Meanwhile I know Mensa since 1985. At that time I have got the german version of the riddle book "A Mensa Puzzle Book or Problems, Posers, Puzzles and Pastimes for the Superintelligent" by Victor Serebriakoff. Otherwise I was not very interested in Mensa at first. Later, in the year 1987, I also got the german version of the second riddle book "A Second Mensa Quzzle Book" by Victor Serebriakoff, but at this point of time, as well, Mensa itself stayed rather less fascinating for me. After all, during a school trip to Frankfurt in 1989 I also have bought the book "The Mensa Genius Quiz Book", yet. During my time with the german army and the beginning of the years of study Mensa almost completely fell into oblivion of me.

In principle I just again got in contakt with Mensa due to a television show of Günther Jauch in the year 2001. Then went one time to a mensa's regulars' table in Düsseldorf and have found there a very enjoyable and funny "bunch" of people. Even for a longer time it was clear in my mind, that I am carrying around a little bit of intelligence with me, but now I finally wanted to know what's up. A three hour lasting group test of Mensa at a day, that actually was very much to warm (to the end the focussing suffered), showed, that it was good enough for a membership, which I appied after that.

The regulars' table in Düsseldorf at present always takes place every 18. from about 19:30 o'clock in the Abraxas at the Merowinger Str. 16.
Besides there normally is a Trivial Pursuit group every last friday from about 19:30 o'clock in the China-Garten at the Unterrather Str. 99.

general informations about Mensa in Germany.

Websites of Mensa:

Mensa International

Mensa Deutschland

Mensa Düsseldorf

This nice caricature aready caused a long row of discussions about elegant as possible solutions. It is not even implicitly trivial to find easy solutions for this small problem, at all.
The best two proposals, that I know at present, are 22 and 16.

22							to a general examination of solution in the form F(n)=F(n-1)*(a+n*b)+(c+n*d) with a given F(0) as default
4
4	=	4	x	2	-	4
7	=	4	x	3	-	5
22	=	7	x	4	-	6
103	=	22	x	5	-	7

16							to a general examination of solution in the form F(2n+1)=F(2n)*a+b ; F(2n)=F(2n-1)*c+d with a given F(0) as default
4
4	=	4	x	4	-	12
7	=	4	x	8	-	25
16	=	7	x	4	-	12
103	=	16	x	8	-	25

Furthermore you obviously can also find solutions in form of a polynom [F(n)=a+b*n+c*n²+d*n³+e*n⁴] or with structures similar to Fibonacci-rows [F(n)=a*(F(n-2)+F(n-1))+b+n*c ; F(n)=a*F(n-2)+b*F(n-1)+c] with given F(0) and F(1) as default.

Likewise interesting it additionally is, that there exist four special values for X, which deliver whole-numbered continuations in each five alternatives I examined here:
10, (+3) 13, (+6) 19, (+12) 31.

Who knows further possibilities of solutions? Please send it to me.

In case of questions or suggestions just write an Email to me.