OFFSET
1,1
COMMENTS
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
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
LINKS
Robert G. Wilson v and T. D. Noe, Table of n, a(n) for n=1..1000
FORMULA
a(n) = MIN{k in A001358 and n|(k-1)}.
MATHEMATICA
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 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Jonathan Vos Post, Sep 22 2007
STATUS
approved