A262827 -id:A262827 - cp's OEIS frontend

A270969 Number of ways to write n as w^4 + x^2 + y^2 + z^2, where w, x, y and z are nonnegative integers with x <= y <= z.

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

Offset: 0

Zhi-Wei Sun, Mar 27 2016

nonn

Theorem: a(n) > 0 for all n = 0,1,2,.... In other words, any nonnegative integer can be written as the sum of a fourth power and three squares.
This is stronger than Lagrange's four-square theorem, and it can be proved by induction on n. It is easy to check that a(n) > 0 for all n = 0..16. Now let n be an integer greater than 16, and assume that a(m) > 0 for all m = 0..n-1. If 16|n, then n/16 can be written as w^4+x^2+y^2+z^2 with w,x,y,z integers, and hence n = (2w)^4+(4x)^2+(4y)^2+(4z)^2. If n == 8 (mod 16), then n is not of the form 4^k*(8q+7) and hence n = 0^4+x^2+y^2+z^2 for some integers x,y,z. If n == 4 (mod 8), then n-1^4 can be written as the sum of three squares. If n == 2 (mod 4), then n-0^4 is a sum of three squares. If n == 7 (mod 8), then n-1^4 can be written as the sum of three squares. If n is odd but not congruent to 7 modulo 8, then n-0^4 can be expressed as the sum of three squares.
We have a(n) = 1 if n has the form 16^k*q with k a nonnegative integer and q among 7, 8, 15, 23, 31, 47, 71, 79. In fact, if n = 16*m with m > 0, and 16*m = w^4+x^2+y^2+z^2 with w,x,y,z integers, then w,x,y,z are all even and hence m = (w/2)^4+(x/2)^2+(y/2)^2+(z/2)^2. Therefore a(16*m) = a(m) for all m > 0. It is easy to check that a(q) = 1 for every q = 7, 8, 15, 23, 31, 47, 71, 79.
For (a,b,c) = (1,1,2),(1,1,3),(1,1,4),(1,1,6),(1,2,2),(1,2,3),(1,2,4),(1,2,5), we are also able to show that any natural number can be written as w^4+a*x^2+b*y^2+c*z^2 with w,x,y,z integers.
Conjecture: For each triple (a,b,c) = (1,2,11),(1,2,12),(1,2,13),(2,3,5), any natural number can be written as w^4+a*x^2+b*y^2+c*z^2 with w,x,y,z integers.

			a(7) = 1 since 7 = 1^4 + 1^2 + 1^2 + 2^2.
a(8) = 1 since 8 = 0^4 + 0^2 + 2^2 + 2^2.
a(15) = 1 since 15 = 1^4 + 1^2 + 2^2 + 3^2.
a(23) = 1 since 23 = 1^4 + 2^2 + 3^2 + 3^2.
a(31) = 1 since 31 = 1^4 + 1^2 + 2^2 + 5^2.
a(47) = 1 since 47 = 1^4 + 1^2 + 3^2 + 6^2.
a(71) = 1 since 71 = 1^4 + 3^2 + 5^2 + 6^2.
a(79) = 1 since 79 = 1^4 + 2^2 + 5^2 + 7^2.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, A result similar to Lagrange's theorem, J. Number Theory 162(2016), 190-211.

Cf. A000290, A000583, A262827, A270516, A270533, A270559, A270566.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[SQ[n-w^4-x^2-y^2],r=r+1],{w,0,n^(1/4)},{x,0,Sqrt[(n-w^4)/3]},{y,x,Sqrt[(n-w^4-x^2)/2]}];Print[n," ",r];Continue,{n,0,80}]

A270566 Number of ordered ways to write n as x^4 + y(3y+1)/2 + z(7z+1)/2, where x, y and z are integers with x nonnegative.

Original entry on oeis.org

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

Offset: 0

Zhi-Wei Sun, Mar 19 2016

nonn

Conjecture: a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 24, 47, 63, 78, 79, 143, 153, 325, 494, 949, 1079, 3328, 4335, 5609, 7949, 7967, 8888, 9665.

Conjecture verified for n up to 10^11. - Mauro Fiorentini, Jul 24 2023

			a(24) = 1 since 24 = 2^4 + (-2)*(3*(-2)+1)/2 + (-1)*(7*(-1)+1)/2.
a(78) = 1 since 78 = 1^4 + 7*(3*7+1)/2 + 0*(7*0+1)/2.
a(143) = 1 since 143 = 1^4 + 6*(3*6+1)/2 + (-5)*(7*(-5)+1)/2.
a(494) = 1 since 494 = 4^4 + (-7)*(3*(-7)+1)/2 + (-7)*(7*(-7)+1)/2.
a(949) = 1 since 949 = 4^4 + 0*(3*0+1)/2 + 14*(7*14+1)/2.
a(1079) = 1 since 1079 = 0^4 + 25*(3*25+1)/2 + 6*(7*6+1)/2.
a(3328) = 1 since 3328 = 0^4 + 38*(3*38+1)/2 + 18*(7*18+1)/2.
a(4335) = 1 since 4335 = 2^4 + 49*(3*49+1)/2 + 14*(7*14+1)/2.
a(5609) = 1 since 5609 = 0^4 + (-61)*(3*(-61)+1)/2 + 4*(7*4+1)/2.
a(7949) = 1 since 7949 = 3^4 + 43*(3*43+1)/2 + 38*(7*38+1)/2.
a(7967) = 1 since 7967 = 7^4 + (-61)*(3*(-61)+1)/2 + 2*(7*2+1)/2.
a(8888) = 1 since 8888 = 0^4 + (-77)*(3*(-77)+1)/2 + 3*(7*3+1)/2.
a(9665) = 1 since 9665 = 3^4 + 73*(3*73+1)/2 + 21*(7*21+1)/2.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113.
Zhi-Wei Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), no. 7, 1367-1396.
Zhi-Wei Sun, On x(ax+1)+y(by+1)+z(cz+1) and x(ax+b)+y(ay+c)+z(az+d), J. Number Theory 171(2017), 275-283.

Cf. A001318, A000583, A262813, A262815, A262816, A262827, A262941, A262944, A262945, A262954, A262955, A262956, A270469, A270488, A270516, A270533, A270559.

(* From Zhi-Wei Sun, Start *)
pQ[n_] := pQ[n] = IntegerQ[Sqrt[24 n + 1]];
Do[r = 0; Do[If[pQ[n - x^4 - y (7 y + 1)/2], r = r + 1], {x, 0, n^(1/4)}, {y, -Floor[(Sqrt[56 (n - x^4) + 1] + 1)/14], (Sqrt[56 (n - x^4) + 1] - 1)/14}]; Print[n, " ", r]; Continue, {n, 0, 80}]
(* From Zhi-Wei Sun, End *)
A270566[n_] := Length@Solve[x >= 0 && n == x^4 + y*(3 y + 1)/2 + z*(7 z + 1)/2, {x, y, z}, Integers];
Array[A270566, 25, 0] (* JungHwan Min, Mar 19 2016 *)

A270533 Number of ordered ways to write n = x^4 + x^3 + y^2 + z*(3z-1)/2, where x and y are nonnegative integers, and z is an integer.

Original entry on oeis.org

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

Offset: 0

Zhi-Wei Sun, Mar 18 2016

nonn

Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 20, 22, 70, 87, 167, 252, 388, 562, 636, 658, 873, 2598, 14979, 18892, 20824.
(ii) Each n = 0,1,2,... can be written as x*P(x) + y^2 + z*(3z-1)/2 with x and y nonnegative integers, and z an integer, where P(x) is either of the polynomials x^3+2, x^3+3, x^3+2x+8, x^3+x^2+4x+2, x^3+x^2+7x+6.
(iii) Any nonnegative integer can be expressed as x*(x^3+3) + y*(5y+4) + z*(3z-1)/2, where x is an nonnegative integer, and y and z are integers.
See also A270516 for a similar conjecture.

			a(20) = 1 since 20 = 1^4 + 1^3 + 4^2 + (-1)*(3*(-1)-1)/2.
a(22) = 1 since 22 = 0^4 + 0^3 + 0^2 + 4*(3*4-1)/2.
a(873) = 1 since 873 = 5^4 + 5^3 + 11^2 + (-1)*(3*(-1)-1)/2.
a(2598) = 1 since 2598 =  4^4 + 4^3 + 4^2 + 39*(3*39-1)/2.
a(14979) = 1 since 14979 = 1^4 + 1^3 + 51^2 + 91*(3*91-1)/2.
a(18892) = 1 since 18892 = 3^4 + 3^3 + 137^2 + (-3)*(3*(-3)-1)/2.
a(20824) = 1 since 20824 = 1^4 + 1^3 + 115^2 + (-71)*(3*(-71)-1)/2.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), no. 7, 1367-1396.

Cf. A000290, A000578, A000583, A001318, A262813, A262815, A262816, A262827, A262941, A262944, A262945, A262954, A262955, A262956, A270469, A270488, A270516.

pQ[x_]:=pQ[x]=IntegerQ[Sqrt[24x+1]]
Do[r=0;Do[If[pQ[n-y^2-x^3*(x+1)],r=r+1],{y,0,Sqrt[n]},{x,0,(n-y^2)^(1/4)}];Print[n," ",r];Continue,{n,0,90}]

A262941 Number of ordered pairs (x,y) with x >= 0 and y > 0 such that n - x^4 - y*(y+1)/2 is an even square or twice a square.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Oct 04 2015

nonn

Conjecture: a(n) > 0 for all n > 0. In other words, any positive integer n can be written as x^4 + 2^k*y^2 + z*(z+1)/2, where k is 1 or 2, and x,y,z are integers with z > 0.
This has been verified for n up to 2*10^6. We also guess that a(n) = 1 only for n = 1, 2, 13, 16, 50, 59, 239, 493, 1156, 1492, 1984, 3332.
See also A262944, A262945, A262954, A262955, A262956 for similar conjectures.

			a(1)    = 1 since    1 = 0^4 +    0^2 +  1*2/2  with  0 even.
a(2)    = 1 since    2 = 1^4 +    0^2 +  1*2/2  with  0 even.
a(13)   = 1 since   13 = 1^4 + 2* 1^2 +  4*5/2.
a(16)   = 1 since   16 = 1^4 +    0^2 +  5*6/2  with  0 even.
a(50)   = 1 since   50 = 1^4 +    2^2 +  9*10/2 with  2 even.
a(59)   = 1 since   59 = 0^4 +    2^2 + 10*11/2 with  2 even.
a(239)  = 1 since  239 = 0^4 + 2* 2^2 + 21*22/2 with  2 even.
a(493)  = 1 since  493 = 2^4 +   18^2 + 17*18/2 with 18 even.
a(1156) = 1 since 1156 = 1^4 + 2*24^2 +  2*3/2  with 24 even.
a(1492) = 1 since 1492 = 2^4 + 2* 7^2 + 52*53/2.
a(1984) = 1 since 1984 = 5^4 +   18^2 + 45*46/2 with 18 even.
a(3332) = 1 since 3332 = 5^4 +   52^2 +  2*3/2  with 52 even.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113.
Zhi-Wei Sun, On universal sums ax^2+by^2+f(z), aT_x+bT_y+f(z) and aT_x+by^2+f(z), arXiv:1502.03056 [math.NT], 2015.

Cf. A000217, A000290, A000583, A254885, A262813, A262827, A262944, A262945, A262954, A262955, A262956.

SQ[n_]:=IntegerQ[Sqrt[n/2]]||IntegerQ[Sqrt[n/4]]
Do[r=0;Do[If[SQ[n-x^4-y(y+1)/2],r=r+1],{x,0,n^(1/4)},{y,1,(Sqrt[8(n-x^4)+1]-1)/2}];Print[n," ",r];Continue,{n,1,100}]

A270559 Number of ordered ways to write n as x^4 + x^3 + y^2 + z*(z+1)/2, where x, y and z are integers with x nonzero, y nonnegative and z positive.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Mar 18 2016

nonn

Conjecture: (i) a(n) > 0 for all n > 0. In other words, for each n = 1,2,3,... there are integers x and y such that n-(x^4+x^3+y^2) is a positive triangular number.
(ii) a(n) = 1 only for n = 1, 2, 8, 20, 62, 97, 296, 1493, 4283, 4346, 5433.
In contrast, the author conjectured in A262813 that any positive integer can be expressed as the sum of a nonnegative cube, a square and a positive triangular number.

			a(1) = 1 since 1 = (-1)^4 + (-1)^3 + 0^2 + 1*2/2.
a(2) = 1 since 2 = (-1)^4 + (-1)^3 + 1^2 + 1*2/2.
a(8) = 1 since 8 = 1^4 + 1^3 + 0^2 + 3*4/2.
a(20) = 1 since 20 = (-2)^4 + (-2)^3 + 3^2 + 2*3/2.
a(62) = 1 since 62 = (-2)^4 + (-2)^3 + 3^2 + 9*10/2.
a(97) = 1 since 97 = 1^4 + 1^3 + 2^2 + 13*14/2.
a(296) = 1 since 296 = (-4)^4 + (-4)^3 + 7^2 + 10*11/2.
a(1493) = 1 since 1493 = (-2)^4 + (-2)^3 + 0^2 + 54*55/2.
a(4283) = 1 since 4283 = (-6)^4 + (-6)^3 + 50^2 + 37*38/2.
a(4346) = 1 since 4346 = (-3)^4 + (-3)^3 + 49^2 + 61*62/2.
a(5433) = 1 since 5433 = (-8)^4 + (-8)^3 + 14^2 + 57*58/2.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113.

Cf. A000217, A000290, A000578, A000583, A262813, A262815, A262816, A262827, A262941, A262944, A262945, A262954, A262955, A262956, A270469, A270488, A270516, A270533.

TQ[n_]:=TQ[n]=n>0&&IntegerQ[Sqrt[8n+1]]
Do[r=0;Do[If[x!=0&&TQ[n-y^2-x^4-x^3],r=r+1],{y,0,Sqrt[n]},{x,-1-Floor[(n-y^2)^(1/4)],(n-y^2)^(1/4)}];Print[n," ",r];Continue,{n,1,10000}]

A266152 Least positive integer y such that n = x^4 - y^3 + z^2 for some positive integers x and z, or 0 if no such y exists.

Original entry on oeis.org

8, 1, 2, 17, 1, 3, 139, 19, 37, 1, 3, 9, 2, 7, 3, 1411, 1, 2, 2, 1, 5, 4, 387, 3, 1, 1, 4, 7, 9, 2, 35, 1, 33, 2, 6, 5, 1, 4, 3, 11, 1, 6, 2, 429, 2, 5, 11, 179, 73, 1, 15, 1, 4, 3, 11, 3, 5, 2, 3, 15, 5, 6, 7, 3, 1, 6, 4, 6337, 8, 16, 3

Offset: 0

Zhi-Wei Sun, Dec 22 2015

nonn

Conjecture: Any integer m can be written as x^4 - y^3 + z^2, where x, y and z are positive integers.
This is slightly stronger than the conjecture in A266003.
See also A266153 for a related sequence, and A266212 for a stronger conjecture.
If n is a positive square, then a(n) = 1. - Altug Alkan, Dec 23 2015

			a(0) = 8 since 0 = 4^4 - 8^3 + 16^2.
a(6) = 139 since 6 = 36^4 - 139^3 + 1003^2.
a(15) = 1411 since 15 = 119^4 - 1411^3 + 51075^2.
a(11019) = 71383 since 11019 = 4325^4 - 71383^3 + 3719409^2.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, Checking the conjecture for integers m with |m| <= 10^5
Zhi-Wei Sun, New conjectures on representations of integers (I), Nanjing Univ. J. Math. Biquarterly 34(2017), no. 2, 97-120.

Cf. A000290, A000578, A000583, A262827, A266003, A266004, A266153, A266212.

SQ[n_]:=SQ[n]=n>0&&IntegerQ[Sqrt[n]]
Do[y=1;Label[bb];Do[If[SQ[n+y^3-x^4],Print[n," ",y];Goto[aa]],{x,1,(n+y^3)^(1/4)}];y=y+1;Goto[bb];Label[aa];Continue,{n,0,70}]

A266153 Least positive integer y such that -n = x^4 - y^3 + z^2 for some positive integers x and z, or 0 if no such y exists.

Original entry on oeis.org

3, 3, 2, 6, 13, 2, 3, 5, 5, 3, 28, 4, 15, 4, 10, 33, 3, 7, 5, 238, 31, 3, 4, 5, 3, 11, 4, 5, 21, 11, 6, 4, 17, 11, 5, 98, 7, 4, 4, 5, 147, 19, 5, 4, 5, 6, 4, 29, 75, 1011, 7, 9, 7, 4, 8, 6, 59, 47, 4, 5, 71, 4, 17, 45, 13, 7, 18, 9, 175, 8

Offset: 1

Zhi-Wei Sun, Dec 22 2015

nonn

The conjecture in A266152 implies that a(n) > 0 for all n > 0.

It seems that a(n) < n*(n+4)/2 for all n > 1.

			a(1) = 3 since -1 = 1^4 - 3^3 + 5^2.
a(2) = 3 since -2 = 2^4 - 3^3 + 3^2.
a(11) = 28 since -11 = 5^4 - 28^3 + 146^2.
a(20) = 238 since -20 = 32^4 - 238^3 + 3526^2.

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

Cf. A000290, A000578, A000583, A262827, A266004, A266152.

SQ[n_]:=SQ[n]=n>0&&IntegerQ[Sqrt[n]]
Do[y=Floor[n^(1/3)]+1;Label[bb];Do[If[SQ[-n+y^3-x^4],Print[n," ",y];Goto[aa]],{x,1,(-n+y^3)^(1/4)}];y=y+1;Goto[bb];Label[aa];Continue,{n,1,70}]

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}]

A266212 Positive integers x such that x^3 = y^4 + z^2 for some positive integers y and z.

Original entry on oeis.org

8, 13, 20, 25, 40, 125, 128, 193, 200, 208, 225, 313, 320, 328, 400, 500, 605, 640, 648, 1000, 1053, 1156, 1521, 1620, 1625, 1681, 1700, 2000, 2025, 2048, 2125, 2465, 2493, 2873, 2920, 3025, 3088, 3185, 3200, 3240, 3328, 3400, 3600, 3656, 3748, 3816, 4225, 4625, 4913, 5000, 5008, 5120, 5248, 6400, 6728, 6760, 6793, 6845, 7225, 8000

Offset: 1

Zhi-Wei Sun, Dec 23 2015

nonn

If x^3 = y^4 + z^2, then (a^(4k)*x)^3 = (a^(3k)*y)^4 + (a^(6k)*z)^2 for all a = 1,2,3,... and k = 0,1,2,... So the sequence has infinitely many terms.
Conjecture: For any integer m, there are infinitely many triples (x,y,z) of positive integers with x^4 - y^3 + z^2 = m.
This is stronger than the conjecture in A266152.

			a(1) = 8 since 8^3 = 4^4 + 16^2.
a(2) = 13 since 13^3 = 3^4 + 46^2.
a(3) = 20 since 20^3 = 4^4 + 88^2.
a(8) = 193 since 193^3 = 6^4 + 2681^2.
a(12) = 313 since 313^3 = 66^4 + 3419^2.
a(20) = 1000 since 1000^3 = 100^4 + 30000^2.

Zhi-Wei Sun and Chai Wah Wu, Table of n, a(n) for n = 1..698 n = 1..100 from Zhi-Wei Sun
Zhi-Wei Sun, New conjectures on representations of integers (I), Nanjing Univ. J. Math. Biquarterly 34(2017), no. 2, 97-120.

Cf. A000290, A000578, A000583, A262827, A266003, A266004, A266152, A266153.

SQ[n_]:=SQ[n]=n>0&&IntegerQ[Sqrt[n]]
n=0;Do[Do[If[SQ[x^3-y^4],n=n+1;Print[n," ",x];Goto[aa]],{y,1,x^(3/4)}];Label[aa];Continue,{x,1,8000}]

A262824 Number of ordered ways to write n as w^2 + x^3 + 2y^3 + 3z^3, where w, x, y and z are nonnegative integers.

Original entry on oeis.org

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

Offset: 0

Zhi-Wei Sun, Oct 03 2015

nonn

Conjecture: (i) For any m = 3, 4, 5, 6 and n >= 0, there are nonnegative integers w, x, y, z such that n = w^2 + x^3 + 2*y^3 + m*z^3.
(ii) For P(w,x,y,z) = w^2 + x^3 + 2*y^3 + z^4, w^2 + x^3 + 2*y^3 + 3*z^4, w^2 + x^3 + 2*y^3 + 6*z^4, 2*w^2 + x^3 + 4*y^3 + z^4, we have {P(w,x,y,z): w,x,y,z = 0,1,2,...} ={0,1,2,...}.
Conjectures (i) and (ii) verified up to 10^11. - Mauro Fiorentini, Jul 22 2023
See also A262827 and A262857 for similar conjectures.

			a(0) = 1 since 0 = 0^2 + 0^3 + 2*0^3 + 3*0^3.
a(8) = 2 since 8 = 2^2 + 1^3 + 2*0^3 + 3*1^3 = 0^2 + 2^3 + 2*0^3 + 3*0^3.
a(23) = 1 since 23 = 2^2 + 0^3 + 2*2^3 + 3*1^3.
a(37) = 1 since 37 = 6^2 + 1^3 + 2*0^3 + 3*0^3.
a(72) = 1 since 72 = 8^2 + 2^3 + 2*0^3 + 3*0^3.

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

Cf. A000290, A000578, A262813, A262815, A262816, A262827, A262857.

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

cp's OEIS Frontend

A270969 Number of ways to write n as w^4 + x^2 + y^2 + z^2, where w, x, y and z are nonnegative integers with x <= y <= z.

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A270566 Number of ordered ways to write n as x^4 + y*(3y+1)/2 + z*(7z+1)/2, where x, y and z are integers with x nonnegative.

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A270533 Number of ordered ways to write n = x^4 + x^3 + y^2 + z*(3z-1)/2, where x and y are nonnegative integers, and z is an integer.

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A262941 Number of ordered pairs (x,y) with x >= 0 and y > 0 such that n - x^4 - y*(y+1)/2 is an even square or twice a square.

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A270559 Number of ordered ways to write n as x^4 + x^3 + y^2 + z*(z+1)/2, where x, y and z are integers with x nonzero, y nonnegative and z positive.

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A266152 Least positive integer y such that n = x^4 - y^3 + z^2 for some positive integers x and z, or 0 if no such y exists.

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A266153 Least positive integer y such that -n = x^4 - y^3 + z^2 for some positive integers x and z, or 0 if no such y exists.

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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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A270566 Number of ordered ways to write n as x^4 + y(3y+1)/2 + z(7z+1)/2, where x, y and z are integers with x nonnegative.

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.

A262824 Number of ordered ways to write n as w^2 + x^3 + 2y^3 + 3z^3, where w, x, y and z are nonnegative integers.