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A308342
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Number of ways to write 2*n as phi(x^2) + phi(y^2) + phi(z^2), where x,y,z are positive integers with x <= y <= z, and phi(.) is Euler's totient function (A000010).
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1
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0, 1, 1, 1, 2, 1, 2, 2, 2, 2, 3, 3, 2, 2, 2, 2, 3, 3, 1, 2, 3, 4, 4, 3, 4, 5, 4, 6, 5, 5, 5, 5, 5, 5, 4, 5, 4, 4, 2, 5, 5, 3, 6, 6, 3, 7, 6, 6, 6, 5, 6, 6, 4, 5, 5, 5, 5, 6, 4, 5, 8, 7, 5, 9, 6, 7, 8, 8, 7, 6, 6, 8, 5, 7, 7, 6, 5, 6, 8, 8, 8, 10, 6, 10, 13, 10, 10, 9, 6, 11, 9, 7, 3, 9, 6, 6, 9, 7, 5, 12
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OFFSET
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1,5
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COMMENTS
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Conjecture 1: a(n) > 0 for all n > 1. In other words, the set {phi(x^2) + phi(y^2) + phi(z^2): x,y,z = 1,2,3,...} contains all even numbers greater than two.
Conjecture 2: For any integer n > 3, we can write 2*n+1 as phi(x^2) + phi(y^2) + sigma(z^2) with x,y,z positive integers, where the function sigma(.) is given by A000203.
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LINKS
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EXAMPLE
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a(2) = 1 with 2*2 = phi(1^2) + phi(1^2) + phi(2^2).
a(3) = 1 with 2*3 = phi(2^2) + phi(2^2) + phi(2^2).
a(4) = 1 with 2*4 = phi(1^2) + phi(1^2) + phi(3^2).
a(6) = 1 with 2*6 = phi(2^2) + phi(2^2) + phi(4^2).
a(19) = 1 with 2*19 = phi(3^2) + phi(5^2) + phi(6^2).
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MATHEMATICA
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f[n_]:=f[n]=n*EulerPhi[n]
T={}; Do[If[f[n]<=200, T=Append[T, f[n]]], {n, 1, 200}];
tab={}; Do[r=0; Do[If[f[k]>2n/3, Goto[cc]]; Do[If[f[m]<f[k]||f[m]>(2n-f[k])/2, Goto[bb]]; If[MemberQ[T, 2n-f[k]-f[m]], r=r+1]; Label[bb], {m, 1, (2n-f[k])/2}]; Label[cc], {k, 1, 2n/3}]; tab=Append[tab, r], {n, 1, 100}]; Print[tab]
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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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