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Search: id:A082897
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| A082897 |
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Perfect totient numbers. |
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+0 6
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| 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
(list; graph; listen)
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OFFSET
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1,1
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COMMENT
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It is trivial that perfect totient numbers must be odd. It is easy to show that powers of 3 are perfect totient numbers.
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REFERENCES
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L. Perez Cacho, "Sobre la suma de indicadores de ordenes sucesivos", Revista Matematica Hispano-Americana, 5.3 (1939), 45-50.
A. L. Mohan and D. Suryanarayana, "Perfect totient numbers", in: Number Theory (Proc. Third Matscience Conf., Mysore, 1981) Lecture Notes in Math. 938 (Springer-Verlag, New York, 1982) pp. 101-105.
Igor E. Shparlinski, On the Sum of Iterations of the Euler Function, Journal of Integer Sequences, Vol. 9 (2006), Article 06.1.6.
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LINKS
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Douglas E. Iannucci, Deng Moujie and Graeme L. Cohen, On Perfect Totient Numbers, J. Integer Sequences, 6 (2003), #03.4.5.
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FORMULA
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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 2*n = A053478(n). - Vladeta Jovovic (vladeta(AT)Eunet.yu), Jul 02 2004
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EXAMPLE
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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.
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MATHEMATICA
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kMax=57395631; a=Table[0, {kMax}]; lst={}; Do[e=EulerPhi[k]; a[[k]]=e+a[[e]]; If[k==a[[k]], AppendTo[lst, k]], {k, 2, kMax}]; lst
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CROSSREFS
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Cf. A092693 (sum of iterated phi(n)). See also A091847.
Adjacent sequences: A082894 A082895 A082896 this_sequence A082898 A082899 A082900
Sequence in context: A055927 A087031 A089632 this_sequence A131822 A131801 A122819
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KEYWORD
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nonn
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AUTHOR
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Douglas E. Iannucci (diannuc(AT)uvi.edu), Jul 21 2003
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EXTENSIONS
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Corrected by T. D. Noe (noe(AT)sspectra.com), Mar 11 2004
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