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A262747
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Number of ordered ways to write n as x^2 + y^2 + phi(z^2) with 0 <= x <= y, z > 0, 2 | x*y*z, and phi(k^2) < n for all 0 < k < z, where phi(.) is Euler's totient function given by A000010.
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5
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1, 2, 1, 1, 1, 3, 2, 1, 2, 4, 2, 2, 1, 2, 2, 1, 2, 3, 2, 3, 4, 3, 1, 2, 3, 4, 2, 4, 2, 2, 3, 1, 4, 2, 1, 2, 5, 4, 1, 2, 2, 6, 3, 2, 4, 4, 3, 3, 4, 3, 3, 5, 3, 3, 4, 2, 6, 7, 3, 4, 4, 5, 2, 2, 5, 6, 6, 1, 5, 4, 4, 4, 6, 6, 1, 4, 4, 2, 4, 3, 5, 6, 3, 4, 5, 5, 4, 2, 2, 6, 5, 4, 6, 3, 3, 1, 5, 3, 2, 3
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OFFSET
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1,2
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COMMENTS
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Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 3, 4, 5, 8, 13, 16, 23, 32, 35, 39, 68, 75, 96, 215, 219, 243, 363, 471, 723, 759, 923, 1443, 1551, 1839, 2739, 2883.
It is easy to see that all the numbers phi(n^2) = n*phi(n) (n = 1,2,3,...) are pairwise distinct. We have verified that a(n) > 0 for all n = 1,...,10^7.
See also A262311 for a related conjecture.
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LINKS
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EXAMPLE
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a(5) = 1 since 5 = 0^2 + 2^2 + phi(1^2).
a(8) = 1 since 8 = 0^2 + 0^2 + phi(4^2) with 0*0*4 even, and phi(1^2) = 1, phi(2^2) = 2, phi(3^2) = 6 all smaller than 8.
a(13) = 1 since 13 = 1^2 + 2^2 + phi(4^2).
a(32) = 1 since 32 = 2^2 + 4^2 + phi(6^2).
a(68) = 1 since 68 = 0^2 + 6^2 + phi(8^2).
a(96) = 1 since 96 = 0^2 + 8^2 + phi(8^2).
a(363) = 1 since 363 = 0^2 + 19^2 + phi(2^2).
a(471) = 1 since 471 = 0^2 + 19^2 + phi(11^2).
a(723) = 1 since 723 = 17^2 + 18^2 + phi(11^2).
a(759) = 1 since 759 = 9^2 + 26^2 + phi(2^2).
a(923) = 1 since 923 = 16^2 + 25^2 + phi(7^2).
a(1443) = 1 since 1443 = 19^2 + 24^2 +phi(23^2).
a(1551) = 1 since 1551 = 18^2 + 35^2 + phi(2^2).
a(1839) = 1 since 1839 = 3^2 + 30^2 + phi(31^2).
a(2739) = 1 since 2739 = 1^2 + 24^2 + phi(47^2).
a(2883) = 1 since 2883 = 21^2 + 44^2 + phi(23^2).
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MATHEMATICA
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f[n_]:=EulerPhi[n^2]
SQ[n_]:=IntegerQ[Sqrt[n]]
Do[r=0; Do[If[f[x]>n, Goto[aa]]; Do[If[SQ[n-f[x]-y^2]&&(Mod[x*y, 2]==0||Mod[Sqrt[n-f[x]-y^2], 2]==0), r=r+1], {y, 0, Sqrt[(n-f[x])/2]}]; Continue, {x, 1, n}]; Label[aa]; Print[n, " ", r]; Continue, {n, 1, 100}]
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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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