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%I #7 Oct 01 2017 13:00:12
%S 104,52,26,13,59,67,41,73,89,97,101,103,74,37,26,13,17,67,41,73,89,82,
%T 41,103,104,52,26,13,29,67,41,73,74,37,101,103,104,52,26,13,59,67,41,
%U 73,89,67,86,43,104,52,26,13,59,37,41,73,89,97,101,103,104,52,26,13,59,67
%N Smallest positive k such that phi(1+k*2^n) <= phi(k*2^n), where phi is Euler's totient function.
%C Newman proves that k always exists for all n. Surprisingly, it appears that only 19 values of k suffice for all n. Note that a(n) = 26 when n = 2 (mod 12), a(n) = 13 when n = 3 (mod 12), a(n) = 41 when n = 6 (mod 12) and a(n) = 73 when n = 7 (mod 12). Is this sequence periodic?
%C A091025 shows why this sequence has only a finite number of distinct terms.
%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.
%t Table[k=1; While[EulerPhi[1+k*2^n] > EulerPhi[k*2^n], k++ ]; k, {n, 100}]
%Y Cf. A090851 (least k such that phi(2n*k+1) < phi(2n*k)).
%K nonn
%O 0,1
%A _T. D. Noe_, Dec 09 2003