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%I #9 Nov 21 2013 12:48:23
%S 1,3,5,7,9,11,13,15,17,19,12,12,14,15,16,18,18,21,20,24,22,27,24,30,
%T 26,33,28,36,30,39,33,32,35,35,37,38,39,41,41,44,43,47,45,50,47,53,49,
%U 56,51,59,54,52,56,55,58,58,60,61,62,64,64,67,66,70,68,73,70,76,72,79,75
%N Consider the sequence {b(n), n >= 1} of digits of the integers: 0 1 2 3 4 5 6 7 8 9 1 0 1 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 2 0... (A033307); a(n) = b(n) + n.
%C Add its rank to each digit of the counting numbers (beginning with 0).
%e The sequence of digits of the integers is
%e 0 1 2 3 4 5 6 7 8 9 1 0 1 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 2 0...
%e The 15th term, for instance, is a 1. Thus 1+15=16 is the 15th term of this sequence.
%e Next one is 2 (the digit) + 16 (the rank) = 18
%t dgs = Flatten[IntegerDigits/@Range[0, 50]]; Plus@@@Partition[Riffle[dgs, Range[Length[dgs]]], 2] (* _Harvey P. Dale_, Jan 15 2011 *)
%K base,nonn,easy
%O 1,2
%A _Alexandre Wajnberg_, Sep 30 2004
%E More terms from _Joshua Zucker_, May 18 2006