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• 
  [deleted]
  Nov 30, 2011

  Take the sequence of all prime numbers p greater or equal than 5.

  that is p_0 = 5 , p_1 = 7 , p_2 = 11 , p_3 = 13 ... and so on...

  Now consider ALL the possible combinations of the type p_i x p_j.

  Now arrange the couples (p_i, p_j) in order of increasing magnitude with respect to such product. Done.

  If you just computed "your terms" you will find, in fact, the following sequence, 25 , 35 , 49 , 55 , 65 , 77 , 85 , 91, 95 , 115 .. and so on and so forth.

  Dealing with primes, It is unlikely that a closed formula of the type "the n-th term of the series is equal something" exists.

  FOLLOW-UP (after RichardPhd comment) .

  I admit that I forgot to include the standard "diagonal inductive" argument (which turns out to be an effective, even if not really fast, algorithm) which is used to order the required set.

  1) Consider the standard (n,m) plane in N^2 ( i.e. take some standard rectangular sheet of graph paper, draw the positive quadrant of the cartesian coordinates, and put the natural numbers on the axes)

  2) Restrict yourself to the elements (n,m) such that n is less or equal to m. ( that is take the "diagonal" and the elements "above it" ).

  3)Take an empty set "S". Take the first diagonal element and compute its product ( i.e. 5x5 = 25 ) . Add the element (5,5) to your ordered set S.

  "Inductive loop"

  4) Take the smallest diagonal element not in "S" and compute "p_i x p_i". (i.e. begin with 7x7 ). Add it to your ordered set S "after" "p_{i-1} x p_{i - 1} ".

  5) Go back to to the previous diagonal element (and compute all the elements of the form "p__{i-1} x p_j" for all the "j" as long as such product is less than p_i x p_i and greater than p_{i-1} x p_{i-1}.
  Repeat by computing and choosing all the "p_{i-2} x ..." until you get to " 5 times .... " for which it holds "less than p_i x p_i and greater than p_{i-1} x p_{i-1}". Order all these couples by their magnitude and add them to your ordered set "between p_{i-1} times p_{i-1} and p_i times p_i.

  Repeat the two steps above.

  Note that there are faster ways (i.e. requiring a smaller number of operations) to order the set of couples.

  Last I suggest you take a look at http://mathworld.wolfram.com/Semiprime.html and references therein if you are interested in the subject.

  Cheers!

• 
  richardphd
  Nov 30, 2011

  Along the lines of the previous solution by giulietti ...

  The difficult part is the order of the results. The problem is easily stated in set theory if we discard the sorting of the results into ascending order.

  Let P be the set of prime mumbers. Let P5 be the set of elements p[i] of P which satisfy p[i] >= 5. Then the required set is P5 x P5, i.e. the set whose elements are generated by choosing two elements (not necessarily distinct) of P5 and multiplying them together.

  So we can write (let S be the required solution set):

  S = {s[i,j]: s[i,j] = p[i]*p[j] such that for all k, p[k] is prime and p[k] >= 5}

  Note: there is no problem with duplications caused by s[i,j] = s[j,i]; duplicates count only for a single set membership.

  As giulietti stated, it is unlikely that there will be a closed formula for the enumeration of this set.