0,1

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).

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

Greg Martin, The smallest solution of phi(30n+1) < phi(30n) is ..., Amer. Math. Monthly, Vol. 106, No. 5 (1999), pp. 449-451.

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

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

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

Sequence in context: A097014 A106297 A090849 this_sequence A054904 A117845 A045208

Adjacent sequences: A091022 A091023 A091024 this_sequence A091026 A091027 A091028

fini,full,nonn

T. D. Noe (noe(AT)sspectra.com), Dec 15 2003