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A131016
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Smallest semiprime == 1 (mod n).
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1
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4, 9, 4, 9, 6, 25, 15, 9, 10, 21, 34, 25, 14, 15, 46, 33, 35, 55, 39, 21, 22, 111, 93, 25, 26, 183, 55, 57, 146, 91, 94, 33, 34, 35, 106, 145, 38, 39, 118, 121, 206, 85, 87, 133, 46, 93, 95, 49, 295, 51, 205, 209, 213, 55, 111, 57, 58, 291, 119, 121, 62, 187, 253, 65, 326
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OFFSET
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1,1
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COMMENTS
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It is easy to prove that the first N>1 terms contain every semiprime < N. Assume the opposite: there is some semiprime pq that does not appear in the first pq-1 terms. This implies that n does not divide pq-1 for any n<pq. However, this is false because we can take n=pq-1. We see this behavior in the sequence quite often: a(5)=6, a(9)=10, a(13)=14, etc. - T. D. Noe, Sep 26 2007
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LINKS
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FORMULA
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a(n) = MIN{k in A001358 and n|(k-1)}.
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MATHEMATICA
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semiPrimeQ[x_] := Plus @@ Last /@ FactorInteger@ x == 2; sp = Select[ Range@ 346, semiPrimeQ@ # &]; f[1] = sp[[1]]; f[n_] := Block[{k = 1}, While[ Mod[ sp[[k]], n] != 1, k++ ]; sp[[k]] ]; Array[f, 65] (* Robert G. Wilson v *)
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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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