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A262979
Number of ordered ways to write n as x^4 + phi(y^2) + z*(3*z-1)/2 with x >= 0 and y > 0, where phi(.) is Euler's totient function given by A000010.
2
1, 3, 4, 3, 1, 2, 4, 5, 5, 3, 2, 2, 5, 6, 3, 2, 4, 5, 4, 4, 4, 5, 5, 7, 5, 2, 4, 6, 6, 3, 2, 4, 6, 5, 5, 4, 5, 4, 4, 4, 5, 7, 9, 6, 3, 4, 6, 9, 5, 6, 2, 4, 7, 6, 8, 6, 6, 8, 7, 7, 4, 4, 8, 6, 4, 4, 3, 5, 5, 6, 7, 5, 4, 3, 5, 5, 5, 5, 6, 4, 3, 5, 8, 7, 6, 4, 5, 5, 8, 8, 5, 5
OFFSET
1,2
COMMENTS
Conjecture: (i) a(n) > 0 for all n > 0.
(ii) Any positive integer n can be written as x^4 + phi(y^2) + pi(z^2) (or x^4 + pi(y^2) + pi(z^2)) with y > 0 and z > 0, where pi(m) denotes the number of primes not exceeding m.
EXAMPLE
a(5) = 1 since 5 = 1^4 + phi(2^2) + (-1)*(3*(-1)-1)/2.
a(6) = 2 since 6 = 0^4 + phi(1^2) + 2*(3*2-1)/2 = 0^4 + phi(3^2) + 0*(3*0-1)/2.
a(16) = 2 since 16 = 0^4 + phi(1^2) + (-3)*(3*(-3)-1)/2
= 1^4 + phi(4^2) + (-2)*(3*(-2)-1)/2.
MATHEMATICA
f[n_]:=EulerPhi[n^2]
PenQ[n_]:=IntegerQ[Sqrt[24n+1]]
Do[r=0; Do[If[f[x]>n, Goto[aa]]; Do[If[PenQ[n-f[x]-y^4], r=r+1], {y, 0, (n-f[x])^(1/4)}]; Label[aa]; Continue, {x, 1, n}]; Print[n, " ", r]; Continue, {n, 1, 100}]
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Oct 06 2015
STATUS
approved