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A281747
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Smallest b > 1 such that p = prime(n) satisfies b^(p-1) == 1 (mod p^p).
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1
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5, 26, 1068, 82681, 5392282366, 11356596271444, 34451905517028761171, 340625514346676110671584, 308318432223607315018221180590, 8566187045843934976180705488213013173127, 1099862052702774330481800364074681495062836757, 8170421001593885871548404108552563632485969048059688187
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OFFSET
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1,1
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COMMENTS
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a(n) is the element in row prime(n), column n of the table in A257833.
Is the sequence always nondecreasing, or stronger, is it always increasing?
For odd primes p, if c is a primitive root mod p^p then b == c^(p^(p-1)) (mod p^p) satisfies this. Thus a(n) < prime(n)^prime(n) for n > 1. - Robert Israel, Jan 30 2017
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LINKS
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MAPLE
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f:= proc(p) local c, j;
c:= numtheory:-primroot(p^p);
min(seq(c &^ (j*p^(p-1)) mod p^p, j=1..p-2))
end proc:
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MATHEMATICA
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Table[b = 2; While[PowerMod[b, (# - 1), #^#] &@ Prime@ n != 1, b++]; b, {n, 4}] (* Michael De Vlieger, Jan 30 2017 *)
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PROG
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(PARI) a(n) = my(p=prime(n), b=2); while(Mod(b, p^p)^(p-1)!=1, b++); b
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CROSSREFS
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KEYWORD
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nonn,hard,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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