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A261015 Irregular triangle read by rows: T(n,k) (0 <= k <= 2^n-1) = number of binary strings of length n such that the smallest number whose binary representation is not visible in the string is k.

%I A261015 #27 Feb 21 2024 08:19:45
%S A261015 1,1,1,1,1,1,1,1,2,3,1,0,0,0,1,1,3,6,4,1,0,0,0,0,0,0,0,0,0,0,1,1,4,11,
%T A261015 10,5,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,5,19,21,
%U A261015 15,0,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0
%N A261015 Irregular triangle read by rows: T(n,k) (0 <= k <= 2^n-1) = number of binary strings of length n such that the smallest number whose binary representation is not visible in the string is k.
%C A261015 Suggested by A260273.
%H A261015 Alois P. Heinz, <a href="/A261015/b261015.txt">Rows n = 1..15, flattened</a>
%e A261015 Triangle begins:
%e A261015   1,1,
%e A261015   1,1,1,1,
%e A261015   1,1,2,3,1,0,0,0,
%e A261015   1,1,3,6,4,1,0,0,0,0,0,0,0,0,0,0,
%e A261015 ...
%e A261015 For row 3, here are the 8 strings of length 3 and for each one, the smallest missing number k:
%e A261015  000 1
%e A261015  001 2
%e A261015  010 3
%e A261015  011 2
%e A261015  100 3
%e A261015  101 3
%e A261015  110 4
%e A261015  111 0
%t A261015 notVis[bits_] := For[i = 0, True, i++, If[SequencePosition[bits, IntegerDigits[i, 2]] == {}, Return[i]]];
%t A261015 T[n_, k_] := Select[Rest[IntegerDigits[#, 2]]& /@ Range[2^n, 2^(n+1)-1], notVis[#] == k&] // Length;
%t A261015 Table[T[n, k], {n, 1, 6}, {k, 0, 2^n-1}] // Flatten (* _Jean-François Alcover_, Aug 02 2018 *)
%Y A261015 Cf. A260273, A261016, A261017.
%Y A261015 See A261019 for a more compact version (which has further information about the columns).
%K A261015 nonn,tabf
%O A261015 1,9
%A A261015 _N. J. A. Sloane_, Aug 16 2015
%E A261015 More terms from _Alois P. Heinz_, Aug 17 2015