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A263429
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Smallest prime p such that binomial(2*p-1, p-1) == 1 (mod p^n), or 0 if no such p exists.
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1
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OFFSET
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1,1
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COMMENTS
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For n > 1, smallest p = prime(i) such that A244919(i) = n.
Conjecture: a(n) = 0 for n > 4 (McIntosh, 1995, p. 387).
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LINKS
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PROG
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(PARI) a(n) = my(p=2); while(Mod(binomial(2*p-1, p-1), p^n)!=1, p=nextprime(p+1)); p
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CROSSREFS
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KEYWORD
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nonn,hard,more,bref
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AUTHOR
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STATUS
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approved
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