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 1-8 of 8 results.

A248007 Least positive integer m such that m + n divides phi(m)*phi(n), where phi(.) is Euler's totient function.

Original entry on oeis.org

5, 8, 3, 14, 9, 20, 11, 10, 9, 16, 7, 18, 5, 12, 3, 38, 21, 8, 15, 58, 9, 20, 11, 18, 14, 32, 7, 14, 13, 12, 35, 22, 9, 24, 7, 46, 13, 31, 3, 42, 45, 16, 11, 30, 13, 44, 19, 27, 25, 40, 15, 26, 28, 36, 35, 28, 9, 64, 7, 54, 21, 28, 19, 26
Offset: 7

Views

Author

Zhi-Wei Sun, Sep 29 2014

Keywords

Comments

Conjecture: a(n) exists for any n > 6. - Zhi-Wei Sun, Sep 29 2014
Numbers n for which a(n) > n: 10, 12, 22, 26, 42, 78, 166, 266, 290. The next term in this mini-sequence, if it exists, is greater than 3*10^4. I conjecture this list is finite. - Derek Orr, Sep 29 2014
a(2^n) <= 2^n for all n > 2. Also, if a(i) = j, then a(j) <= i. - Derek Orr, Sep 29 2014

Examples

			a(10) = 14 since 10 + 14 divides phi(10)*phi(14) = 4*6 = 24.

Links

Crossrefs

Cf. A000010, A248004, A248006, A248008.

Programs

  • Mathematica
    Do[m=1; Label[aa]; If[Mod[EulerPhi[m]*EulerPhi[n], m+n]==0, Print[n, " ", m]; Goto[bb]]; m=m+1; Goto[aa]; Label[bb]; Continue, {n, 7, 70}]
  • PARI
    a(n)=m=1;while((eulerphi(m)*eulerphi(n))%(m+n),m++);m
vector(100,n,a(n+6)) \\ Derek Orr, Sep 29 2014

A248036 Least positive integer m such that m + n divides sigma(m)^2 + sigma(n)^2, where sigma(k) denotes the number of positive divisors of k.

Original entry on oeis.org

1, 3, 2, 1, 10, 6, 3, 50, 1, 5, 34, 28, 7, 6, 10, 18, 3, 16, 33, 5, 20, 14, 83, 24, 1, 10, 10, 12, 56, 6, 33, 2, 15, 11, 93, 13, 204, 27, 52, 38, 17, 6, 7, 6, 15, 14, 5, 944, 1, 8, 17, 39, 32, 33, 5, 24, 7, 59, 58, 15
Offset: 1

Views

Author

Zhi-Wei Sun, Sep 29 2014

Keywords

Comments

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

Examples

			a(5) = 10 since 10 + 5 = 15 divides sigma(10)^2 + sigma(5)^2 = 18^2 + 6^2 = 360.
a(1024) = 2098177 since 2098177 + 1024 = 2099201 divides sigma(2098177)^2 + sigma(1024)^2 = 2103300^2 + 2047^2 = 4423875080209 = 2099201*2107409.

Links

Crossrefs

Cf. A000203, A247975, A248008, A248035.

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}]
lpi[n_]:=Module[{m=1,dsn=DivisorSigma[1,n]^2},While[ !Divisible[ DivisorSigma[ 1,m]^2+ dsn, m+n], m++];m]; Array[lpi, 60] (* Harvey P. Dale, May 07 2016 *)

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}]

A248588 Least positive integer m such that m + n divides sigma(m), where sigma(m) is the sum of all positive divisors of m.

Original entry on oeis.org

2, 12, 4, 9, 40, 6, 8, 10, 15, 14, 21, 112, 27, 22, 16, 12, 39, 289, 65, 34, 18, 20, 57, 60, 95, 46, 69, 28, 115, 96, 32, 58, 45, 62, 93, 24, 155, 340, 217, 44, 63, 30, 50, 82, 123, 52, 129, 204, 75, 40, 141, 228, 235, 42, 36, 106, 99, 68, 265, 120
Offset: 1

Views

Author

Zhi-Wei Sun, Oct 09 2014

Keywords

Comments

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

Examples

			a(5) = 40 since 40 + 5 = 45 divides sigma(40) =  90.
a(1162) = 24031232 since 24031232 + 1162 = 24032394 divides sigma(24031232) =  48064788 = 2*24032394.

Links

Crossrefs

Cf. A000203, A248007, A248008, A248029, A248030, A248568, A248590.

Programs

  • Mathematica
    Do[m=1;Label[aa];If[Mod[DivisorSigma[1,m],m+n]==0,Print[n," ",m];Goto[bb]];m=m+1;Goto[aa];Label[bb];Continue,{n,1,60}]
lpi[n_]:=Module[{m=1},While[!Divisible[DivisorSigma[1,m],m+n],m++];m]; Array[lpi,60] (* Harvey P. Dale, Feb 21 2020 *)
  • PARI
    a(n) = my(m = 1); while(sigma(m) % (m+n), m++); m; \\ Michel Marcus, Aug 08 2017

A248006 Least positive integer m such that m + n divides phi(m*n), where phi(.) is Euler's totient function.

Original entry on oeis.org

3, 4, 3, 6, 5, 8, 9, 6, 9, 4, 11, 7, 5, 16, 7, 9, 5, 12, 7, 18, 21, 8, 15, 13, 27, 14, 11, 10, 14, 32, 7, 14, 5, 12, 35, 10, 13, 24, 7, 14, 13, 11, 9, 42, 45, 16, 11, 30, 13, 12, 19, 27, 33, 8, 15, 22, 28, 4, 35, 28, 18, 64, 7, 14, 21, 28, 19, 10
Offset: 3

Views

Author

Zhi-Wei Sun, Sep 29 2014

Keywords

Comments

Conjecture: For any n > 2, a(n) exists and a(n) <= n.
See also A248007 and A248008 for similar conjectures. - Zhi-Wei Sun, Sep 29 2014
The conjecture is true: One can show that 2*n divides phi(n^2) for all n > 2. So, a(n) is at most n. - Derek Orr, Sep 29 2014
a(n) >= 3 for all n. - Robert Israel, Sep 29 2014

Examples

			a(5) = 3 since 3 + 5 divides phi(3*5) = 8.

Links

Crossrefs

Cf. A000010, A248004, A248007, A248008.

Programs

  • Maple
    f:= proc(n)
local m;
for m from 3 do
  if numtheory:-phi(m*n) mod (m+n) = 0 then return m fi
od
end proc;
seq(f(n),n=3..100); # Robert Israel, Sep 29 2014
  • Mathematica
    Do[m=1;Label[aa];If[Mod[EulerPhi[m*n],m+n]==0,Print[n," ",m];Goto[bb]];m=m+1;Goto[aa];Label[bb];Continue,{n,3,70}]
  • PARI
    a(n)=m=1;while(eulerphi(m*n)%(m+n),m++);m
vector(100,n,a(n+2)) \\ Derek Orr, Sep 29 2014

A248029 Least positive integer m such that m + n divides phi(m)*sigma(n), where phi(.) and sigma(.) are given by A000010 and A000203.

Original entry on oeis.org

1, 1, 3, 1, 6, 1, 7, 4, 8, 1, 2, 1, 10, 9, 15, 1, 8, 1, 1, 11, 14, 1, 6, 6, 16, 5, 14, 1, 6, 1, 10, 15, 11, 13, 16, 1, 7, 9, 5, 1, 6, 1, 12, 7, 26, 1, 14, 8, 12, 21, 46, 1, 6, 17, 4, 23, 32, 1, 24, 1, 34, 41, 63, 7, 6, 1, 16, 11, 2
Offset: 2

Views

Author

Zhi-Wei Sun, Sep 29 2014

Keywords

Comments

Conjecture: For any n > 1, we have a(n) <= n.
The existence of a(n) is easy; in fact, for m = sigma(n) - n, obviously m + n divides phi(m)*sigma(n). - Zhi-Wei Sun, Oct 02 2014

Examples

			a(8) = 7 since 7 + 8 = 15 divides phi(7)*sigma(8) = 6*15 = 90.

Links

Crossrefs

Cf. A000010, A000203, A248004, A248007, A248008, A248030.

Programs

  • Mathematica
    Do[m=1;Label[aa];If[Mod[EulerPhi[m]*DivisorSigma[1,n],m+n]==0,Print[n," ",m];Goto[bb]];m=m+1;Goto[aa];Label[bb];Continue,{n,2,70}]
  • PARI
    a(n)=m=1;while((eulerphi(m)*sigma(n))%(m+n),m++);m
vector(100,n,a(n)) \\ Derek Orr, Sep 29 2014

A248030 Least positive integer m such that m + n divides sigma(m)*phi(n), where sigma(.) and phi(.) are given by A000203 and A000010.

Original entry on oeis.org

2, 12, 4, 2, 3, 6, 2, 10, 3, 2, 21, 8, 3, 22, 13, 8, 9, 6, 8, 12, 3, 8, 10, 4, 5, 10, 21, 8, 20, 26, 4, 8, 7, 14, 13, 12, 8, 4, 33, 8, 23, 6, 20, 12, 3, 16, 22, 72, 7, 10, 13, 4, 27, 42, 5, 24, 15, 26, 57, 18, 11, 38, 27, 20, 31, 4, 21, 36, 19, 2
Offset: 1

Views

Author

Zhi-Wei Sun, Sep 29 2014

Keywords

Comments

Conjecture: a(n) < 2*n for any n > 2.
Numbers n such that a(n) > n: 1, 2, 3, 8, 11, 14, 48, 227, 908, 4478, ... The next number, if it exists, is greater than 10^5. - Derek Orr, Sep 29 2014

Examples

			a(2) = 12 since 12 + 2 = 14 divides sigma(12)*phi(2) = 28.

Links

Crossrefs

Cf. A000010, A000203, A248004, A248007, A248008, A248029.

Programs

  • Mathematica
    Do[m=1;Label[aa];If[Mod[DivisorSigma[1,m]*EulerPhi[n],m+n]==0,Print[n," ",m];Goto[bb]];m= m+1; Goto[aa];Label[bb];Continue,{n,1,70}]
lpim[n_]:=Module[{m=1,ephn=EulerPhi[n]},While[Mod[ephn*DivisorSigma[1,m],m+n]!=0, m++]; m]; Array[lpim,70] (* Harvey P. Dale, Feb 14 2024 *)
  • PARI
    a(n)=m=1;while((eulerphi(n)*sigma(m))%(m+n),m++);m
vector(100,n,a(n)) \\ Derek Orr, Sep 29 2014

A248144 Least positive integer m such that m + n divides p(m*n), where p(.) is the partition function given by A000041.

Original entry on oeis.org

4, 5, 3, 1, 2, 14, 2, 10, 1, 4, 17, 5, 9, 1, 1, 6, 4, 10, 16, 357, 1, 197, 14, 1, 3, 9, 6, 1123, 15, 93, 4, 1, 8, 46, 77, 99, 18, 53, 10, 76, 4, 2, 15, 152, 4, 3, 10, 29, 6, 12, 4, 1, 25, 1, 252, 64, 106, 11, 11, 136
Offset: 1

Views

Author

Zhi-Wei Sun, Oct 02 2014

Keywords

Comments

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

Examples

			 a(6) = 14 since 6 + 14 = 20 divides p(6*14) = 26543660.

Links

Crossrefs

Cf. A000041, A248004, A248008, A248143, A248175.

Programs

  • Mathematica
    Do[m=1;Label[aa];If[Mod[PartitionsP[m*n],m+n]==0,Print[n," ",m];Goto[bb]];m=m+1;Goto[aa];Label[bb];Continue,{n,1,60}]
Showing 1-8 of 8 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.