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A180871
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Index of term in Sylvester's sequence A000058 divisible by prime A007996(n).
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7
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0, 1, 2, 4, 3, 11, 4, 9, 6, 6, 6, 29, 64, 42, 9, 59, 10, 80, 39, 103, 140, 41, 137, 53, 69, 146, 104, 14, 92, 15, 117, 199, 75, 98, 316, 233, 28, 92, 281, 44, 136, 26, 258, 7, 38, 6, 176, 126, 74, 59, 89, 61, 45, 79, 13, 448, 119, 180, 290, 184, 348, 502, 508, 161, 7, 265, 229
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OFFSET
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1,3
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COMMENTS
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Because all terms of Sylvester's sequence are coprime to each other, each prime in A007996 divides only one term of A000058. The Mathematica program computes both the primes in A007996 and the terms in this sequence. Using modular arithmetic, it is easy to see that if prime p divides A000058(k) for some k, then we must have k < p. In practice, k < 5*sqrt(p).
An open problem is to prove that all terms of Sylvester's sequence are squarefree or to find a counterexample. Using the p from A007996 and k found here, it is simple to determine whether A000058(k) = 0 (mod p^2). No p < 10^10 was found to have this property.
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LINKS
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FORMULA
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EXAMPLE
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MATHEMATICA
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t={}; p=1; While[Length[t]<100, p=NextPrime[p]; s=Mod[2, p]; k=0; modSet={}; While[s>0 && !MemberQ[modSet, s], AppendTo[modSet, s]; k++; s=Mod[s^2-s+1, p]]; If[s==0, AppendTo[t, {p, k}]]]; Transpose[t][[2]]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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