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A254444 Largest k such that p = prime(n) satisfies b^(p-1) == 1 (mod p^k) for some base b with 1 < b < p. 3
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

2,4

COMMENTS

a(n) > 1 iff p is in A134307.

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?

LINKS

Felix Fröhlich, Table of n, a(n) for n = 2..10000

W. Keller and J. Richstein, Solutions of the congruence a^p-1 == 1 (mod p^r), Math. Comp., 74 (2005), 927-936.

EXAMPLE

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.

PROG

(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, ", "))

CROSSREFS

Cf. A134307.

Sequence in context: A279497 A138330 A128591 * A102005 A051700 A337006

Adjacent sequences:  A254441 A254442 A254443 * A254445 A254446 A254447

KEYWORD

nonn

AUTHOR

Felix Fröhlich, May 04 2015

STATUS

approved

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Last modified July 24 02:05 EDT 2021. Contains 346269 sequences. (Running on oeis4.)