A125734 -id:A125734 - cp's OEIS frontend

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

Douglas E. Iannucci, Jul 21 2003

nonn

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

			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.

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).

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.

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

Cf. A051179, A125734.

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

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 *)

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

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

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

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

Corrected by T. D. Noe, Mar 11 2004

A005537 Numbers m such that 4*3^m + 1 is prime.

Original entry on oeis.org

0, 1, 2, 3, 6, 14, 15, 39, 201, 249, 885, 1005, 1254, 1635, 3306, 3522, 9602, 19785, 72698, 233583, 328689, 537918, 887535, 980925, 1154598, 1499606, 1936890, 2016951, 2143374

Offset: 1

N. J. A. Sloane, Chris K. Caldwell

a(27) > 1.5*10^6. - Matthias Baur, Jan 16 2020
a(20) > 2*10^5. - Robert Price, Nov 23 2013
Primes resulting from a(1)-a(19) are confirmed primes (not probable primes) using BLS (N-1/N+1) test in pfgw. - Robert Price, Nov 23 2013
From Matthias Baur, Jan 16 2020: (Start)
Double checked to n=2*10^5, tested further to n=1.5*10^6 using the sieve programs newpgen and srsieve and using Jean Penné's LLR application (BLS (N-1/N+1) test).
a(20) was already known in 2005, but was not listed here until 2018 (see Prime Pages link). (End)
Because of the factorization 4*x^4 + 1 = (2*x^2 - 2*x + 1)*(2*x^2 + 2*x + 1), the only term divisible by 4 is 0. - Jeppe Stig Nielsen, Sep 12 2024

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

C. K. Caldwell, The Prime Pages
Douglas E. Iannucci, Deng Moujie and Graeme L. Cohen, On Perfect Totient Numbers, J. Integer Sequences, 6 (2003), #03.4.5.
P. Loomis, M. Plytage and J. Polhill, Summing up the Euler 'phi' function, The College Mathematics Journal, vol. 39 (2008), pp. 34-42.
H. C. Williams and C. R. Zarnke, Some prime numbers of the forms 2*3^n+1 and 2*3^n-1, Math. Comp., 26 (1972), 995-998.

Cf. A005540, A125734.

a[n_]:=If[PrimeQ[4*3^n + 1 ], n]; DeleteCases[Array[a, 40, 0], Null] (* Stefano Spezia, Nov 12 2018 *)

is_a(m) = isprime(4*3^m + 1) \\ Michel Marcus, Jul 12 2013

a(15)-a(17) from Douglas Burke (dburke(AT)nevada.edu)
a(18) from Mohammed Bouayoun (Mohammed.Bouayoun(AT)sanef.com), Jan 26 2004
a(19) from Robert Price, Nov 23 2013
a(20)-a(21) from Matthias Baur, Nov 07 2018
a(22) from Matthias Baur, Dec 06 2018
a(23)-a(24) from Matthias Baur, Jul 23 2019
a(25) from Matthias Baur, Dec 07 2019
a(26) from Matthias Baur, Jan 16 2020
a(27)-a(29) from Ryan Propper, May 08 2020

cp's OEIS Frontend

A082897 Perfect totient numbers.

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