, 298, 307, 325, 325, 325, 334, 352, 352, 361, 370

Offset: 1

Zhining Yang, Jun 26 2024

			a(6) = 28 because 28 is the largest digital sum encountered among all 6-digit sixth powers (117649, 262144, 531441).

Kevin Ryde, C Code

Cf. A001014, A348300, A373727, A373914, A374025.

C
```
/* See links. */
```

Table[Max@Map[Total@IntegerDigits[#^6] &, Range[Ceiling[10^((n - 1)/6)], Floor[(10^n-1)^(1/6)]]], {n, 42}]

A069473 First differences of (n+1)^6-n^6 (A022522).

Original entry on oeis.org

62, 602, 2702, 8162, 19502, 39962, 73502, 124802, 199262, 303002, 442862, 626402, 861902, 1158362, 1525502, 1973762, 2514302, 3159002, 3920462, 4812002, 5847662, 7042202, 8411102, 9970562, 11737502, 13729562, 15965102, 18463202

Offset: 0

Eli McGowan (ejmcgowa(AT)mail.lakeheadu.ca), Mar 26 2002

Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A001014, A022522, A069474, A069475, A069476, A068236.

[30*n^4+120*n^3+210*n^2+180*n+62: n in [0..30]]; // Bruno Berselli, Feb 25 2015

Differences[Table[(n + 1)^6 - n^6, {n, 0, 30}]] (* Harvey P. Dale, Dec 27 2011 *)

a(n) = 30*n^4 + 120*n^3 + 210*n^2 + 180*n + 62.

G.f.: 2*(31 + 146*x + 156*x^2 + 26*x^3 + x^4)/(1 - x)^5. [Bruno Berselli, Feb 25 2015]

Offset changed from 1 to 0 and added a(0)=62 by Bruno Berselli, Feb 25 2015

A069476 First differences of A069475, successive differences of (n+1)^6-n^6.

Original entry on oeis.org

1800, 2520, 3240, 3960, 4680, 5400, 6120, 6840, 7560, 8280, 9000, 9720, 10440, 11160, 11880, 12600, 13320, 14040, 14760, 15480, 16200, 16920, 17640, 18360, 19080, 19800, 20520, 21240, 21960, 22680, 23400, 24120, 24840, 25560, 26280, 27000

Offset: 0

Eli McGowan (ejmcgowa(AT)mail.lakeheadu.ca), Mar 26 2002

Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A001014, A020735, A022522, A069473, A069474, A069475.

Table[720 n + 1800, {n, 0, 40}] (* Bruno Berselli, Feb 25 2015 *)

a(n) = 720*n + 1800 = 360*A020735(n+1).

G.f.: 360*(5 - 3*x)/(1 - x)^2. [Bruno Berselli, Feb 25 2015]

Offset changed from 1 to 0 and added a(0)=1800 by Bruno Berselli, Feb 25 2015

cp's OEIS Frontend

A194769 Sum of distinct nonzero sixth powers.

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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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A343384 Number of ways to write n as [a^3/3] + [b^3/4] + [c^3/5] + [d^6/6] with a,b,c,d positive integers, where [.] is the floor function.

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A349957 Number of ways to write n as x^4 + y^2 + (z^2 + 11*16^w)/60, where x,y,z are nonnegative integers, and w is 0 or 1.

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A369823 S is a "boomerang sequence": replace each digit d of S by its sixth power: the sequence S remains identical to itself if we follow each result with a comma.

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A369824 S is a "boomerang sequence": replace each digit d of S by its eighth power: the sequence S remains identical to itself if we follow each result with a comma.

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A371344 a(n)/144 is the minimum squared volume > 0 of a tetrahedron with integer edge lengths whose largest is n.

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PARI

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A373994 a(n) is the largest digit sum of all n-digit sixth powers.

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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.