

A263992


Number of ordered ways to write n as x^2 + 2*y^2 + phi(z^2) (x >= 0, y >= 0 and z > 0) such that y or z has the form p1 with p prime, where phi(.) is Euler's totient function.


2



1, 2, 2, 2, 2, 1, 1, 3, 3, 4, 3, 4, 3, 4, 2, 2, 4, 4, 4, 5, 3, 1, 4, 4, 2, 5, 2, 4, 4, 4, 2, 2, 3, 5, 6, 2, 4, 5, 5, 4, 4, 4, 3, 9, 5, 4, 2, 5, 6, 7, 6, 7, 6, 3, 3, 9, 6, 6, 8, 5, 3, 5, 5, 4, 8, 7, 6, 5, 5, 3, 3, 5, 6, 8, 6, 6, 4, 8, 2, 6, 5, 5, 8, 8, 2, 5, 7, 4, 9, 7, 5, 5, 6, 5, 4, 4, 5, 6, 7, 6
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OFFSET

1,2


COMMENTS

Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 6, 7, 22, 3447.
This is similar to the conjecture in A262311, and we have verified it for n up to 10^6.


LINKS

ZhiWei Sun, Table of n, a(n) for n = 1..10000
ZhiWei Sun, Some mysterious representations of integers, a message to Number Theory Mailing List, Oct. 25, 2015.


EXAMPLE

a(1) = 1 since 1 = 0^2 + 2*0^2 + phi(1^2) with 1 + 1 = 2 prime.
a(6) = 1 since 6 = 2^2 + 2*0^2 + phi(2^2) with 2 + 1 = 3 prime.
a(7) = 1 since 7 = 2^2 + 2*1^2 + phi(1^2) with 1 + 1 = 2 prime.
a(22) = 1 since 22 = 0^2 + 2*1^2 + phi(5^2) with 1 + 1 = 2 prime.
a(3447) = 1 since 3447 = 42^2 + 2*29^2 + phi(1^2) with 1 + 1 = 2 prime.


MATHEMATICA

phi[n_]:=phi[n]=EulerPhi[n]
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0; Do[If[nz*phi[z]<0, Goto[aa]]; Do[If[SQ[nz*phi[z]2y^2]&&(PrimeQ[y+1]PrimeQ[z+1]), r=r+1], {y, 0, Sqrt[(nz*phi[z])/2]}]; Label[aa]; Continue, {z, 1, n}]; Print[n, " ", r]; Continue, {n, 1, 100}]


CROSSREFS

Cf. A000010, A000290, A006093, A262311, A263998.
Sequence in context: A057155 A037812 A037200 * A180174 A172363 A181877
Adjacent sequences: A263989 A263990 A263991 * A263993 A263994 A263995


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Oct 31 2015


STATUS

approved



