%I A154714
%S A154714 3,4,8,2048
%N A154714 Let f(1,x)=x+1; f(n+1,x)=f(n,f(n,...f(n,x)...)), the formula contains 
               x applications of f. This sequence is a(n)=f(n,2).
%C A154714 The next term is too large to include.
%F A154714 f(1,x)=x+1; f(n+1,x)=f(n,f(n,...f(n,x)...)), the formula contains x applications 
               of f; a(n)=f(n,2)
%F A154714 f(2,x)=2.x; f(3,x)=2^x.x; f(4,x)=product(b(k),k=0..x) where b(0)=x, b(m+1)=2^product(b(l),
               l=0..m). - Benoit Jubin, Jan 15 2009
%e A154714 a(1)=f(1,2)=1+2=3; a(2)=f(2,2)=f(1,f(1,2))=f(1,3)=3+1=4; a(3)=f(3,2)=f(2,
               f(2,2))=f(2,4)=f(1,f(1,f(1,f(1,4))=4+1+1+1+1=8; a(4)=f(4,2)=f(3,f(3,
               2))=f(3,8)=f(2,f(2,f(2,f(2,f(2,f(2,f(2,f(2,8))))))))=...=2048;
%e A154714 a(5) is obtained by applying the operation f(3,.) 2048 times to 2048, 
               where f(3, 2^p) = 2^( 2^p+p ). Thus a(5) is larger than 2^(2^(....2^(2^11+11)...)), 
               with 2049(?) occurrences of "2^" = M. F. Hasler, Jan 15 2009
%t A154714 f[1, x_] := x + 1; f[n_, x_] := Nest[f[n - 1, # ]&, x, x]; Table[f[n, 
               2], {n, 1, 4}]
%Y A154714 Sequence in context: A119529 A096847 A011993 this_sequence A001695 A019676 
               A019900
%Y A154714 Adjacent sequences: A154711 A154712 A154713 this_sequence A154715 A154716 
               A154717
%K A154714 nonn
%O A154714 1,1
%A A154714 Vladimir Reshetnikov (v.reshetnikov(AT)gmail.com), Jan 14 2009