A271099 -id:A271099 - cp's OEIS frontend

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.

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}]

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}]

A267826 Numbers not of the form w^3 + 2x^3 + 3y^3 + 4*z^3, where w, x, y and z are nonnegative integers.

Original entry on oeis.org

18, 22, 39, 60, 63, 74, 76, 77, 100, 103, 106, 107, 117, 126, 178, 180, 201, 215, 228, 230, 245, 271, 289, 291, 295, 315, 341, 356, 357, 393, 413, 419, 420, 480, 481, 523, 559, 606, 616, 671, 673, 705, 854, 855, 963, 980, 981, 998, 1103, 1121, 1130, 1298, 1484, 1510, 1643, 1729, 1849, 1916, 1934, 1946

Offset: 1

Zhi-Wei Sun, Apr 07 2016

nonn

Conjecture: The sequence has exactly 122 terms the last of   which is a(122) = 41405.
We have verified that there are no terms between 41406 and 2*10^5.
The conjecture implies that {P(v)+w^3+2*x^3+3*y^3+4*z^3: w,x,y,z = 0,1,2,...} = {0,1,2,...} whenever P(v) is among the polynomials a*v^3 (a = 1,5,6,7,9,10,12,15,18), b*v^4 (b = 1,2,3,5,6,12,18), c*v^5 (c = 1,2,5,12) and d*v^k (d = 5,12; k = 6,7). Moreover, it also implies that {8*t+w^3+2*x^3+3*y^3+4*z^3: t = 0,1; w,x,y,z = 0,1,2,...} = {0,1,2,...}. If a,b,c,d and m are positive integers with {m*t+a*w^3+b*x^3+c*y^3+d*z^3: t = 0,1; w,x,y,z = 0,1,2,...} = {0,1,2,...}, then we must have m = 8 and {a,b,c,d} = {1,2,3,4}.

			a(1) = 18 since it is the first nonnegative integer not in the set {w^3 + 2*x^3 + 3*y^3 + 4*z^3: w,x,y,z = 0,1,2,...}.

Zhi-Wei Sun, Table of n, a(n) for n = 1..122
Zhi-Wei Sun, Universal sums u^3+a*v^3+b*x^3+c*y^3+d*z^3 with u, v, x, y, z nonnegative integers, a message to Number Theory Mailing List, April 3, 2016.

Cf. A000578, A002804, A022566, A267861, A271099, A271169, A271237.

CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)]
n=0;Do[Do[If[CQ[m-4*z^3-3y^3-2x^3],Goto[aa]],{z,0,(m/4)^(1/3)},{y,0,((m-4z^3)/3)^(1/3)},{x,0,((m-4z^3-3y^3)/2)^(1/3)}];n=n+1;Print[n," ",m];Label[aa];Continue,{m,0,1946}]

A267861 Number of ways to write n as 2t + u^4 + v^4 + 2w^4 + 3x^4 + 4y^4 + 6*z^4, where t is 0 or 1, and u, v, w, x, y, z are nonnegative integers with u <= v and v > 0.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Apr 07 2016

nonn

Conjecture: a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 1, 2, 111, 127, 143, 158, 221, 223, 240, 460, 463, 480, 545, 560, 561, 1455, 1695, 1776, 2175. Moreover, any integer n > 10^4 not among 10543, 17935, 37583, 40383, 78543 can be written as u^4 + v^4 + 2*w^4 + 3*x^4 + 4*y^4 + 6*z^4 with u,v,w,x,y,z nonnegative integers.

If a(1),...,a(7) are positive integers with a(1) <= a(2) <= ... <= a(7) and a(1)+...+a(7) = g(4) = 19 such that {a(1)*x(1)^4+...+a(7)*x(7)^4: x(1),...,x(7) = 0,1,2,...} = {0,1,2,...}, then the tuple (a(1),...,a(7)) must be (1,1,2,2,3,4,6) or (1,1,2,2,3,3,7). Similarly, if a(1),...,a(8) are positive integers with a(1) <= a(2) <= ... <= a(8) and a(1)+...+a(8) = g(5) = 37 such that {a(1)*x(1)^5+...+a(8)*x(8)^5: x(1),...,x(8) = 0,1,2,...} = {0,1,2,...}, then (a(1),...,a(8)) must be (1,1,2,3,4,6,8,12) or (1,1,2,3,4,5,7,14).

			a(111) = 1 since 111 = 2*1 + 2^4 + 3^4 + 2*1^4 + 3*0^4 + 4*1^4 + 6*1^4.
a(240) = 1 since 240 = 2*0 + 2^4 + 2^4 + 2*0^4 + 3*2^4 + 4*2^4 + 6*2^4.
a(1776) = 1 since 1776 = 2*0 + 4^4 + 5^4 + 2*3^4 + 3*3^4 + 4*1^4 + 6*3^4.
a(2175) = 1 since 2175 = 2*1 + 0^4 + 4^4 + 2*2^4 + 3*5^4 + 4*1^4 + 6*1^4.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Upgrade Waring's Problem, a message to Number Theory Mailing List, April 2, 2016.
Zhi-Wei Sun, New conjectures on representations of integers (I), Nanjing Univ. J. Math. Biquarterly 34(2017), no. 2, 97-120.

Cf. A000583, A000584, A002804, A267826, A271099, A271169, A271237.

QQ[n_]:=QQ[n]=n>0&&IntegerQ[n^(1/4)]
Do[r=0;Do[If[QQ[n-2t-6*z^4-4y^4-3x^4-2w^4-u^4],r=r+1],{t,0,Min[1,n/2]},{z,0,((n-2t^8)/6)^(1/4)},{y,0,((n-2t-6z^4)/4)^(1/4)},{x,0,((n-2t-6z^4-4y^4)/3)^(1/4)},
{w,0,((n-2t-6z^4-4y^4-3x^4)/2)^(1/4)},{u,0,((n-2t-6z^4-4y^4-3x^4-2w^4)/2)^(1/4)}];Print[n," ",r];Continue,{n,1,70}]

A271076 Number of ordered ways to write n as u^5 + v^4 + x^3 + 2y^3 + 3z^3, where u, v , x, y and z are nonnegative integers with v > 0.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Apr 07 2016

nonn

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 8, 9, 10, 11, 13, 14, 15, 16, 23, 24, 38, 39, 61, 76, 77, 104, 118, 188, 214, 229.
(ii) We have {P(u,v,x,y,z): u,v,x,y,z = 0,1,2,...} = {0,1,2,...} whenever P(u,v,x,y,z) is among the following polynomials: u^5+v^4+x^3+2*y^3+5*z^3, u^5+2*v^4+x^3+2*y^3+3*z^3, u^5+3*v^4+x^3+2*y^3+3*z^3, 2*u^5+v^4+x^3+y^3+4*z^3, 2*u^5+v^4+x^3+2*y^3+4*z^3, 3*u^5+v^4+x^3+2*y^3+4*z^3, 5*u^5+v^4+x^3+2*y^3+4*z^3, u^4+2*v^4+x^3+y^3+4*z^3, u^4+2*v^4+x^3+2*y^3+3*z^3, u^4+2*v^4+x^3+2*y^3+4*z^3, u^4+2*v^4+x^3+2*y^3+6*z^3, u^4+2*v^4+x^3+3*y^3+4*z^3, u^4+2*v^4+x^3+4*y^3+5*z^3, u^4+2*v^4+x^3+4*y^3+6*z^3, u^4+2*v^4+x^3+4*y^3+10*z^3, u^4+3*v^4+x^3+2*y^3+3*z^3, u^4+3*v^4+x^3+2*y^3+4*z^3, u^4+3*v^4+x^3+2*y^3+6*z^3, u^4+4*v^4+x^3+y^3+2*z^3, u^4+4*v^4+x^3+2*y^3+3*z^3, u^4+4*v^4+x^3+2*y^3+4*z^3, u^4+5*v^4+x^3+2*y^3+4*z^3, u^4+6*v^4+x^3+2*y^3+3*z^3, u^4+7*v^4+x^3+2*y^3+3*z^3, u^4+9*v^4+x^3+2*y^3+4*z^3, 2*u^4+4*v^4+x^3+2*y^3+3*z^3,2*u^4+6*v^4+x^3+2*y^3+4*z^3, 3*u^4+6*v^4+x^3+2*y^3+4*z^3.
(iii) We have {P(u,v,x,y,z): u,v,x,y,z = 0,1,2,...} = {0,1,2,...} whenever P(u,v,x,y,z) is among the following polynomials: u^5+v^3+x^3+2*y^3+4*z^3, u^5+v^3+2*x^3+3*y^3+c*z^3 (c = 6,9), a*u^5+v^3+2*x^3+4*y^3+5*z^3 (a = 1,2,6), b*u^5+v^3+2*x^3+4*y^3+6*z^3 (b = 2,3), u^5+v^3+2*x^3+4*y^3+d*z^3 (d = 9,13).
The listed polynomials in part (ii), together with u^5+v^4+x^3+2*y^3+3*z^3, should essentially exhaust all those polynomials P(u,v,x,y,z) = s*u^k+t*v^j+a*x^3+b*y^3+c*z^3 with s,t,a,b,c positive integers and k >= j > 3, such that {P(u,v,x,y,z): u,v,x,y,z = 0,1,2,...} = {0,1,2,...}.
There are also finitely many (but quite a lot) polynomials P(u,v,x,y,z) of the form m*u^4+a*v^3+b*x^3+c*y^3+d*z^3 with a,b,c,d and m positive integers such that {P(u,v,x,y,z): u,v,x,y,z = 0,1,2,...} = {0,1,2,...}.
See also A267826, A271099 and A271237 for related comments.
Conjectures (i), (ii) and (iii) verified for n up to 10^11 for all polynomials. - Mauro Fiorentini, Sep 20 2023

			a(16) = 1 since 16 = 0^5 + 2^4 +0^3 + 2*0^3 + 3*0^3.
a(104) = 1 since 104 = 0^5 + 2^4 + 4^3 + 2*0^3 + 3*2^3.
a(188) = 1 since 188 = 2^5 + 1^4 + 3^3 + 2*4^3 + 3*0^3.
a(229) = 1 since 229 = 1^5 + 3^4 + 4^3 + 2*1^3 + 3*3^3.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Cf. A000578, A000583, A000584, A266968, A267826, A271099, A271237.

CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)]
Do[r=0;Do[If[CQ[n-u^5-v^4-3z^3-2y^3],r=r+1],{u,0,(n-1)^(1/5)},{v,1,(n-u^5)^(1/4)},{z,0,((n-u^5-v^4)/3)^(1/3)},{y,0,((n-u^5-v^4-3z^3)/2)^(1/3)}];Print[n," ",r];Continue,{n,1,80}]

A352503 Number of ways to write n as w^3 + 2x^3 + 4y^3 + 5*z^3 + t^6, where w is a positive integer, and x,y,z,t are nonnegative integers.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Mar 28 2022

nonn

Conjecture: a(n) > 0 for all n > 0.
This has been verified for n = 1..10^6.
It seems that a(n) = 1 only for n = 1..5, 16, 19, 20, 21, 23, 24, 25, 26, 31, 37, 51, 58, 63, 77, 78, 79, 94, 108, 207, 208, 218, 316, 487, 490, 559.

			a(20) = 1 with 20 = 2^3 + 2*1^3 + 4*1^3 + 5*1^3 + 1^6.
a(79) = 1 with 79 = 2^3 + 2*1^3 + 4*0^3 + 5*1^3 + 2^6.
a(316) = 1 with 316 = 1^3 + 2*3^3 + 4*4^3 + 5*1^3 + 0^6.
a(487) = 1 with 487 = 5^3 + 2*!^3 + 4*4^3 + 5*2^3 + 2^6.
a(490) = 1 with 490 = 2^3 + 2*3^3 + 4*3^3 + 5*4^3 + 0^6.
a(559) = 1 with 559 = 8^3 + 2*1^3 + 4*1^3 + 5*2^3 + 1^6.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, New conjectures on representations of integers (I), Nanjing Univ. J. Math. Biquarterly 34 (2017), no. 2, 97-120. (See Conjecture 3.4(i).)

Cf. A000578, A001014, A271099, A271237.

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

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

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A267826 Numbers not of the form w^3 + 2*x^3 + 3*y^3 + 4*z^3, where w, x, y and z are nonnegative integers.

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A267861 Number of ways to write n as 2*t + u^4 + v^4 + 2*w^4 + 3*x^4 + 4*y^4 + 6*z^4, where t is 0 or 1, and u, v, w, x, y, z are nonnegative integers with u <= v and v > 0.

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A271076 Number of ordered ways to write n as u^5 + v^4 + x^3 + 2*y^3 + 3*z^3, where u, v , x, y and z are nonnegative integers with v > 0.

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A352503 Number of ways to write n as w^3 + 2*x^3 + 4*y^3 + 5*z^3 + t^6, where w is a positive integer, and x,y,z,t are nonnegative integers.

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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.

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.

A267826 Numbers not of the form w^3 + 2x^3 + 3y^3 + 4*z^3, where w, x, y and z are nonnegative integers.

A267861 Number of ways to write n as 2t + u^4 + v^4 + 2w^4 + 3x^4 + 4y^4 + 6*z^4, where t is 0 or 1, and u, v, w, x, y, z are nonnegative integers with u <= v and v > 0.

A271076 Number of ordered ways to write n as u^5 + v^4 + x^3 + 2y^3 + 3z^3, where u, v , x, y and z are nonnegative integers with v > 0.

A352503 Number of ways to write n as w^3 + 2x^3 + 4y^3 + 5*z^3 + t^6, where w is a positive integer, and x,y,z,t are nonnegative integers.