cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

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.

Original entry on oeis.org

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
Offset: 1

Views

Author

Zhi-Wei Sun, Oct 01 2015

Keywords

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.

Examples

			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.

Links

Crossrefs

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

Programs

  • Mathematica
    SQ[n_]:=IntegerQ[Sqrt[n]]
f[n_]:=EulerPhi[n^2]
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}]

A262781 Number of ordered ways to write n as x^2 + phi(y^2) + phi(z^2) (x >= 0 and 0 < y <= z) with y or z prime, where phi(.) is Euler's totient function given by A000010.

Original entry on oeis.org

0, 0, 1, 2, 1, 0, 2, 3, 1, 1, 2, 3, 2, 3, 2, 2, 1, 3, 3, 1, 2, 3, 4, 1, 1, 3, 2, 3, 2, 4, 1, 3, 2, 2, 3, 1, 3, 3, 4, 2, 2, 3, 5, 5, 1, 4, 4, 4, 2, 6, 4, 4, 4, 6, 3, 4, 5, 4, 5, 4, 4, 3, 6, 4, 2, 3, 3, 5, 4, 4, 4, 3, 1, 4, 5, 4, 3, 6, 3, 1, 2, 3, 4, 4, 5, 5, 3, 3, 2, 8, 5, 3, 4, 2, 4, 4, 2, 3, 7, 2
Offset: 1

Views

Author

Zhi-Wei Sun, Oct 01 2015

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 6, and a(n) = 1 only for n = 3, 5, 9, 10, 17, 20, 24, 25, 31, 36, 45, 73, 80, 101, 136, 145, 388, 649.
(ii) For any integer n > 4, we can write 2*n as phi(p^2) + phi(x^2) + phi(y^2) with p prime and p <= x <= y.
See also A262311 for a similar conjecture.

Links

Crossrefs

Cf. A000010, A000040, A000290, A002618, A262311, A262746, A262747.

Programs

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

Formula

a(3) = 1 since 3 = 0^2 + phi(1^2) + phi(2^2) with 2 prime.
a(5) = 1 since 5 = 1^2 + phi(2^2) + phi(2^2) with 2 prime.
a(9) = 1 since 9 = 1^2 + phi(2^2) + phi(3^2) with 2 and 3 both prime.
a(10) = 1 since 10 = 0^2 + phi(2^2) + phi(4^2) with 2 prime.
a(17) = 1 since 17 = 3^2 + phi(2^2) + phi(3^2) with 2 and 3 both prime.
a(20) = 1 since 20 = 4^2 + phi(2^2) + phi(2^2) with 2 prime.
a(24) = 1 since 24 = 4^2 + phi(2^2) + phi(3^2) with 2 and 3 both prime.
a(25) = 1 since 25 = 2^2 + phi(1^2) + phi(5^2) with 5 prime.
a(31) = 1 since 31 = 3^2 + phi(2^2) + phi(5^2) with 2 and 5 both prime.
a(36) = 1 since 36 = 2^2 + phi(5^2) + phi(6^2) with 5 prime.
a(45) = 1 since 45 = 1^2 + phi(2^2) + phi(7^2) with 2 and 7 both prime.
a(73) = 1 since 73 = 5^2 + phi(3^2) + phi(7^2) with 3 and 7 both prime.
a(80) = 1 since 80 = 6^2 + phi(2^2) + phi(7^2) with 2 and 7 both prime.
a(101) = 1 since 101 = 7^2 + phi(5^2) + phi(8^2) with 5 prime.
a(136) = 1 since 136 = 5^2 + phi(1^2) + phi(11^2) with 11 prime.
a(145) = 1 since 145 = 7^2 + phi(7^2) + phi(9^2) with 7 prime.
a(388) = 1 since 388 = 2^2 + phi(7^2) + phi(19^2) with 7 and 19 both prime.
a(649) = 1 since 649 = 11^2 + phi(7^2) + phi(27^2) with 7 prime.

A262750 Least positive integer z such that n - phi(z^2) = x^2 + y^2 for some integers x and y with x*y*z even and phi(k^2) < n for all 0 < k < z, or 0 if no such z exists, where phi(.) is Euler's totient function given by A000010.

Original entry on oeis.org

1, 1, 2, 2, 1, 1, 2, 4, 1, 1, 2, 2, 4, 1, 2, 4, 1, 1, 2, 2, 1, 2, 3, 4, 4, 1, 2, 2, 5, 1, 2, 6, 1, 2, 3, 2, 1, 1, 2, 4, 1, 1, 2, 4, 4, 1, 2, 4, 4, 1, 2, 2, 1, 1, 2, 5, 4, 3, 3, 2, 4, 1, 2, 6, 1, 1, 2, 8, 1, 2, 3, 4, 1, 1, 2, 2, 6, 3, 3, 4, 1, 1, 2, 2, 5, 1, 2, 4, 4, 1, 2, 2, 4, 6, 3, 8, 4, 1, 2, 2
Offset: 1

Views

Author

Zhi-Wei Sun, Sep 30 2015

Keywords

Comments

Conjecture: a(n) <= sqrt(n) except for n = 3, 8, 13, 32.
The conjecture in A262747 implies that a(n) > 0 for all n > 0.

Examples

			a(68) = 8 since 68 - phi(8^2) = 68 - 32 = 36 = 0^2 + 6^2 with 0*6*8 even and all those phi(k^2) (k = 1,...,7) smaller than 68.
a(5403) = 67 since 5403 - phi(67^2) = 5403 - 4422 = 981 = 9^2 + 30^2 with 9*30*67 even and all those phi(k^2) (k = 1,...,5403) smaller than 5403.

Links

Crossrefs

Cf. A000010, A001481, A002618, A233867, A262747.

Programs

  • Mathematica
    f[n_]:=EulerPhi[n^2]
SQ[n_]:=IntegerQ[Sqrt[n]]
Do[Do[If[f[x]>n,Goto[aa]]; Do[If[SQ[n-f[x]-y^2]&&(Mod[x*y, 2]==0||Mod[n-f[x]-y^2, 2]==0),Print[n," ",x];Goto[bb]], {y, 0, Sqrt[(n-f[x])/2]}]; Continue, {x, 1, n}]; Label[aa];Print[n," ",0];Label[bb]; Continue, {n,1,100}]

A262766 Number of positive integers z such that n - phi(z^2) = x^2 + y^2 for some integers x and y with x*y*z even and phi(k^2) < n for all 0 < k < z.

Original entry on oeis.org

1, 2, 1, 1, 1, 3, 2, 1, 2, 4, 2, 2, 1, 2, 2, 1, 2, 3, 2, 3, 4, 3, 1, 2, 3, 3, 1, 4, 2, 2, 2, 1, 3, 2, 1, 2, 4, 4, 1, 2, 2, 6, 3, 2, 3, 4, 3, 3, 4, 3, 3, 4, 3, 3, 4, 2, 5, 6, 3, 4, 4, 4, 2, 2, 4, 5, 4, 1, 5, 4, 3, 4, 5, 6, 1, 4, 3, 2, 3, 3, 5, 5, 3, 4, 4, 4, 3, 2, 2, 5, 4, 4, 5, 3, 3, 1, 3, 3, 2, 3
Offset: 1

Views

Author

Zhi-Wei Sun, Sep 30 2015

Keywords

Comments

The conjecture in A262747 implies that a(n) > 0 for all n > 0.

Examples

			a(4) = 1 since 4 = 1^2 + 1^2 + phi(2^2) with 2*1*1 even and phi(1^2) < 4.
a(9) = 2 since 9 - phi(1^2) = 2^2 + 2^2 with 2*2*1 even, and 9 - phi(4^2) = 0^2 + 1^2 with 0*1*4 even and phi(k^2) < 9 for all k = 1..3.
a(35) = 1 since 35 - phi(3^2) = 2^2 + 5^2 with 2*5*3 even and phi(1^2) < phi(2^2) < 35.
a(96) = 1 since 96 - phi(8^2) = 0^2 + 8^2 with 0*8*8 even and phi(k^2) < 96 for all k = 1..7.

Crossrefs

Cf. A000010, A001481, A002618, A262747, A262750.

Programs

  • Mathematica
    f[n_]:=EulerPhi[n^2]
SQ[n_]:=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[f[x]>n,Goto[aa]];Do[If[(Mod[x*y,2]==0||Mod[Sqrt[n-f[x]-y^2],2]==0)&&SQ[n-f[x]-y^2],r=r+1;Goto[bb]],{y,0,Sqrt[(n-f[x])/2]}];Label[bb];Continue,{x,1,n}]; Label[aa];Print[n," ",r];Continue,{n,1,100}]
Showing 1-4 of 4 results.

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