1,2

More generally, a(n)=a(n-1)*(B^K +1) ;a(0)=1;B integer; K=floor(log_B a(n-1)) + 1.

B=2 gives A051179; B=3 gives A059918

a(n)=[B^2^n]-1/(B-1)

a(n)=a(n-1)*(10^K +1) ;a(0)=1; K=floor(log_10 a(n-1)) + 1.

Cf. A051179, A059918.

Sequence in context: A130602 A099814 A068053 this_sequence A075600 A078568 A015095

Adjacent sequences: A136305 A136306 A136307 this_sequence A136309 A136310 A136311

easy,nonn

Ctibor O. ZIZKA (ctibor.zizka(AT)seznam.cz), Mar 22 2008