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

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

Author

T. D. Noe, Aug 14 2008

Keywords

• nonn

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

• T. D. Noe, Table of n, a(n) for n=1..10000
• T. D. Noe, Numbers Like 16842752
• 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).