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.

A020491 Numbers k such that sigma_0(k) divides phi(k).

Original entry on oeis.org

1, 3, 5, 7, 8, 9, 10, 11, 13, 15, 17, 18, 19, 21, 23, 24, 26, 28, 29, 30, 31, 33, 34, 35, 37, 39, 40, 41, 43, 45, 47, 49, 51, 52, 53, 55, 56, 57, 58, 59, 61, 63, 65, 67, 69, 70, 71, 72, 73, 74, 76, 77, 78, 79, 82, 83, 84, 85, 87, 88, 89, 90, 91, 93, 95, 97, 98, 99, 101, 102, 103, 104
Offset: 1

Views

Author

David W. Wilson

Keywords

Comments

In other words, numbers k such that d(k) divides phi(k).
From Enrique Pérez Herrero, Aug 11 2010: (Start)
sigma_0(k) divides phi(k) when:
k is an odd prime: A065091;
k is an odd squarefree number: A056911;
k = 2^m, where m <> 1 is a Mersenne number (A000225).
If d divides (p-1), with p prime, then p^(d-1) is in this sequence, as are p^(p-1), p^(p-2) and p^(-1+p^n).
(End)
phi(n) and d(n) are multiplicative functions, so if m and n are coprime and both of them are in this sequence then m*n is also in this sequence. - Enrique Pérez Herrero, Sep 05 2010
From Bernard Schott, Aug 14 2020: (Start)
The corresponding quotients are in A289585.
About the 3rd case of Enrique Pérez Herrero's comment: if k = 2^M_m, where M_m = 2^m - 1 is a Mersenne number >= 3 (A000225), then the corresponding quotient phi(k)/d(k) is the integer 2^(2^m-m-2) = A076688(m); hence, these numbers k, A058891 \ {2}, form a subsequence. (End)

Links

Crossrefs

Cf. A000005, A000010.
Complement of A015733. [Enrique Pérez Herrero, Aug 11 2010]
Cf. A058891, A076688, A289585.

Programs

  • Mathematica
    Select[ Range[ 105 ], IntegerQ[ EulerPhi[ # ]/DivisorSigma[ 0, # ] ]& ]
  • PARI
    isok(k) = !(eulerphi(k) % numdiv(k)); \\ Michel Marcus, Aug 10 2020

A076688 a(n) = 2^(2^n-n-2).

Original entry on oeis.org

1, 8, 1024, 33554432, 72057594037927936, 664613997892457936451903530140172288, 113078212145816597093331040047546785012958969400039613319782796882727665664
Offset: 2

Views

Author

Benoit Cloitre, Oct 25 2002

Keywords

Comments

Integer values of 1/(2-Sum_{i=1..m} i/2^i).
The next term a(9) has 151 digits, and is too large to include in data. - Bernard Schott, Aug 27 2020

Crossrefs

Cf. A036799.
Subsequence of A289585.
Cf. A000225, A020491, A058891.

Programs

  • Maple
    Data := [seq(2^(2^n-n-2),  n = 2..8)]; \\ Bernard Schott, Aug 26 2020
  • Mathematica
    Table[2^(2^n - n - 2), {n, 2, 8}] (* Amiram Eldar, Aug 27 2020 *)

A341939 Numbers m such that phi(m)/tau(m) is a square of an integer where phi is the Euler totient function (A000010) and tau is the number of divisors function (A000005).

Original entry on oeis.org

1, 3, 8, 10, 18, 19, 24, 30, 34, 45, 52, 57, 73, 74, 85, 102, 125, 135, 140, 152, 153, 156, 163, 182, 185, 190, 202, 219, 222, 252, 255, 333, 342, 360, 375, 394, 416, 420, 436, 451, 455, 456, 459, 476, 489, 505, 514, 546, 555, 570, 584, 606, 625, 629, 640, 646, 679, 680, 730
Offset: 1

Views

Author

Bernard Schott, Feb 24 2021

Keywords

Comments

The first 11 terms of this sequence are also the first 11 terms of A341938: m such that phi(m)*tau(m) is a square, then, a(12) = 57 while A341938(12) = 54. Indeed, if phi(m)/tau(m) is a perfect square then phi(m)*tau(m) is also a square, but the converse is false. These counterexamples are in A341940, the first one is 54 (last example).
Some subsequences (see examples):
-> The seven terms that satisfy also tau(m) = phi(m) form the subsequence A020488 with phi(m)/tau(m) = 1^2.
-> Primes p of the form 2*k^2 + 1 (A090698) form another subsequence because tau(p) = 2 and phi(p) = p-1 = 2*k^2, so phi(p)/tau(p) = k^2.
-> Cubes p^3 where p is a prime of the form k^2+1 (A002496) form another subset because if p = 2, phi(8)/tau(8)=1, and if p odd, phi(p^3)/tau(p^3) = (k*p/2)^2 with k even.

Examples

			phi(30) = 8, tau(30) = 8 so phi(30)/tau(30) = 1^2, and 30 is a term.
phi(45) = 24, tau(45) = 6, so phi(45)/tau(45) = 4 = 2^2, and 85 is a term.
phi(125) = 100, tau(125) = 4, so phi(125)/tau(125) = 25 = 5^2, and 125 is a term.
phi(54) = 18, tau(54) = 8, and phi(54)/tau(54) = 18/8 = 9/4 = (3/2)^2 and 54 is not a term while phi(54)*tau(54) = 12^2.

Crossrefs

Intersection of A020491 and A341938.
Similar for: A144695 (sigma(n)/tau(n) perfect square), A293391 (sigma(n)/phi(n) perfect square).
Subsequences: A090698, A020488.
Cf. A069237, A289585, A341940.
Cf. A000005 (phi), A000010(tau).

Programs

  • Maple
    with(numtheory): filter:= q -> phi(q)/tau(q) = floor(phi(q)/tau(q)) and issqr(phi(q)/tau(q)) : select(filter, [$1..750]);
  • Mathematica
    Select[Range[1000], IntegerQ @ Sqrt[EulerPhi[#]/DivisorSigma[0, #]] &] (* Amiram Eldar, Feb 24 2021 *)
  • PARI
    isok(m) = my(x=eulerphi(m)/numdiv(m)); (denominator(x)==1) && issquare(x); \\ Michel Marcus, Feb 24 2021

A290634 Positive integers which are never the quotient of phi(n)/tau(n).

Original entry on oeis.org

17, 19, 31, 38, 47, 59, 61, 62, 71, 85, 91, 101, 103, 107, 109, 118, 121, 133, 137, 149, 151, 157, 167, 181, 187, 197, 211, 217, 218, 223, 227, 229, 241, 247, 257, 259, 263, 266, 269, 271, 283, 289, 305, 311, 313, 314, 317, 327, 331, 334, 337, 347, 349, 353, 355, 361, 367
Offset: 1

Views

Author

Bernard Schott, Aug 08 2017

Keywords

Comments

For phi(n)/tau(n) see A279287/A279288.
Numbers that do not appear in A175667.
The first nine terms of this sequence are exactly A119480(3) through A119480(11), and many other terms are common to these two sequences.

Crossrefs

Cf. A000010, A000005, A020491, A015733, A020488, A062516, A175667, A112954, A112955, A119480, A289585, A279287/A279288.
Showing 1-4 of 4 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.