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a(0) = 0; a(n) is obtained by incrementing each digit of a(n-1) by 2.
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%I #9 Jun 24 2016 12:50:35

%S 0,2,4,6,8,10,32,54,76,98,1110,3332,5554,7776,9998,11111110,33333332,

%T 55555554,77777776,99999998,1111111111111110,3333333333333332,

%U 5555555555555554,7777777777777776,9999999999999998,11111111111111111111111111111110,33333333333333333333333333333332

%N a(0) = 0; a(n) is obtained by incrementing each digit of a(n-1) by 2.

%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 Every term > 8 is made up of only two different consecutive digits, the smaller of which occurs only as the least significant digit.

%C Otherwise said, these are one less than the odd repdigits (A010785) of length 2^k, cf. formula. - _M. F. Hasler_, Jun 24 2016

%F a(n) = A061512(n)-1 = (10^2^floor(n/5)-1)/9*(n%5*2+1) - 1, where n%5 means the remainder (in {0..4}) of n divided by 5. - _M. F. Hasler_, Jun 24 2016

%e Following 32; 3+2 = 5 and 2+2 = 4, hence the next term is 54.

%t NestList[FromDigits[Flatten[IntegerDigits/@(IntegerDigits[#]+2)]]&,0,30] (* _Harvey P. Dale_, Jul 07 2012 *)

%o (PARI) A061513(n)=10^2^(n\5)\9*(n%5*2+1)-1 \\ _M. F. Hasler_, Jun 24 2016

%Y Cf. A061511 - A061522, A061746 - A061750; A061581 - A061587.

%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