A272979 -id:A272979 - cp's OEIS frontend

A274274 Number of ordered ways to write n as x^3 + y^2 + z^2, where x,y,z are nonnegative integers with y <= z.

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

Offset: 0

Zhi-Wei Sun, Jul 14 2016

nonn

Conjecture: Let n be any nonnegative integer.
(i) Either a(n) > 0 or a(n-2) > 0. Also, a(n) > 0 or a(n-6) > 0. Moreover, if n has the form 2^k*(4m+1) with k and m nonnegative integers, then a(n) > 0 except for n = 813, 4404, 6420, 28804.
(ii) Either n or n-3 can be written as x^3 + y^2 + 3*z^2 with x,y,z nonnegative integers.
(iii) For each d = 4, 5, 11, 12, either n or n-d can be written as x^3 + y^2 + 2*z^2 with x,y,z nonnegative integers.
We have verified that a(n) or a(n-2) is positive for every n = 0..2*10^6. Note that for each n = 0,1,2,... either n or n-2 can be written as x^2 + y^2 + z^2 with x,y,z nonnegative integers, which follows immediately from the Gauss-Legendre theorem on sums of three squares.

			a(6) = 1 since 6 = 1^3 + 1^2 + 2^2.
a(14) = 1 since 14 = 1^3 + 2^2 + 3^2.
a(31) = 1 since 31 = 3^3 + 0^2 + 2^2.

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

Cf. A000290, A000578, A022551, A022552, A262857, A272979.

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

A275297 Number of ordered ways to write n as x^2 + y^2 + z^2 + w^3 with x + 2*y a square, where x,y,z,w are nonnegative integers with z >= w.

Original entry on oeis.org

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

Offset: 0

Zhi-Wei Sun, Jul 22 2016

nonn

Conjecture: a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 3, 7, 8, 11, 12, 15, 23, 24, 32, 39, 47, 71, 103, 120, 136, 159, 176, 183, 218, 359, 463.
Compare this conjecture with Conjecture 5.1 of the author's preprint arXiv:1604.06723. See also A275298 and A275299 for similar conjectures.
By Theorem 1.1 of arXiv:1604.06723, any natural number can be written as the sum of three squares and a sixth power.
Let c be 1 or 2. By the conjecture in A272979, any n = 0,1,2,... can be written as x^2 + 2*y^2 + z^3 + 2*c^2*w^4 with x,y,z,w nonnegative integers, and hence n = x^2 + (y+c*w^2)^2 + (y-c*w^2)^2 + z^3 with (y+c*w^2)-(y-c*w^2) = 2*c*w^2. If n > 0 is not among the 174 terms in the b-file of A275169, then the conjecture in A275169 implies that n can be written as x^2 + y^2 + z^2 + w^3 with x - y = 0^2, where x,y,z,w are nonnegative integers. If n is among the 174 terms in the b-file of A275169, then we may use a computer to verify that n can be written as x^2 + y^2 + z^2 + w^3 with c*(x-y) a square, where x,y,z,w are nonnegative integers.

			a(0) = 1 since 0 = 0^2 + 0^2 + 0^2 + 0^3 with 0 + 2*0 = 0^2 and 0 = 0.
a(1) = 2 since 1 = 0^2 + 0^2 + 1^2 + 0^3 with 0 + 2*0 = 0^2 and 1 > 0, and also 1 = 1^2 + 0^2 + 0^2 + 0^2 with 1 + 2*0 = 1^2 and 0 = 0.
a(3) = 1 since 3 = 1^2 + 0^2 + 1^2 + 1^3 with 1 + 2*0 = 1^2 and 1 = 1.
a(7) = 1 since 7 = 2^2 + 1^2 + 1^2 + 1^3 with 2 + 2*1 = 2^2 and 1 = 1.
a(8) = 1 since 8 = 0^2 + 2^2 + 2^2 + 0^3 with 0 + 2*2 = 2^2 and 2 > 0.
a(11) = 1 since 11 = 1^2 + 0^2 + 3^2 + 1^3 with 1 + 2*0 = 1^2 and 3 > 1.
a(12) = 1 since 12 = 0^2 + 0^2 + 2^2 + 2^3 with 0 + 2*0 = 0^2 and 2 = 2.
a(15) = 1 since 15 = 2^2 + 1^2 + 3^2 + 1^3 with 2 + 2*1 = 2^2 and 3 > 1.
a(23) = 1 since 23 = 3^2 + 3^2 + 2^2 + 1^3 with 3 + 2*3 = 3^2 and 2 > 1.
a(24) = 1 since 24 = 0^2 + 0^2 + 4^2 + 2^3 with 0 + 2*0 = 0^2 and 4 > 2.
a(32) = 1 since 32 = 4^2 + 0^2 + 4^2 + 0^3 with 4 + 2*0 = 2^2 and 4 > 0.
a(39) = 1 since 39 = 5^2 + 2^2 + 3^2 + 1^3 with 5 + 2*2 = 3^2 and 3 > 1.
a(47) = 1 since 47 = 0^2 + 2^2 + 4^2 + 3^3 with 0 + 2*2 = 2^2 and 4 > 3.
a(71) = 1 since 71 = 6^2 + 5^2 + 3^2 + 1^3 with 6 + 2*5 = 4^2 and 3 > 1.
a(103) = 1 since 103 = 2^2 + 7^2 + 7^2 + 1^3 with 2 + 2*7 = 4^2 and 7 > 1.
a(120) = 1 since 120 = 5^2 + 2^2 + 8^2 + 3^3 with 5 + 2*2 = 3^2 and 8 > 3.
a(136) = 1 since 136 = 0^2 + 8^2 + 8^2 + 2^3 with 0 + 2*8 = 4^2 and 8 > 2.
a(159) = 1 since 159 = 10^2 + 3^2 + 7^2 + 1^3 with 10 + 2*3 = 4^2 and 7 > 1.
a(176) = 1 since 176 = 2^2 + 1^2 + 12^2 + 3^3 with 2 + 2*1 = 2^2 and 12 > 3.
a(183) = 1 since 183 = 6^2 + 5^2 + 11^2 + 1^3 with 6 + 2*5 = 4^2 and 11 > 1.
a(218) = 1 since 218 = 5^2 + 2^2 + 8^2 + 5^3 with 5 + 2*2 = 3^2 and 8 > 5.
a(359) = 1 since 359 = 11^2 + 7^2 + 8^2 + 5^3 with 11 + 2*7 = 5^2 and 8 > 5.
a(463) = 1 since 463 = 2^2 + 17^2 + 13^2 + 1^3 with 2 + 2*17 = 6^2 and 13 > 1.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, arXiv:1604.06723 [math.GM], 2016.

Cf. A000290, A000578, A271518, A272979, A273404, A273429, A275169, A275298, A275299.

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

A275083 Positive integers congruent to 0 or 1 modulo 4 that cannot be written as x^3 + y^2 + z^2 with x,y,z nonnegative integers.

Original entry on oeis.org

120, 312, 813, 2136, 2680, 3224, 4404, 5340, 6420, 10060, 11320, 11824, 14008, 15856, 26544, 28804, 34392, 47984

Offset: 1

Zhi-Wei Sun, Jul 15 2016

nonn

Conjecture: (i) The sequence has totally 18 terms as listed.
(ii) For each r = 2,3 there are infinitely many positive integers n == r (mod 4) not in the form x^3 + y^2 + z^2 with x,y,z nonnegative integers.
Our computation indicates that the sequence has no other terms below 10^6.
Let d be 2 or 6. Clearly, n-d is congruent to 0 or 1 modulo 4 if n is congruent to 2 or 3 modulo 4. So part (i) of the conjecture essentially implies that for each n = 0,1,2,... either n or n-d can be written as x^3 + y^2 + z^2 with x,y,z nonnegative integers.

			a(1) = 120 since all those positive integers congruent to 0 or 1 modulo 4 and smaller than 120 can be written as x^3 + y^2 + z^2 with x,y,z nonnegative integers but 120 (divisible by 4) cannot be written in this way.

Cf. A000290, A000578, A022551, A022552, A262813, A262857, A272979, A274274.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
n=0;Do[If[Mod[m,4]>1,Goto[aa]];Do[If[SQ[m-x^3-y^2],Goto[aa]],{x,0,m^(1/3)},{y,0,Sqrt[(m-x^3)/2]}];n=n+1;Print[n," ",m];Label[aa];Continue,{m,1,50000}]

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A274274 Number of ordered ways to write n as x^3 + y^2 + z^2, where x,y,z are nonnegative integers with y <= z.

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A275297 Number of ordered ways to write n as x^2 + y^2 + z^2 + w^3 with x + 2*y a square, where x,y,z,w are nonnegative integers with z >= w.

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A275083 Positive integers congruent to 0 or 1 modulo 4 that cannot be written as x^3 + y^2 + z^2 with x,y,z nonnegative integers.

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