A091025 Smallest positive k such that phi(1+k*2^m) <= phi(k*2^m) for all m = n (mod 12), where phi is Euler's totient function.

104, 52, 26, 13, 59, 82, 41, 73, 89, 97, 101, 103

Offset

0,1

Comments

This sequence shows why A090849 has only a small number of distinct terms. It can be shown that 1+a(n) 2^m has factors of 3, 5 and 7 for all m = n (mod 12). Using that fact and the fact that (1-1/3)*(1-1/5)*(1-1/7) < 1/2, it is easy to verify that phi(1+k*2^m) <= phi(k*2^m) for all m = n (mod 12). Note that each successive term can be obtained by dividing by 2 (mod 105).

Links

Table of n, a(n) for n=0..11.

Greg Martin, The smallest solution of phi(30n+1) < phi(30n) is ..., arXiv:math/9804025 [math.NT], 1998; Amer. Math. Monthly, Vol. 106, No. 5 (1999), pp. 449-451.

D. J. Newman, Euler's phi function on arithmetic progressions, Amer. Math. Monthly, Vol. 104, No. 3 (Mar. 1997), pp. 256-257.

Herman te Riele, On the size of solutions of the inequality phi(ax+b) < phi(ax)

Mathematica

Table[k=1; While[Mod[2^n k, 105] != 104, k++ ]; k, {n, 0, 11}]

Crossrefs

Cf. A090849 (least k such that phi(1+k*2^n) <= phi(k*2^n)).

Sequence in context: A097014 A106297 A090849 * A054904 A117845 A286812

Adjacent sequences: A091022 A091023 A091024 * A091026 A091027 A091028

Keyword

nonn,fini,full

Author

T. D. Noe, Dec 15 2003

Extensions

Replaced arXiv URL by non-cached version - R. J. Mathar, Oct 30 2009

Status

approved