A262824 -id:A262824 - cp's OEIS frontend

A262827 Number of ordered ways to write n as w^2 + x^3 + y^4 + 2*z^4, where w, x, y and z are nonnegative integers.

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

Offset: 0

Zhi-Wei Sun, Oct 03 2015

nonn

Conjecture: a(n) > 0 for all n >= 0. Also, a(n) = 1 only for the following 41 values of n: 0, 14, 15, 21, 22, 23, 62, 71, 78, 87, 136, 216, 405, 437, 448, 477, 535, 583, 591, 623, 671, 696, 885, 950, 1046, 1135, 1206, 1208, 1248, 1317, 2288, 2383, 2543, 3167, 3717, 3974, 6847, 7918, 8328, 9096, 21935.
We have verified that a(n) > 0 for all n = 0..10^7.
Conjecture verified up to 10^11. - Mauro Fiorentini, Jul 07 2023
We also conjecture that if f(w,x,y,z) is one of the 8 polynomials 2w^2+x^3+4y^3+z^4, w^2+x^3+2y^3+c*z^3 (c = 3,4,5,6) and w^2+x^3+2y^3+d*z^4 (d = 1,3,6) then each n = 0,1,2,... can be written as f(w,x,y,z) with w,x,y,z nonnegative integers. - Zhi-Wei Sun, Dec 30 2017
Conjecture verified up to 10^11 for all 8 polynomials. - Mauro Fiorentini, Jul 07 2023

			a(14) = 1 since 14 = 2^2 + 2^3 + 0^4 + 2*1^4.
a(87) = 1 since 87 = 2^2 + 0^3 + 3^4 + 2*1^4.
a(216) = 1 since 216 = 0^2 + 6^3 + 0^4 + 2*0^4.
a(405) = 1 since 405 = 18^2 + 0^3 + 3^4 + 2*0^4.
a(1248) = 1 since 1248 = 31^2 + 5^3 + 0^4 + 2*3^4.
a(1317) = 1 since 1317 = 23^2 + 1^3 + 5^4 + 2*3^4.
a(2288) = 1 since 2288 = 44^2 + 4^3 + 4^4 +2*2^4.
a(2383) = 1 since 2383 = 1462 + 9^3 + 6^4 + 2*3^4.
a(2543) = 1 since 2543 = 50^2 + 3^3 + 2^4 + 2*0^4.
a(3167) = 1 since 3167 = 54^2 + 2^3 + 3^4 + 2*3^4.
a(3717) = 1 since 3717 = 18^2 + 15^3 + 2^4 + 2*1^4.
a(3974) = 1 since 3974 = 39^2 + 13^3 + 4^4 + 2*0^4.
a(6847) = 1 since 6847 = 52^2 + 15^3 + 4^4 + 2*4^4.
a(7918) = 1 since 7918 = 46^2 + 10^3 + 0^4 + 2*7^4.
a(8328) = 1 since 8328 = 42^2 + 1^3 + 9^4 + 2*1^4.
a(9096) = 1 since 9096 = 44^2 + 18^3 + 6^4 + 2*2^4.
a(21935) = 1 since 21935 = 66^2 + 26^3 + 1^4 + 2*1^4.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190. (See Remark 1.1.)
Zhi-Wei Sun, New conjectures on representations of integers (I), Nanjing Univ. J. Math. Biquarterly 34(2017), no. 2, 97-120. (See Theorem 1.1 and Conjecture 1.1.)

Cf. A000290, A000578, A000583, A262813, A262815, A262816, A262824.

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

A262857 Number of ordered ways to write n as w^3 + 2x^3 + y^2 + 2z^2, where w, x, y and z are nonnegative integers.

Original entry on oeis.org

1, 2, 3, 4, 4, 3, 3, 2, 3, 5, 5, 6, 6, 3, 4, 1, 4, 6, 7, 10, 7, 5, 4, 2, 5, 8, 8, 9, 9, 6, 6, 2, 6, 10, 8, 13, 9, 6, 7, 5, 5, 8, 6, 9, 10, 6, 9, 4, 5, 9, 6, 13, 10, 7, 11, 6, 8, 10, 8, 10, 12, 9, 9, 7, 8, 13, 10, 16, 12, 6, 12, 8, 10, 13, 12, 13, 12, 8, 11, 7, 10, 16, 15, 17, 16, 6, 11, 7, 12, 16, 11, 16, 9, 10, 5, 6, 10, 15, 17, 18, 16

Offset: 0

Zhi-Wei Sun, Oct 03 2015

nonn

Conjecture: We have {a*w^3+b*x^3+c*y^2+d*z^2: w,x,y,z = 0,1,2,...} = {0,1,2,...} if (a,b,c,d) is among the following 63 quadruples:
(1,1,1,2),(1,1,2,4),(1,2,1,1),(1,2,1,2),(1,2,1,3),(1,2,1,4),(1,2,1,6),(1,2,1,13),(1,2,2,3),(1,2,2,4),(1,2,2,5),(1,3,1,1),(1,3,1,2),(1,3,1,3),(1,3,1,5),(1,3,1,6),(1,3,2,3),(1,3,2,4),(1,3,2,5),(1,4,1,1),(1,4,1,2),(1,4,1,3),(1,4,2,2),(1,4,2,3),(1,4,2,5),(1,5,1,1),(1,5,1,2),(1,6,1,1),(1,6,1,3),(1,7,1,2),(1,8,1,2),(1,9,1,2),(1,9,2,4),(1,10,1,2),(1,11,1,2),(1,11,2,4),(1,12,1,2),(1,14,1,2),(1,15,1,2),(2,3,1,1),(2,3,1,2),(2,3,1,3),(2,3,1,4),(2,4,1,1),(2,4,1,2),(2,4,1,6),(2,4,1,8),(2,4,1,10),(2,5,1,3),(2,6,1,1),(2,7,1,3),(2,8,1,1),(2,8,1,4),(2,10,1,1),(2,13,1,1),(3,4,1,2),(3,5,1,2),(3,7,1,2),(3,9,1,2),(4,5,1,2),(4,6,1,2),(4,8,1,2),(4,11,1,2).
Conjecture verified up to 10^11 for all quadruples. - Mauro Fiorentini, Jul 18 2023

			a(7) = 2 since 7 = 1^3 + 2*0^3 + 2^2 + 2*1^2 = 1^3 + 2*1^3 + 2^2 + 2*0^2.
a(15) = 1 since 15 = 1^3 + 2*1^3 + 2^2 + 2*2^2.

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

Cf. A000290, A000578, A262813, A262815, A262816, A262824, A262827.

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

A271106 Number of ordered ways to write n as x^6 + 3y^3 + z^3 + w(w+1)/2, where x and y are nonnegative integers, and z and w are positive integers.

Original entry on oeis.org

0, 0, 1, 1, 1, 2, 1, 2, 2, 1, 2, 3, 3, 1, 3, 3, 1, 2, 2, 2, 1, 1, 2, 2, 1, 1, 3, 2, 2, 4, 3, 3, 4, 5, 3, 2, 4, 4, 3, 2, 4, 3, 2, 2, 1, 2, 3, 4, 3, 2, 1, 1, 2, 4, 4, 2, 3, 3, 2, 2, 2, 3, 2, 2, 2, 1, 5, 5, 5, 3, 4

Offset: 0

Zhi-Wei Sun, Mar 30 2016

nonn

Conjecture: a(n) > 0 for all n > 1, and a(n) = 1 only for n = 2, 3, 4, 6, 9, 13, 16, 20, 21, 24, 25, 44, 50, 51, 65, 84, 189, 290, 484, 616, 664, 680, 917, 1501, 1639, 3013.
Based on our computation, we also formulate the following general conjecture.
General Conjecture: Let T(w) = w*(w+1)/2. We have {P(x,y,z,w): x,y,z,w = 0,1,2,...} = {0,1,2,...} for any of the following polynomials P(x,y,z,w): x^3+y^3+c*z^3+T(w) (c = 2,3,4,6), x^3+y^3+c*z^3+2*T(w) (c = 2,3), x^3+b*y^3+3z^3+3*T(w) (b = 1,2), x^3+2y^3+3z^3+w(5w-1)/2, x^3+2y^3+3z^3+w(5w-3)/2, x^3+2y^3+c*z^3+T(w) (c = 2,3,4,5,6,7,12,20,21,34,35,40), x^3+2y^3+c*z^3+2*T(w) (c = 3,4,5,6,11), x^3+2y^3+c*z^3+w^2 (c = 3,4,5,6), x^3+2y^3+4z^3+w(3w-1)/2, x^3+2y^3+4z^3+w(3w+1)/2, x^3+2y^3+4z^3+w(2w-1), x^3+2y^3+6z^3+w(3w-1)/2, x^3+3y^3+c*z^3+T(w) (c = 3,4,5,6,10,11,13,15,16,18,20), x^3+3y^3+c*z^3+2*T(w) (c = 5,6,11), x^3+4y^3+c*z^3+T(w) (c = 5,10,12,16), x^3+4y^3+5z^3+2*T(w), x^3+5y^3+10z^3+T(w), 2x^3+3y^3+c*z^3+T(w) (c = 4,6), 2x^3+4y^3+8z^3+T(w), x^4+y^3+3z^3+w(3w-1)/2, x^4+y^3+c*z^3+T(w) (c = 2,3,4,5,7,12,13), x^4+y^3+c*z^3+2*T(w) (c = 2,3,4,5), x^4+y^3+2z^3+w^2, x^4+y^3+4z^3+2w^2, x^4+2y^3+c*z^3+T(w) (c = 4,5,12), x^4+2y^3+3z^3+2*T(w), 2x^4+y^3+2z^3+w(3w-1)/2, 2x^4+y^3+c*z^3+T(w) (c = 1,2,3,4,5,6,10,11), 2x^4+y^3+c*z^3+2*T(w) (c = 2,3,4), 2x^4+2y^3+c*z^3+T(w) (c = 3,5), 3x^4+y^3+c*z^3+T(w) (c = 1,2,3,4,5,11), 3x^4+y^3+2z^3+2*T(w), 3x^4+y^3+2z^3+w^2, 3x^4+y^3+2z^3+w(3w-1)/2, 4x^4+y^3+c*z^3+T(w) (c = 2,3,4,6), 4x^4+y^3+2z^3+2*T(w), 5x^4+y^3+c*z^3+T(w) (c = 2,4), a*x^4+y^3+2z^3+T(w) (a = 6,20,28,40), 6x^4+y^3+2z^3+2*T(w), 6x^4+y^3+2z^3+w^2, a*x^4+y^3+3z^3+T(w) (a = 6,8,11), 8x^4+2y^3+4z^3+T(w), x^5+y^3+c*z^3+T(w) (c = 2,3,4), x^5+2y^3+c*z^3+T(w) (c = 3,6,8), 2x^5+y^3+4z^3+T(w), 3x^5+y^3+2z^3+T(w), 5x^5+y^3+c*z^3+T(w) (c = 2,4), x^6+y^3+3z^3+T(w), x^7+y^3+4z^3+T(w), x^4+2y^4+z^3+w^2, x^4+2y^4+2z^3+T(w),  x^4+b*y^4+z^3+T(w) (b = 2,3,4), 2x^4+3y^4+z^3+T(w), a*x^5+y^4+z^3+T(w) (a = 1,2), x^5+2y^4+z^3+T(w).
The polynomials listed in the general conjecture should exhaust all those polynomials P(x,y,z,w) = a*x^i+b*y^j+c*z^k+w*(s*w+/-t)/2 with {P(x,y,z,w): x,y,z,w = 0,1,2,...} = {0,1,2,...}, where a,b,c,s > 0, 0 <= t <= s, s == t (mod 2), i >= j >= k >= 3, a <= b if i = j, and b <= c if j = k.

			a(9) = 1 since 9 = 0^6 + 3*0^6 + 2^3 + 1*2/2.
a(24) = 1 since 24 = 1^6 + 3*0^6 + 2^3 + 5*6/2.
a(1501) = 1 since 1501 = 2^6 + 3*5^3 + 3^3 + 45*46/2.
a(1639) = 1 since 1639 = 0^6 + 3*6^3 + 1^3 + 44*45/2.
a(3013) = 1 since 3013 = 3^6 + 3*3^3 + 13^3 + 3*4/2.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Z.-W. Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113.
Z.-W. Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), 1367-1396.

Cf. A000217, A000290, A000326, A000578, A000583, A000584, A001014, A005449, A262813, A262824, A262827, A262857, A262880, A266968, A270469, A270488, A270516, A270533, A270559, A270566, A270920, A271026.

TQ[n_]:=TQ[n]=n>0&&IntegerQ[Sqrt[8n+1]]
Do[r=0;Do[If[TQ[n-x^6-3*y^3-z^3],r=r+1],{x,0,n^(1/6)},{y,0,((n-x^6)/3)^(1/3)},{z,1,(n-x^6-3y^3)^(1/3)}];Print[n," ",r];Continue,{n,0,70}]

A272979 Number of ways to write n as x^2 + 2y^2 + 3z^3 + 4*w^4 with x,y,z,w nonnegative integers.

Original entry on oeis.org

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

Offset: 0

Zhi-Wei Sun, Jul 13 2016

nonn

Conjecture: For positive integers a,b,c,d, any natural number can be written as a*x^2 + b*y^2 + c*z^3 + d*w^4 with x,y,z,w nonnegative integers, if and only if (a,b,c,d) is among the following 49 quadruples: (1,2,1,1), (1,3,1,1), (1,6,1,1), (2,3,1,1), (2,4,1,1), (1,1,2,1), (1,4,2,1), (1,2,3,1), (1,2,4,1), (1,2,12,1), (1,1,1,2), (1,2,1,2), (1,3,1,2), (1,4,1,2), (1,5,1,2), (1,11,1,2), (1,12,1,2), (2,4,1,2), (3,5,1,2), (1,1,4,2), (1,1,1,3), (1,2,1,3), (1,3,1,3), (1,2,4,3), (1,2,1,4), (1,3,1,4), (2,3,1,4), (1,1,2,4), (1,2,2,4), (1,8,2,4), (1,2,3,4), (1,1,1,5), (1,2,1,5), (2,3,1,5), (2,4,1,5), (1,3,2,5), (1,1,1,6), (1,3,1,6), (1,1,2,6), (1,2,1,8), (1,2,4,8), (1,2,1,10), (1,1,2,10), (1,2,1,11), (2,4,1,11), (1,2,1,12), (1,1,2,13), (1,2,1,14),(1,2,1,15).

See also A262824, A262827, A262857 and A273917 for similar conjectures.

			a(0) = 1 since 0 = 0^2 + 2*0^2 + 3*0^3 + 4*0^4.
a(1) = 1 since 1 = 1^2 + 2*0^2 + 3*0^3 + 4*0^4.
a(2) = 1 since 2 = 0^2 + 2*1^2 + 3*0^3 + 4*0^4.
a(14) = 1 since 14 = 3^2 + 2*1^2 + 3*1^3 + 4*0^4.
a(17) = 1 since 17 = 3^2 + 2*2^2 + 3*0^3 + 4*0^4.
a(59) = 1 since 59 = 3^2 + 2*5^2 + 3*0^3 + 4*0^4.
a(63) = 1 since 63 = 3^2 + 2*5^2 + 3*0^2 + 4*1^4.
a(287) = 1 since 287 = 11^2 + 2*9^2 + 3*0^2 + 4*1^4.

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

Cf. A000290, A000578, A000583, A262824, A262827, A262857, A270969, A273429, A273915, A273917.

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

A262880 Number of ordered ways to write n as w(w+1)/2 + x^3 + y^3 + 2z^3 with w > 0, 0 <= x <= y and z >= 0.

Original entry on oeis.org

1, 1, 3, 2, 3, 2, 2, 2, 2, 3, 3, 4, 2, 3, 2, 2, 5, 3, 6, 2, 4, 3, 4, 4, 3, 4, 2, 5, 3, 6, 7, 4, 5, 2, 3, 4, 5, 8, 6, 4, 1, 2, 2, 5, 7, 6, 6, 2, 3, 3, 1, 5, 5, 5, 5, 5, 8, 5, 4, 4, 5, 3, 6, 6, 7, 8, 3, 6, 6, 5, 9, 6, 9, 3, 7, 5, 7, 3, 5, 9, 3, 11, 6, 9, 5, 3, 7, 4, 4, 7, 9, 8, 5, 8, 7, 7, 2, 6, 7, 4

Offset: 1

Zhi-Wei Sun, Oct 04 2015

nonn

Conjecture: (i) Any positive integer can be written as w*(w+1)/2 + x^3 + b*y^3 + c*z^3 with w > 0 and x,y,z >= 0, provided that (b,c) is among the following ordered pairs: (1,2),(1,3),(1,4),(1,6),(2,2),(2,3),(2,4),(2,5),(2,6),(2,7),(2,20),(2,21),(2,34),(3,3),(3,4),(3,5),(3,6),(4,10).
(ii) For (b,c) = (3,4),(3,6),(4,8), we have {w*(w+1)/2 + 2*x^3 + b*y^3 + c*z^3: w,x,y,z = 0,1,2,...} = {0,1,2,...}.
See also A262813, A262824 and A262857 for similar conjectures.

			a(2) = 1 since 2 = 1*2/2 + 0^3 + 1^3 + 2*0^3.
a(34) = 2 since 34 = 4*5/2 + 0^3 + 2^3 + 2*2^3 = 3*4/2 + 1^3 + 3^3 + 2*0^3.
a(41) = 1 since 41 = 3*4/2 + 2^3 + 3^3 + 2*0^3.
a(51) = 1 since 51 = 6*7/2 + 1^3 + 3^3 + 2*1^3.
a(104) = 1 since 104 = 5*6/2 + 2^3 + 3^3 + 2*3^3.

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

Cf. A000217, A000578, A262813, A262824, A262857.

TQ[n_]:=n>0&&IntegerQ[Sqrt[8n+1]]
Do[r=0;Do[If[TQ[n-x^3-y^3-2*z^3],r=r+1],{x,0,(n/2)^(1/3)},{y,x,(n-x^3)^(1/3)},{z,0,((n-x^3-y^3)/2)^(1/3)}];Print[n," ",r];Continue,{n,1,100}]

A271365 Number of ordered ways to write n as u^2 + v^3 + x^4 + y^5 + z^6, where u is a positive integer, and v, x, y, z are nonnegative integers.

Original entry on oeis.org

1, 4, 6, 5, 5, 6, 4, 1, 2, 7, 9, 6, 4, 3, 1, 1, 6, 12, 10, 4, 3, 3, 1, 1, 6, 12, 11, 7, 6, 4, 2, 4, 9, 12, 8, 5, 10, 12, 6, 2, 5, 9, 8, 8, 10, 6, 2, 3, 8, 13, 10, 8, 11, 8, 2, 1, 6, 10, 8, 7, 6, 2, 2, 7, 15, 20, 14, 9, 13, 11

Offset: 1

Zhi-Wei Sun, Apr 05 2016

nonn

Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 8, 15, 16, 23, 24, 56. Moreover, the only positive integers not represented by u^2+v^3+x^4+y^5 (u > 0 and v,x,y >= 0) are 8, 15, 23, 55, 62, 71, 471, 478, 510, 646, 806, 839, 879, 939, 1023, 1063, 1287, 2127, 5135, 6811, 7499, 9191, 26471.

Note that 1/2+1/3+1/4+1/5+1/6 = 29/20 < 3/2.

			a(1) = 1 since 1 = 1^2 + 0^3 + 0^4 + 0^5 + 0^6.
a(8) = 1 since 8 = 2^2 + 1^3 + 1^4 + 1^5 + 1^6.
a(15) = 1 since 15 = 2^2 + 2^3 + 1^4 + 1^5 + 1^6.
a(16) = 1 since 16 = 4^2 + 0^3 + 0^4 + 0^5 + 0^6.
a(23) = 1 since 23 = 2^2 + 1^3 + 2^4 + 1^5 + 1^6.
a(24) = 1 since 24 = 4^2 + 2^3 + 0^4 + 0^5 + 0^6.
a(56) = 1 since 56 = 4^2 + 2^3 + 0^4 + 2^5 + 0^6.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, A result similar to Lagrange's theorem, J. Number Theory 162(2016), 190-211.

Cf. A000290, A000578, A000583, A000584, A001014, A262813, A262824, A262827, A262857, A266968, A270488, A270516, A270533, A270559, A270566, A270920, A271026, A271106, A271325.

SQ[n_]:=SQ[n]=n>0&&IntegerQ[Sqrt[n]]
Do[r=0;Do[If[SQ[n-x1^6-x2^5-x3^4-x4^3],r=r+1],{x1,0,n^(1/6)},{x2,0,(n-x1^6)^(1/5)},{x3,0,(n-x1^6-x2^5)^(1/4)},{x4,0,(n-x1^6-x2^5-x3^4)^(1/3)}];Print[n," ",r];Continue,{n,1,70}]

A271381 Number of ordered ways to write n as u^2 + v^2 + x^3 + y^3, where u, v, x, y are nonnegative integers with 2 | u*v, u <= v and x <= y.

Original entry on oeis.org

1, 2, 2, 1, 1, 2, 2, 1, 2, 4, 3, 1, 1, 3, 2, 1, 3, 5, 3, 1, 2, 3, 2, 0, 2, 5, 3, 3, 3, 4, 1, 2, 4, 5, 3, 2, 5, 4, 3, 2, 4, 6, 2, 3, 4, 5, 2, 2, 4, 3, 2, 2, 5, 6, 4, 3, 2, 3, 2, 2, 4, 4, 3, 3, 5, 7, 4, 5, 5, 6, 4, 2, 6, 9, 6, 2, 4, 5, 1, 3, 8

Offset: 0

Zhi-Wei Sun, Apr 06 2016

nonn

Conjecture: (i) a(n) > 0 except for n = 23, and a(n) = 1 only for n = 0, 3, 4, 7, 11, 12, 15, 19, 23, 30, 78, 203, 219.

(ii) Any natural number can be written as the sum of two squares and three fourth powers.

			a(3) = 1 since 3 = 0^2 + 1^2 + 1^3 + 1^3 with 0 even.
a(4) = 1 since 4 = 0^2 + 2^2 + 0^3 + 0^3 with 0 and 2 even.
a(7) = 1 since 7 = 1^2 + 2^2 + 1^3 + 1^3 with 2 even.
a(11) = 1 since 11 = 0^2 + 3^2 + 1^3 + 1^3 with 0 even.
a(12) = 1 since 12 = 0^2 + 2^2 + 0^3 + 2^3 with 0 and 2 even.
a(15) = 1 since 15 = 2^2 + 3^2 + 1^3 + 1^3 with 2 even.
a(19) = 1 since 19 = 1^2 + 4^2 + 1^3 + 1^3 with 4 even.
a(30) = 1 since 30 = 2^2 + 5^2 + 0^3 + 1^3 with 2 even.
a(78) = 1 since 78 = 2^2 + 3^2 + 1^3 + 4^3 with 2 even.
a(203) = 1 since 203 = 7^2 + 10^2 + 3^3 + 3^3 with 10 even.
a(219) = 1 since 219 = 8^2 + 8^2 + 3^3 + 4^3 with 8 even.

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

Cf. A000290, A000578, A000583, A262813, A262824, A262827, A262857, A262880, A270488, A270559, A270969, A271325.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[SQ[n-x^3-y^3-u^2]&&(Mod[u*Sqrt[n-x^3-y^3-u^2],2]==0),r=r+1],{x,0,(n/2)^(1/3)},{y,x,(n-x^3)^(1/3)},{u,0,((n-x^3-y^3)/2)^(1/2)}];Print[n," ",r];Continue,{n,0,80}]

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A262827 Number of ordered ways to write n as w^2 + x^3 + y^4 + 2*z^4, where w, x, y and z are nonnegative integers.

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A262857 Number of ordered ways to write n as w^3 + 2*x^3 + y^2 + 2*z^2, where w, x, y and z are nonnegative integers.

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A271106 Number of ordered ways to write n as x^6 + 3*y^3 + z^3 + w*(w+1)/2, where x and y are nonnegative integers, and z and w are positive integers.

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A272979 Number of ways to write n as x^2 + 2*y^2 + 3*z^3 + 4*w^4 with x,y,z,w nonnegative integers.

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A262880 Number of ordered ways to write n as w*(w+1)/2 + x^3 + y^3 + 2*z^3 with w > 0, 0 <= x <= y and z >= 0.

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A271365 Number of ordered ways to write n as u^2 + v^3 + x^4 + y^5 + z^6, where u is a positive integer, and v, x, y, z are nonnegative integers.

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A271381 Number of ordered ways to write n as u^2 + v^2 + x^3 + y^3, where u, v, x, y are nonnegative integers with 2 | u*v, u <= v and x <= y.

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A262857 Number of ordered ways to write n as w^3 + 2x^3 + y^2 + 2z^2, where w, x, y and z are nonnegative integers.

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A272979 Number of ways to write n as x^2 + 2y^2 + 3z^3 + 4*w^4 with x,y,z,w nonnegative integers.

A262880 Number of ordered ways to write n as w(w+1)/2 + x^3 + y^3 + 2z^3 with w > 0, 0 <= x <= y and z >= 0.