A262311 - OEIS

%I #40 Oct 01 2015 16:52:23

%S 0,1,1,1,1,3,2,1,1,3,3,2,1,2,2,3,2,2,3,3,3,4,1,2,2,3,3,2,1,3,3,1,2,2,

%T 2,3,4,4,1,3,2,6,3,2,4,4,3,1,3,4,5,5,3,3,3,4,4,8,4,3,5,4,2,2,3,6,6,1,

%U 2,5,3,2,4,5,2,2,2,3,3,2,3,6,3,2,3,3,4,4,3,3,4,5,4,3,3,1,4,3,2,3

%N Number of ordered ways to write n = x^2 + y^2 + phi(z^2) (0 <= x <= y and z > 0) with y or z prime, where phi(.) is Euler's totient function given by A000010.

%C Conjecture: a(n) > 0 for all n > 1.

%C I have verified this for n up to 10^6, and found that a(n) = 1 for the following 67 values of n: 2, 3, 4, 5, 8, 9, 13, 23, 29, 32, 39, 48, 68, 96, 108, 140, 144, 215, 264, 268, 324, 328, 384, 396, 404, 460, 471, 476, 500, 503, 684, 716, 759, 764, 768, 788, 860, 908, 936, 1032, 1076, 1112, 1148, 1164, 1259, 1344, 1399, 1443, 1484, 1503, 1551, 1839, 1868, 2088, 2723, 2883, 3744, 4296, 5963, 6804, 8328, 9680, 10331, 11948, 21524, 39716, 94415. It seems that a(n) = 1 for no other values of n.

%C It is easy to see that all the numbers phi(n^2) = n*phi(n) (n = 1,2,3,...) are pairwise distinct.

%C See also A262781 for a similar conjecture.

%H Zhi-Wei Sun, <a href="/A262311/b262311.txt">Table of n, a(n) for n = 1..10000</a>

%e a(2) = 1 since 2 = 0^2 + 0^2 + phi(2^2) with 2 prime.

%e a(5) = 1 since 5 = 0^2 + 2^2 + phi(1^2) with 2 prime.

%e a(23) = 1 since 23 = 1^2 + 4^2 + phi(3^2) with 3 prime.

%e a(29) = 1 since 29 = 0^2 + 3^2 + phi(5^2) with 3 and 5 both prime.

%e a(48) = 1 since 48 = 2^2 + 2^2 + phi(10^2) with 2 prime.

%e a(96) = 1 since 96 = 3^2 + 9^2 + phi(3^2) with 3 prime.

%e a(140) = 1 since 140 = 7^2 + 7^2 + phi(7^2) with 7 prime.

%e a(471) = 1 since 471 = 0^2 + 19^2 + phi(11^2) with 19 and 11 both prime.

%e a(476) = 1 since 476 = 8^2 + 16^2 + phi(13^2) with 13 prime.

%e a(936) = 1 since 936 = 4^2 + 30^2 + phi(5^2) with 5 prime.

%e a(1112) = 1 since 1112 = 23^2 + 23^2 + phi(9^2) with 23 prime.

%e a(1839) = 1 since 1839 = 3^2 + 30^2 + phi(31^2) with 31 prime.

%e a(1868) = 1 since 1868 = 2^2 + 2^2 + phi(62^2) with 2 prime.

%e a(2088) = 1 since 2088 = 15^2 + 39^2 + phi(19^2) with 19 prime.

%e a(2723) = 1 since 2723 = 34^2 + 35^2 + phi(19^2) with 19 prime.

%e a(2883) = 1 since 2883 = 21^2 + 44^2 + phi(23^2) with 23 prime.

%e a(3744) = 1 since 3744 = 4^2 + 54^2 + phi(29^2) with 29 prime.

%e a(4296) = 1 since 4296 = 26^2 + 60^2 + phi(5^2) with 5 prime.

%e a(5963) = 1 since 5963 = 26^2 + 59^2 + phi(43^2) with 59 and 43 both prime.

%e a(6804) = 1 since 6804 = 40^2 + 72^2 + phi(5^2) with 5 prime.

%e a(8328) = 1 since 8328 = 1^2 + 39^2 + phi(83^2) with 83 prime.

%e a(9680) = 1 since 9680 = 68^2 + 70^2 + phi(13^2) with 13 prime.

%e a(10331) = 1 since 10331 = 17^2 + 100^2 + phi(7^2) with 7 prime.

%e a(11948) = 1 since 11948 = 5^2 + 109^2 + phi(7^2) with 109 and 7 both prime.

%e a(21524) = 1 since 21524 = 59^2 + 109^2 + phi(79^2) with 109 and 79 both prime.

%e a(39716) = 1 since 39716 = 5^2 + 17^2 + phi(199^2) with 17 and 199 both prime.

%e a(94415) = 1 since 94415 = 115^2 + 178^2 + phi(223^2) with 223 prime.

%t SQ[n_]:=IntegerQ[Sqrt[n]]

%t f[n_]:=EulerPhi[n^2]

%t Do[r=0;Do[If[f[z]>n,Goto[aa]];Do[If[SQ[n-f[z]-x^2]&&(PrimeQ[z]||PrimeQ[Sqrt[n-f[z]-x^2]]),r=r+1],{x,0,Sqrt[(n-f[z])/2]}];Label[aa];Continue,{z,1,n}];Print[n," ",r];Continue,{n,1,100}]

%Y Cf. A000010, A000040, A000290, A001481, A002618, A262747, A262781.

%K nonn

%O 1,6

%A _Zhi-Wei Sun_, Oct 01 2015