%I #31 Sep 17 2023 15:40:44
%S 1,1,1,1,1,1,1,2,1,1,1,2,3,1,1,1,2,3,4,1,1,1,2,4,5,5,1,1,1,2,4,7,8,6,
%T 1,1,1,2,4,7,12,13,7,1,1,1,2,4,7,12,20,21,8,1,1,1,2,4,7,12,21,33,34,9,
%U 1,1,1,2,4,8,13,20,37,54,55,10,1,1,1,2,4,8,15,24,33,65,88,89,11,1,1
%N Number of binary words of length n avoiding the subword given by the binary expansion of k; square array A(n,k), n>=0, k>=0, read by antidiagonals.
%H Alois P. Heinz, <a href="/A209972/b209972.txt">Antidiagonals n = 0..150, flattened</a>
%e Square array begins:
%e 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
%e 1, 1, 2, 2, 2, 2, 2, 2, 2, ...
%e 1, 1, 3, 3, 4, 4, 4, 4, 4, ...
%e 1, 1, 4, 5, 7, 7, 7, 7, 8, ...
%e 1, 1, 5, 8, 12, 12, 12, 13, 15, ...
%e 1, 1, 6, 13, 20, 21, 20, 24, 28, ...
%e 1, 1, 7, 21, 33, 37, 33, 44, 52, ...
%e 1, 1, 8, 34, 54, 65, 54, 81, 96, ...
%e 1, 1, 9, 55, 88, 114, 88, 149, 177, ...
%t A[n_, k_] := Module[{bb, cnt = 0}, Do[bb = PadLeft[IntegerDigits[j, 2], n]; If[SequencePosition[bb, IntegerDigits[k, 2], 1]=={}, cnt++], {j, 0, 2^n-1 }]; cnt];
%t Table[A[n-k, k], {n, 0, 12}, {k, n, 0, -1}] // Flatten (* _Jean-François Alcover_, Nov 01 2021 *)
%Y Columns give: 0, 1: A000012, 2: A001477(n+1), 3: A000045(n+2), 4, 6: A000071(n+3), 5: A005251(n+3), 7: A000073(n+3), 8, 12, 14: A008937(n+1), 9, 11, 13: A049864(n+2), 10: A118870, 15: A000078(n+4), 16, 20, 24, 26, 28, 30: A107066, 17, 19, 23, 25, 29: A210003, 18, 22: A209888, 21: A152718(n+3), 27: A210021, 31: A001591(n+5), 32: A001949(n+5), 33, 35, 37, 39, 41, 43, 47, 49, 53, 57, 61: A210031.
%Y Main diagonal equals A234005 or column k=0 of A233940.
%K nonn,tabl,base
%O 0,8
%A _Alois P. Heinz_, Mar 16 2012