

A262311


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.


10



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, 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, 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
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OFFSET

1,6


COMMENTS

Conjecture: a(n) > 0 for all n > 1.
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.
It is easy to see that all the numbers phi(n^2) = n*phi(n) (n = 1,2,3,...) are pairwise distinct.
See also A262781 for a similar conjecture.


LINKS

ZhiWei Sun, Table of n, a(n) for n = 1..10000


EXAMPLE

a(2) = 1 since 2 = 0^2 + 0^2 + phi(2^2) with 2 prime.
a(5) = 1 since 5 = 0^2 + 2^2 + phi(1^2) with 2 prime.
a(23) = 1 since 23 = 1^2 + 4^2 + phi(3^2) with 3 prime.
a(29) = 1 since 29 = 0^2 + 3^2 + phi(5^2) with 3 and 5 both prime.
a(48) = 1 since 48 = 2^2 + 2^2 + phi(10^2) with 2 prime.
a(96) = 1 since 96 = 3^2 + 9^2 + phi(3^2) with 3 prime.
a(140) = 1 since 140 = 7^2 + 7^2 + phi(7^2) with 7 prime.
a(471) = 1 since 471 = 0^2 + 19^2 + phi(11^2) with 19 and 11 both prime.
a(476) = 1 since 476 = 8^2 + 16^2 + phi(13^2) with 13 prime.
a(936) = 1 since 936 = 4^2 + 30^2 + phi(5^2) with 5 prime.
a(1112) = 1 since 1112 = 23^2 + 23^2 + phi(9^2) with 23 prime.
a(1839) = 1 since 1839 = 3^2 + 30^2 + phi(31^2) with 31 prime.
a(1868) = 1 since 1868 = 2^2 + 2^2 + phi(62^2) with 2 prime.
a(2088) = 1 since 2088 = 15^2 + 39^2 + phi(19^2) with 19 prime.
a(2723) = 1 since 2723 = 34^2 + 35^2 + phi(19^2) with 19 prime.
a(2883) = 1 since 2883 = 21^2 + 44^2 + phi(23^2) with 23 prime.
a(3744) = 1 since 3744 = 4^2 + 54^2 + phi(29^2) with 29 prime.
a(4296) = 1 since 4296 = 26^2 + 60^2 + phi(5^2) with 5 prime.
a(5963) = 1 since 5963 = 26^2 + 59^2 + phi(43^2) with 59 and 43 both prime.
a(6804) = 1 since 6804 = 40^2 + 72^2 + phi(5^2) with 5 prime.
a(8328) = 1 since 8328 = 1^2 + 39^2 + phi(83^2) with 83 prime.
a(9680) = 1 since 9680 = 68^2 + 70^2 + phi(13^2) with 13 prime.
a(10331) = 1 since 10331 = 17^2 + 100^2 + phi(7^2) with 7 prime.
a(11948) = 1 since 11948 = 5^2 + 109^2 + phi(7^2) with 109 and 7 both prime.
a(21524) = 1 since 21524 = 59^2 + 109^2 + phi(79^2) with 109 and 79 both prime.
a(39716) = 1 since 39716 = 5^2 + 17^2 + phi(199^2) with 17 and 199 both prime.
a(94415) = 1 since 94415 = 115^2 + 178^2 + phi(223^2) with 223 prime.


MATHEMATICA

SQ[n_]:=IntegerQ[Sqrt[n]]
f[n_]:=EulerPhi[n^2]
Do[r=0; Do[If[f[z]>n, Goto[aa]]; Do[If[SQ[nf[z]x^2]&&(PrimeQ[z]PrimeQ[Sqrt[nf[z]x^2]]), r=r+1], {x, 0, Sqrt[(nf[z])/2]}]; Label[aa]; Continue, {z, 1, n}]; Print[n, " ", r]; Continue, {n, 1, 100}]


CROSSREFS

Cf. A000010, A000040, A000290, A001481, A002618, A262747, A262781.
Sequence in context: A172254 A053542 A087284 * A242950 A304972 A152176
Adjacent sequences: A262308 A262309 A262310 * A262312 A262313 A262314


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Oct 01 2015


STATUS

approved



