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.

A248035 Least positive integer m such that m + n divides phi(m)^2 + phi(n)^2, where phi(.) is Euler's totient function.

Original entry on oeis.org

1, 3, 2, 1, 15, 14, 3, 8, 9, 30, 30, 14, 7, 6, 5, 9, 3, 8, 55, 60, 9, 4, 83, 28, 25, 71, 9, 1, 24, 4, 43, 32, 1523, 30, 13, 9, 35, 3, 21, 24, 17, 18, 7, 8, 99, 166, 5, 4, 3, 32, 205, 6, 36, 18, 19, 19, 40, 78, 9, 8
Offset: 1

Views

Author

Zhi-Wei Sun, Sep 29 2014

Keywords

Comments

Conjecture: a(n) exists for any n > 0. Moreover, a(n) <= n^2 except for n = 33.

Examples

			a(5) = 15 since 15 + 5 = 20 divides phi(15)^2 + phi(5)^2 = 8^2 + 4^2 = 80.
a(33) = 1523 since 1523 + 33 = 1556 divides phi(1523)^2 + phi(33)^2 = 1522^2 + 20^2 = 2316884 = 1489*1556.

Links

Crossrefs

Cf. A000010, A247975, A248007, A248036.

Programs

  • Mathematica
    Do[m=1;Label[aa];If[Mod[EulerPhi[m]^2+EulerPhi[n]^2,m+n]==0,Print[n," ",m];Goto[bb]];m=m+1;Goto[aa];Label[bb];Continue,{n,1,60}]
lpim[n_]:=Module[{m=1,p2=EulerPhi[n]^2},While[Mod[p2+EulerPhi[m]^2,m+n]!=0,m++];m]; Array[lpim,60] (* Harvey P. Dale, Nov 19 2020 *)

A248054 Least positive integer m such that m + n divides sigma(m^2) + sigma(n^2), where sigma(k) is the sum of all positive divisors of k.

Original entry on oeis.org

1, 3, 2, 7, 24, 34, 3, 81, 209, 16, 63, 25, 7, 20, 140, 10, 3, 10, 22, 2, 39, 4, 35, 5, 4, 2, 28, 27, 75, 41, 16, 78, 44, 6, 23, 14, 207, 59, 21, 84, 17, 78, 7, 3, 11725, 10, 5, 2, 1669, 361, 134, 10, 141, 310, 21, 73, 21, 33, 38, 121
Offset: 1

Views

Author

Zhi-Wei Sun, Sep 30 2014

Keywords

Comments

Conjecture: a(n) exists for any n > 0.

Examples

			a(4) = 7 since 7 + 4 = 11 divides sigma(7^2) + sigma(4^2) = 57 + 31 = 88.

Links

Crossrefs

Cf. A000203, A248008, A248036, A248052.

Programs

  • Mathematica
    Do[m=1;Label[aa];If[Mod[DivisorSigma[1,m^2]+DivisorSigma[1,n^2],m+n]==0,Print[n," ",m];Goto[bb]];m=m+1;Goto[aa];Label[bb];Continue,{n,1,60}]

A248058 Least positive integer m such that m*n divides phi(m^2+n^2), where phi(.) is Euler's totient function.

Original entry on oeis.org

1, 1, 2, 1, 4, 1, 8, 1, 10, 1, 726, 2, 12, 1, 4, 1, 18, 3, 20, 1, 96, 23, 22, 1, 24, 1, 72, 2, 30, 8, 30, 1, 32, 35, 34, 1, 222, 40, 26, 1, 1312, 43, 42, 46, 360, 44, 48, 2, 588, 1, 50, 2, 5100, 1, 88, 1, 19152, 60, 8, 16
Offset: 1

Views

Author

Zhi-Wei Sun, Sep 30 2014

Keywords

Comments

Conjecture: (i) a(n) exists for any n > 0.
(ii) For each n > 0, there is a positive integer m such that m*n divides sigma(m^2+n^2), where sigma(k) is the sum of all positive divisors of k.
Note that a(n) = 1 if n^2 + 1 is prime. When n^2 + (n+1)^2 is prime, n*(n+1) divides phi(n^2 + (n+1)^2) = n^2 + (n+1)^2 - 1 and hence a(n) <= n + 1.
If (n*q)^2 + 1 is prime for some q > 0, then for m = n^2*q the number phi(m^2+n^2) = phi(n^2)*phi((n*q)^2+1) = phi(n^2)*n^2 *q^2 is divisible by m*n = n^3*q. - Zhi-Wei Sun, Oct 03 2014

Examples

			a(5) = 4 since 4*5 divides phi(4^2 + 5^2) = phi(41) = 40.
a(919) = 37160684 since the product 919*37160684 = 34150668596 divides phi(919^2 + 37160684^2) = phi(1380916436192417) = 1379413805929632 = 40392*34150668596.

Links

Crossrefs

Cf. A000010, A000203, A248035, A248036, A248054.

Programs

  • Mathematica
    Do[m=1;Label[aa];If[Mod[EulerPhi[m^2+n^2],m*n]==0,Print[n," ",m];Goto[bb]];m=m+1;Goto[aa];Label[bb];Continue,{n,1,60}]
  • PARI
    a(n)=m=1;while(eulerphi(m^2+n^2)%(m*n),m++);m
vector(100,n,a(n)) \\ Derek Orr, Oct 01 2014

A248044 Least positive integer m such that m + n divides pi(m)^2 + pi(n)^2, where pi(x) denotes the number of primes not exceeding x.

Original entry on oeis.org

1, 3, 1, 4, 12, 11, 1, 8, 7, 16, 2, 5, 26, 25, 24, 4, 228, 227, 46, 45, 44, 43, 16, 6, 5, 1, 27, 26, 45, 44, 12526, 12525, 12524, 12523, 2970, 502, 351, 350, 46, 45, 236, 235, 10, 9, 8, 4, 1078, 1077, 576, 575, 574, 198, 63, 62, 61, 176, 16, 10, 362, 70
Offset: 1

Views

Author

Zhi-Wei Sun, Sep 30 2014

Keywords

Comments

Conjecture: a(n) exists for any n > 0.

Examples

			a(5) = 12 since 12 + 5 = 17 divides pi(12)^2 + pi(5)^2 = 5^2 + 3^2 = 34.

Links

Crossrefs

Cf. A000720, A247824, A247975, A248035, A248036.

Programs

  • Mathematica
    Do[m=1;Label[aa];If[Mod[PrimePi[m]^2+PrimePi[n]^2,m+n]==0,Print[n," ",m];Goto[bb]];m=m+1;Goto[aa];Label[bb];Continue,{n,1,60}]
Showing 1-4 of 4 results.

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