|
%I #22 Sep 08 2022 08:45:32
%S 1,11,1111,11111111,1111111111111111,11111111111111111111111111111111,
%T 1111111111111111111111111111111111111111111111111111111111111111
%N a(n) = (10^2^n - 1)/9.
%C More generally, reading in base B >= 2: a(n) = (B^2^n - 1)/(B-1).
%C Recurrence: a(n) = a(n-1)*(B^K + 1) and a(0)=1 where K = floor(log_B a(n-1)) + 1.
%C B = 2 gives A051179; B = 3 gives A059918.
%F a(n) = a(n-1)*(10^K + 1) and a(0)=1 where K=floor(log_10 a(n-1)) + 1 = 2^n + 1.
%F a(n) = A000042(A000079(n)) = A007088(A051179(n)) = A007089(A059918(n)).
%t (10^2^Range[0, 10] - 1)/9 (* _G. C. Greubel_, Apr 19 2021 *)
%o (Magma) A136308 := func<n|(10^2^n-1)/9>; [A136308(n):n in[0..7]];
%o (Sage) [(10^2^n -1)/9 for n in (0..10)] # _G. C. Greubel_, Apr 19 2021
%Y Cf. A000042 (repunits).
%Y In other bases: A051179, A059918.
%K easy,nonn
%O 0,2
%A _Ctibor O. Zizka_, Mar 22 2008
%E Edited by _Jason Kimberley_, Dec 18 2012
| 