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A275297
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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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7
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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
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OFFSET
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0,2
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COMMENTS
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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.
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LINKS
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EXAMPLE
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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.
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MATHEMATICA
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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}]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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