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 1-2 of 2 results.

A262747 Number of ordered ways to write n as x^2 + y^2 + phi(z^2) with 0 <= x <= y, z > 0, 2 | x*y*z, and phi(k^2) < n for all 0 < k < z, where phi(.) is Euler's totient function given by A000010.

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

Views

Author

Zhi-Wei Sun, Sep 30 2015

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 3, 4, 5, 8, 13, 16, 23, 32, 35, 39, 68, 75, 96, 215, 219, 243, 363, 471, 723, 759, 923, 1443, 1551, 1839, 2739, 2883.
It is easy to see that all the numbers phi(n^2) = n*phi(n) (n = 1,2,3,...) are pairwise distinct. We have verified that a(n) > 0 for all n = 1,...,10^7.
See also A262311 for a related conjecture.

Examples

			a(5) = 1 since 5 = 0^2 + 2^2 + phi(1^2).
a(8) = 1 since 8 = 0^2 + 0^2 + phi(4^2) with 0*0*4 even, and phi(1^2) = 1, phi(2^2) = 2, phi(3^2) = 6 all smaller than 8.
a(13) = 1 since 13 = 1^2 + 2^2 + phi(4^2).
a(32) = 1 since 32 = 2^2 + 4^2 + phi(6^2).
a(68) = 1 since 68 = 0^2 + 6^2 + phi(8^2).
a(96) = 1 since 96 = 0^2 + 8^2 + phi(8^2).
a(363) = 1 since 363 = 0^2 + 19^2 + phi(2^2).
a(471) = 1 since 471 = 0^2 + 19^2 + phi(11^2).
a(723) = 1 since 723 = 17^2 + 18^2 + phi(11^2).
a(759) = 1 since 759 = 9^2 + 26^2 + phi(2^2).
a(923) = 1 since 923 = 16^2 + 25^2 + phi(7^2).
a(1443) = 1 since 1443 = 19^2 + 24^2 +phi(23^2).
a(1551) = 1 since 1551 = 18^2 + 35^2 + phi(2^2).
a(1839) = 1 since 1839 = 3^2 + 30^2 + phi(31^2).
a(2739) = 1 since 2739 = 1^2 + 24^2 + phi(47^2).
a(2883) = 1 since 2883 = 21^2 + 44^2 + phi(23^2).

Links

Crossrefs

Cf. A000010, A001481, A002618, A262311, A262746, A262750, A262766.

Programs

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

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.