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a(0) = 1; a(n) is obtained by incrementing each digit of a(n-1) by 5.
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%I #5 Dec 05 2013 19:54:51

%S 1,6,11,66,1111,6666,11111111,66666666,1111111111111111,

%T 6666666666666666,11111111111111111111111111111111,

%U 66666666666666666666666666666666

%N a(0) = 1; a(n) is obtained by incrementing each digit of a(n-1) by 5.

%C In A061511-A061522, A061746-A061750 when the incremented digit exceeds 9 it is written as a 2-digit string. So 9+1 becomes the 2-digit string 10, etc.

%C Number of digits of each term is the sequence A016116. [From _Dmitry Kamenetsky_, Jan 17 2009]

%F a(2n) = 6*[10^{2^(n)} - 1]/9 a(2n+1) = [10^(2^n) - 1]/9

%K base,nonn

%O 0,2

%A _Amarnath Murthy_, May 08 2001

%E More terms from Larry Reeves (larryr(AT)acm.org), May 11 2001