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%I #17 Aug 20 2017 10:55:05
%S 0,1,1,0,1,0,0,1,1,1,1,0,0,0,1,0,1,1,1,0,0,0,0,1,0,0,1,0,1,0,1,0,1,1,
%T 0,1,1,0,1,1,0,0,1,1,1,1,0,1,1,1,1,1,1,1,0,0,0,0,0,1,0,0,0,1,1,1,0,0,
%U 1,0,0,1,0,0,1,1,0,1,0,1,0,0,0,1,1,1,0,1,0,1,0,0,0,0,0,0,1,0,0,0,0,1,1,1,0
%N a(0)=0. Concatenate onto the end of the sequence (from left to right) the integer m_n converted into binary (with the most significant digit on the left), where m_n is the smallest integer > A118248(n-1) and whose binary representation does not occur anywhere earlier in the sequence (when the concatenated sequence is read from left to right). A118248(n) then equals m_n when written in decimal.
%C Sequence can be regarded as an irregular number triangle containing the binary digits of A118248(n) in row n. - _Michael De Vlieger_, Aug 19 2017
%H Michael De Vlieger, <a href="/A118247/b118247.txt">Table of n, a(n) for n = 0..11419</a> (Concatenation of A118248(0) through A118248(1000) inclusive).
%e The sequence begins 0,1,1,0,1,0,0,1,1,1,1,0,0,0. Now A118248(5) = 8, which is represented by the 1,0,0,0 at the end of the sequence. The binary representation of 9 (1001 in binary) and (decimal) 10 (1010 in binary) both occur earlier in the sequence. But the binary representation of (decimal) 11 (1011 in binary) does not occur earlier in the sequence, so (1,0,1,1) is added to the end of the sequence. And A118248(6) becomes 11.
%e From _Michael De Vlieger_, Aug 19 2017: (Start)
%e n A118248(n) Binary digits appended to a(n)
%e 0 0 0
%e 1 1 1
%e 2 2 10
%e 3 4 100
%e 4 7 111
%e 5 8 1000
%e 6 11 1011
%e 7 16 10000
%e 8 18 10010
%e 9 21 10101
%e 10 22 10110
%e 11 25 11001
%e 12 29 11101
%e (End)
%t Block[{a = {{0}}, b = {0}, k, d}, Do[k = FromDigits[#, 2] &@ Last@ a + 1; While[SequenceCount[Flatten@ a, Set[d, IntegerDigits[k, 2]]] > 0, k++]; AppendTo[a, d]; AppendTo[b, k], {i, 22}]; Flatten@ a] (* _Michael De Vlieger_, Aug 19 2017 *)
%Y Cf. A118248, A118249, A118251.
%K easy,nonn,base,tabf
%O 0,1
%A _Leroy Quet_, Apr 18 2006
%E More terms from _Joshua Zucker_, Jul 27 2006
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