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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
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, 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, 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
OFFSET
1,4
COMMENTS
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.
a(2698482) = 0. That is, the 2698482nd term of A002202 is 16842752. - T. D. Noe, Aug 19 2008
REFERENCES
R. K. Guy, Unsolved problems in number theory, B39.
LINKS
William P. Wardlaw, L. L. Foster and R. J. Simpson, Problem E3361, Amer. Math. Monthly, Vol. 98, No. 5 (May, 1991), 443-444.
FORMULA
A058277(n) = A143421(n) + A143422(n).
CROSSREFS
Sequence in context: A025893 A296977 A025878 * A137672 A141272 A281527
KEYWORD
nonn
AUTHOR
T. D. Noe, Aug 14 2008
STATUS
approved