login
A262982
Number of ordered ways to write n as x^4 + phi(y^2) + z*(z+1)/2 with x >= 0, y > 0 and z > 0, where phi(.) is Euler's totient function given by A000010.
3
0, 1, 2, 2, 2, 1, 2, 3, 3, 2, 2, 4, 3, 2, 2, 3, 3, 4, 3, 1, 3, 4, 7, 4, 2, 1, 5, 4, 3, 5, 3, 2, 3, 5, 3, 3, 4, 5, 5, 1, 3, 5, 6, 3, 4, 5, 4, 5, 6, 3, 5, 4, 4, 5, 3, 5, 8, 7, 3, 3, 5, 4, 5, 7, 3, 2, 4, 6, 7, 4, 3, 3, 5, 2, 3, 6, 5, 3, 6, 3, 2, 1, 4, 6, 7, 6, 5, 6, 1, 6, 5, 5, 6, 6, 4, 3, 4, 6, 7, 5
OFFSET
1,3
COMMENTS
Conjecture: a(n) > 0 for all n > 1, and a(n) = 1 only for n = 2, 6, 20, 26, 40, 82, 89, 105, 305, 416, 470, 725, 6135, 25430, 90285.
Compare this with the conjectures in A262311, A262785 and A262813.
EXAMPLE
a(2) = 1 since 1 = 0^4 + phi(1^2) + 1*2/2.
a(6) = 1 since 6 = 1^4 + phi(2^2) + 2*3/2.
a(20) = 1 since 20 = 2^4 + phi(1^2) + 2*3/2.
a(26) = 1 since 26 = 0^4 + phi(5^2) + 3*4/2.
a(40) = 1 since 40 = 0^4 + phi(6^2) + 7*8/2.
a(82) = 1 since 82 = 0^4 + phi(9^2) + 7*8/2.
a(89) = 1 since 89 = 3^4 + phi(2^2) + 3*4/2.
a(105) = 1 since 105 = 0^4 + phi(14^2) + 6*7/2.
a(305) = 1 since 305 = 4^4 + phi(12^2) + 1*2/2.
a(416) = 1 since 416 = 4^4 + phi(10^2) + 15*16/2.
a(470) = 1 since 470 = 2^4 + phi(12^2) + 28*29/2.
a(725) = 1 since 725 = 2^4 + phi(3^2) + 37*38/2.
a(6135) = 1 since 6135 = 6^4 + phi(81^2) + 30*31/2.
a(25430) = 1 since 25430 = 5^4 + phi(152^2) + 166*167/2.
a(90285) = 1 since 90285 = 16^4 + phi(212^2) + 73*74/2.
MATHEMATICA
f[n_]:=EulerPhi[n^2]
TQ[n_]:=n>0&&IntegerQ[Sqrt[8n+1]]
Do[r=0; Do[If[f[x]>n, Goto[aa]]; Do[If[TQ[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