1,1

The next term is too large to include.

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(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

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;

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

f[1, x_] := x + 1; f[n_, x_] := Nest[f[n - 1, # ]&, x, x]; Table[f[n, 2], {n, 1, 4}]

Sequence in context: A119529 A096847 A011993 this_sequence A001695 A019676 A019900

Adjacent sequences: A154711 A154712 A154713 this_sequence A154715 A154716 A154717

nonn

Vladimir Reshetnikov (v.reshetnikov(AT)gmail.com), Jan 14 2009