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.

A326835 Numbers whose divisors have distinct values of the Euler totient function (A000010).

Original entry on oeis.org

1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99, 101, 103, 105, 107, 109, 111, 113, 115, 117, 119, 121, 123, 125, 127
Offset: 1

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Author

Amiram Eldar, Oct 28 2019

Keywords

Comments

Since Sum_{d|k} phi(d) = k, these are numbers k such that the set {phi(d) | d|k} is a partition of k into distinct parts.
Includes all the odd prime numbers, since an odd prime p has 2 divisors, 1 and p, whose phi values are 1 and p-1.
If k is a term, then all the divisors of k are also terms. If k is not a term, then all its multiples are not terms. The primitive terms of the complementary sequence are 2, 63, 273, 513, 585, 825, 2107, 2109, 2255, 3069, ....
In particular, all the terms are odd since 2 is not a term (phi(1) = phi(2)).
The number of terms below 10^k for k = 1, 2, ... are 5, 49, 488, 4860, 48598, 485807, 4857394, 48572251, 485716764, 4857144075, ...
Apparently the sequence has an asymptotic density of 0.4857...

Examples

			3 is a term since it has 2 divisors, 1 and 3, and phi(1) = 1 != phi(3) = 2.
15 is a term since the phi values of its divisors, {1, 3, 5, 15}, are distinct: {1, 2, 4, 8}.
		

Crossrefs

Programs

  • Maple
    filter:= proc(n) local D;
      D:=numtheory:-divisors(n);
      nops(D) = nops(map(numtheory:-phi,D))
    end proc:
    select(filter, [seq(i,i=1..200,2)]); # Robert Israel, Oct 29 2019
  • Mathematica
    aQ[n_] := Length @ Union[EulerPhi /@ (d = Divisors[n])] == Length[d];  Select[Range[130], aQ]
  • PARI
    isok(k) = #Set(apply(x->eulerphi(x), divisors(k))) == numdiv(k); \\ Michel Marcus, Oct 28 2019

Formula

Numbers k such that A319696(k) = A000005(k).
Numbers k such that A319695(k) = A032741(k).
Numbers k such that the k-th row of A102190 has distinct terms.