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A252358
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Smallest prime p such that (2^n)^(p-1) == 1 (mod p^2), i.e., smallest Wieferich prime to base 2^n.
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1
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2, 1093, 1093, 3, 1093, 5, 3, 7, 1093, 3, 5, 11, 3, 13, 7, 3, 1093, 17, 3, 19, 5, 3, 11, 23, 3, 5, 13, 3, 7, 29, 3, 31, 1093, 3, 17, 5, 3, 37, 19, 3, 5, 41, 3, 43, 11, 3, 23, 47, 3, 7, 5, 3, 13, 53, 3, 5, 7, 3, 29, 59, 3, 61, 31, 3, 1093, 5, 3, 67, 17, 3, 5
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OFFSET
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0,1
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COMMENTS
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This sequence is bounded above by 1093, i.e., a(n) <= 1093 for any n, since any Wieferich prime to base 2 is also Wieferich to any base that is a power of two.
Do all primes <= 1093 appear in this sequence?
The answer is yes. The smallest n such that a(n) = prime(n) is prime(n). The only exceptions are p = 2 and p = 1093 which first occur at n = 0 and n = 1, respectively.
Apparently, a(n) = 1093 if n is a power of 2 when n < 1093 and a(n) = 1093 if n is prime or a power of 2 when n > 1093. (End)
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LINKS
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MATHEMATICA
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Block[{k}, Table[k = 1; While[PowerMod[2, n (# - 1), #^2] != 1 &@ Prime@ k, k++]; Prime@ k, {n, 0, 70}] ] (* Michael De Vlieger, Dec 10 2020 *)
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PROG
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(PARI) a(n) = forprime(p=1, , if(Mod(2, p^2)^(n*(p-1))==1, return(p)))
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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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