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A254444
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Largest k such that p = prime(n) satisfies b^(p-1) == 1 (mod p^k) for some base b with 1 < b < p.
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3
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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, 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, 2, 2, 2, 2, 2, 2, 1, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1
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OFFSET
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2,4
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COMMENTS
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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).
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).
Do the values of this sequence have an upper bound or, more formally, does this sequence have a supremum?
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LINKS
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EXAMPLE
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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.
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PROG
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(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, ", "))
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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