A091025 - OEIS

%I #15 Aug 06 2024 09:04:41

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

%N 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.

%C 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).

%H Greg Martin, <a href="http://arXiv.org/abs/math/9804025">The smallest solution of phi(30n+1) < phi(30n) is ...</a>, arXiv:math/9804025 [math.NT], 1998; Amer. Math. Monthly, Vol. 106, No. 5 (1999), pp. 449-451.

%H D. J. Newman, <a href="http://www.jstor.org/stable/2974791">Euler's phi function on arithmetic progressions</a>, Amer. Math. Monthly, Vol. 104, No. 3 (Mar. 1997), pp. 256-257.

%H Herman te Riele, <a href="https://citeseerx.ist.psu.edu/pdf/6f75bf159f876d71f4fcfbd1cdcba38dfc358c11">On the size of solutions of the inequality phi(ax+b) < phi(ax)</a>

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

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

%K nonn,fini,full

%O 0,1

%A _T. D. Noe_, Dec 15 2003