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%I #28 May 23 2015 11:47:39
%S 1,1,1,2,1,1,1,1,2,1,2,1,2,1,1,2,1,1,2,1,2,1,1,2,1,2,1,2,3,2,2,2,1,1,
%T 2,1,2,1,1,1,2,2,1,2,2,2,2,1,2,2,1,2,1,2,2,2,1,1,2,2,2,2,1,2,1,2,1,2,
%U 2,2,2,2,2,2,1,1,2,2,1,2,2,1,2,2,1,2,1
%N Largest k such that p = prime(n) satisfies b^(p-1) == 1 (mod p^k) for some base b with 1 < b < p.
%C a(n) > 1 iff p is in A134307.
%C Meyer proved in 1902 that for any prime p exactly p - 1 bases b with b < p^k exist such that b^(p-1) == 1 (mod p^k) (cf. Keller, Richstein, 2005, page 930).
%C a(30) = 3 is the first term with a value > 2, corresponding to prime(30) = 113 (see the comment from 2011 in A134307). This is the first case where A249275(n) < A000040(n).
%C Do the values of this sequence have an upper bound or, more formally, does this sequence have a supremum?
%H Felix Fröhlich, <a href="/A254444/b254444.txt">Table of n, a(n) for n = 2..10000</a>
%H W. Keller and J. Richstein, <a href="http://dx.doi.org/10.1090/S0025-5718-04-01666-7">Solutions of the congruence a^p-1 == 1 (mod p^r)</a>, Math. Comp., 74 (2005), 927-936.
%e With p = 113: For all bases b with 1 < b < 113, p (trivially) satisfies b^112 == 1 (mod 113^k) for k = 1 and for no k > 1, with the single exception of b = 68, where p satisfies the congruence for k = 3 (and hence for k = 1 and k = 2). Since 3 is the largest value of k for all 1 < b < 113, a(30) = 3.
%o (PARI) forprime(p=3, 400, k=1; maxk=0; for(b=2, p-1, while(Mod(b, p^k)^(p-1)==1, k++); if(k-1 > maxk, maxk=k-1)); print1(maxk, ", "))
%Y Cf. A134307.
%K nonn
%O 2,4
%A _Felix Fröhlich_, May 04 2015
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