A082897 Perfect totient numbers.

3, 9, 15, 27, 39, 81, 111, 183, 243, 255, 327, 363, 471, 729, 2187, 2199, 3063, 4359, 4375, 5571, 6561, 8751, 15723, 19683, 36759, 46791, 59049, 65535, 140103, 177147, 208191, 441027, 531441, 1594323, 4190263, 4782969, 9056583, 14348907, 43046721

Offset: 1

Author

Douglas E. Iannucci, Jul 21 2003

Keywords

• nonn

Comments

It is trivial that perfect totient numbers must be odd. It is easy to show that powers of 3 are perfect totient numbers.

The product of the first n Fermat primes (A019434) is also a perfect totient number. There are 57 terms under 10^11. - Jud McCranie, Feb 24 2012

Terms 15, 255, 65535 and 4294967295 also belong to A051179 (see Theorem 4 in Loomis link). - Michel Marcus, Mar 19 2014

For the first 64 terms, a(n) is approximately 1.56^n. - Jud McCranie, Jun 17 2017

These numbers were first studied in 1939 by the Spanish mathematician Laureano Pérez-Cacho Villaverde (1900-1957). The term "perfect totient number" was coined by Venkataraman (1975). - Amiram Eldar, Mar 10 2021

Examples

			327 is a perfect totient number because 327 = 216 + 72 + 24 + 8 + 4 + 2 + 1. Note that 216 = phi(327), 72 = phi(216), 24 = phi(72) and so on.
		

References

• Richard K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, 2004, Section B41, pp. 147-150.
• L. Pérez-Cacho, Sobre la suma de indicadores de órdenes sucesivos (in Spanish), Revista Matematica Hispano-Americana, Vol.5, No. 3 (1939), pp. 45-50.
• József Sándor and Borislav Crstici, Handbook of Number theory II, Kluwer Academic Publishers, 2004, Chapter 3, pp. 240-242.
• D. L. Silverman, Problem 1040, J. Recr. Math., Vol. 14 (1982); Solution by R. I. Hess, ibid., Vol. 15 (1983).

Links

• Jud McCranie, Table of n, a(n) for n = 1..64 (first 51 terms from Robert G. Wilson v)
• Jovele G. Belmonte, On perfect totient numbers, Masteral Thesis, De La Salle University, 2006.
• Li-Xia Dai and Yong-Gao Chen, A note on perfect totient numbers, Journal of Northeast Normal University, Vol. 39, No. 4 (2007), pp. 17-19.
• Tuukka Hyvärinen, Täydelliset totienttiluvut (in Finnish), Master's thesis, Tampere University, 2015; alternative link.
• Douglas E. Iannucci, Deng Moujie and Graeme L. Cohen, On Perfect Totient Numbers, Journal of Integer Sequences, Vol. 6 (2003), Article 03.4.5.
• Paul Loomis, Michael Plytage and John Polhill, Summing up the Euler phi function, The College Mathematics Journal, Vol. 39, No. 1, Jan. 2008.
• Florian Luca, On the Distribution of Perfect Totients, Journal of Integer Sequences, Vol. 9 (2006), Article 06.4.4.
• A. L. Mohan and D. Suryanarayana, Perfect totient numbers, in: K. Alladi (ed.), Number Theory, Proceedings of the Third Matscience Conference Held at Mysore, India, June 3-6, 1981, Lecture Notes in Mathematics, Vol 938, Springer, Berlin, Heidelber, 1982, pp. 101-105.
• Deng Moujie, A Note On Perfect Totient Numbers, Journal of Integer Sequences, Vol. 12 (2009), Article 09.6.2.
• Igor E. Shparlinski, On the sum of iterations of the Euler function, Journal of Integer Sequences, Vol. 9 (2006), Article 06.1.6.
• Hans Sieburg and Michael Kentgens, On Phi-perfect numbers, in: J. Akiyama et al. (eds.), Number Theory and Combinatorics, Japan 1984, World Scientific, 1985, pp. 245-254.
• M. V. Subbarao, On a Function connected with phi(n), The Mathematics Student, Vol. 23 (1955), pp. 178-179; entire volume.
• T. Venkataraman, Perfect totient number, The Mathematics Student, Vol. 43, No. 2 (1975), p. 178; entire issue.
• Wikipedia, Perfect totient number.

Crossrefs

Cf. A092693 (sum of iterated phi(n)). See also A091847.

Cf. A051179, A125734.

Programs

Maple

with(numtheory):
A082897_list := proc(N) local k,p,n,L;
L := NULL;
for n from 3 by 2 to N do
k := 0; p := phi(n);
while 1 < p do k := k + p; p := phi(p) od;
if k + 1 = n then L := L,n fi
od; L end: # Peter Luschny, Nov 01 2010

Mathematica

kMax = 57395631; a = Table[0, {kMax}]; PTNs = {}; Do[e = EulerPhi[k]; a[[k]] = e + a[[e]]; If[k == a[[k]], AppendTo[PTNs, k]], {k, 2, kMax}]; PTNs
perfTotQ[n_] := Plus @@ FixedPointList[ EulerPhi@ # &, n] == 2n + 1; Select[Range[1000], perfTotQ] (* Robert G. Wilson v, Nov 06 2010 *)

PARI

S(n)={n=eulerphi(n);if(n==1,1,n+S(n))}
for(n=2,1e3,if(S(n)==n,print1(n", "))) \\ Charles R Greathouse IV, Mar 29 2012; Corrected by Dana Jacobsen, Dec 16 2018

Perl

use ntheory "euler_phi"; sub S { my $n=euler_phi(shift); return 1 if $n == 1; $n+S($n); }   for (2..1e4) { say if $==S($); } # Dana Jacobsen, Dec 16 2018

Python

from itertools import count, islice
from gmpy2 import digits
from sympy import totient
def A082897_gen(startvalue=3): # generator of terms >= startvalue
    for n in count((k:=max(startvalue,3))+1-(k&1),2):
        t = digits(n,3)
        if t.count('0') == len(t)-1:
            yield n
        else:
            m, s = n, 1
            while (m:=totient(m))>1:
                s += m
            if s == n:
                yield n
A082897_list = list(islice(A082897_gen(),20)) # Chai Wah Wu, Mar 24 2023

Formula

n is a perfect totient number if S(n) = n, where S(n) = phi(n) + phi^2(n) + ... + 1, where phi is Euler's totient function and phi^2(n) = phi(phi(n)), ..., phi^k(n) = phi(phi^(k-1)(n)).

n such that n = A092693(n).

n such that 2n = A053478(n). - Vladeta Jovovic, Jul 02 2004

n log log log log n << a(n) <= 3^n. - Charles R Greathouse IV, Mar 22 2012

Extensions

Corrected by T. D. Noe, Mar 11 2004