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Totient abundant numbers: numbers k such that A092693(k) > k.
7

%I #19 Jul 17 2021 23:31:35

%S 5,7,11,13,17,19,23,25,29,31,33,35,37,41,43,47,49,51,53,55,59,61,65,

%T 67,69,71,73,77,79,83,85,87,89,91,95,97,101,103,107,109,113,115,119,

%U 121,123,125,127,131,133,137,139,141,143,145,149,151,153,155,157

%N Totient abundant numbers: numbers k such that A092693(k) > k.

%H Amiram Eldar, <a href="/A286265/b286265.txt">Table of n, a(n) for n = 1..10000</a>

%H Paul Loomis and Florian Luca, <a href="https://www.emis.de/journals/INTEGERS/papers/i6/i6.Abstract.html">On totient abundant numbers</a>, Electronic Journal of Combinatorial Number Theory, Vol. 8, #A06 (2008).

%e 19 is a totient abundant number since A092693(19) = phi(19) + phi(phi(19)) + ... = 18 + 6 + 2 + 1 = 27 > 19.

%t totAbundantQ[n_] := Plus @@ FixedPointList[ EulerPhi@ # &, n] > 2*n+1; Select[Range[1000],totAbundantQ]

%o (Python)

%o from sympy import totient

%o def a092693(n): return 0 if n==1 else totient(n) + a092693(totient(n))

%o print([n for n in range(1, 201) if a092693(n) > n]) # _Indranil Ghosh_, May 05 2017

%Y Cf. A000010, A082897, A092693.

%K nonn

%O 1,1

%A _Amiram Eldar_, May 05 2017