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A143421 Number of odd numbers k such that phi(k) = n, where n runs through the values (A002202) taken by phi. 2

%I #12 Apr 23 2016 09:50:43

%S 1,1,1,2,1,1,2,1,2,2,1,3,1,1,1,3,3,2,1,1,2,1,1,1,1,4,1,1,1,6,1,2,1,2,

%T 2,1,4,2,1,1,1,4,1,2,1,6,1,1,1,3,1,1,1,5,1,1,3,2,2,1,1,4,1,2,1,5,1,1,

%U 4,1,1,3,1,1,1,1,7,2,1,2,1,1,2,1,10,1,4,1,1,1,3,1,1,2,4,3,1,6,1,1,1,2,1,1,6

%N Number of odd numbers k such that phi(k) = n, where n runs through the values (A002202) taken by phi.

%C The first zero term is for n = 16842752 = 257*2^16. If there are only five Fermat primes, then terms will be zero for n=2^r for all r>31. This is discussed in problem E3361.

%C a(2698482) = 0. That is, the 2698482nd term of A002202 is 16842752. - _T. D. Noe_, Aug 19 2008

%D R. K. Guy, Unsolved problems in number theory, B39.

%H T. D. Noe, <a href="/A143421/b143421.txt">Table of n, a(n) for n=1..10000</a>

%H T. D. Noe, <a href="http://www.sspectra.com/math/16842752.txt">Numbers Like 16842752</a>

%H William P. Wardlaw, L. L. Foster and R. J. Simpson, <a href="http://www.jstor.org/stable/2323869">Problem E3361</a>, Amer. Math. Monthly, Vol. 98, No. 5 (May, 1991), 443-444.

%F A058277(n) = A143421(n) + A143422(n).

%K nonn

%O 1,4

%A _T. D. Noe_, Aug 14 2008

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