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%I #9 Oct 31 2013 12:17:43
%S 4,9,4,9,6,25,15,9,10,21,34,25,14,15,46,33,35,55,39,21,22,111,93,25,
%T 26,183,55,57,146,91,94,33,34,35,106,145,38,39,118,121,206,85,87,133,
%U 46,93,95,49,295,51,205,209,213,55,111,57,58,291,119,121,62,187,253,65,326
%N Smallest semiprime == 1 (mod n).
%C This is to semiprimes A001358 as A034694 is to primes A000040. Conjecture: every semiprime occurs in this sequence. a(21)-a(65) from _Robert G. Wilson v_, Sep 21 2007
%C 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
%H Robert G. Wilson v and T. D. Noe, <a href="/A131016/b131016.txt">Table of n, a(n) for n=1..1000</a>
%F a(n) = MIN{k in A001358 and n|(k-1)}.
%t 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_ *)
%Y Cf. A001358, A034694.
%Y Cf. A085710.
%K nonn
%O 1,1
%A _Jonathan Vos Post_, Sep 22 2007