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A253234
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Smallest prime p such that p^2 divides n^(p-1)-(n-1)^(p-1).
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1
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OFFSET
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2,1
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COMMENTS
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According to the generalized Fermat's little theorem, if a prime p does not divide n or n-1, then p must divide n^(p-1)-(n-1)^(p-1), so we can ask that whether there exists a Wieferich-type prime to the "base" (n, n-1), that is, a prime p such that p^2 divides n^(p-1)-(n-1)^(p-1).
a(n) is currently unknown for n = {11, 36, 49, 52, 66, 70, 84, 89, 102, 112, 133, 142, 148, ...}
a(11) > 6.5*10^10.
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LINKS
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FORMULA
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a(9k+5) = 3.
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MATHEMATICA
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f[n_] := Block[{k = 1}, While[k < 10^8 && PowerMod[n, Prime[k] - 1, Prime[k]^2] != PowerMod[n - 1, Prime[k] - 1, Prime[k]^2], k++ ]; If[k == 10^8, 0, Prime[k]]]; Table[ f[n], {n, 2, 75}]
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PROG
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(PARI) for(n=2, 10, forprime(p=2, 1e8, if(Mod(n, p^2)^(p-1)==Mod(n-1, p^2)^(p-1), print1(p, ", "); next({2}))); print1("--, "))
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CROSSREFS
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KEYWORD
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nonn,more,hard
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AUTHOR
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STATUS
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approved
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