%I #17 May 20 2020 17:56:18
%S 2,2,1,4,4,1,2,2,2,2,3,6,6,3,5,8,2,2,2,2,8,5,6,3,3,6,6,3,3,6,1,1,1,1,
%T 7,10,4,4,4,4,10,7,1,1,1,1,3,9,3,12,3,6,3,6,6,3,6,3,12,3,9,3,1,1,1,1,
%U 10,7,4,4,7,10,4,4,1,1,1,1,1,1,1,1,4,4,10
%N Side length s of the central triangle of zeros in the XOR-triangle T(n).
%C An XOR-triangle T(n) is an inverted 0-1 triangle formed by choosing a top row the binary rendition of n and having each entry in subsequent rows be the XOR of the two values above it, i.e., A038554(n) applied recursively until we reach a single bit.
%C A334556 is the sequence of rotationally symmetrical T(n).
%C A central zero-triangle (CZT) is a field of contiguous 0-bits in T(n) surrounded on all sides by a layer of 1 bits, and generally k > 1 bits of any parity. Alternatively, these might be referred to as "central bubbles".
%H Michael De Vlieger, <a href="/A334770/b334770.txt">Table of n, a(n) for n = 1..10000</a>
%H Michael De Vlieger, <a href="http://vincico.com/seq/a334769.html">Central zero-triangles in rotationally symmetrical XOR-Triangles</a>, 2020.
%H <a href="/index/Bi#binary">Index entries for sequences related to binary expansion of n</a>
%H <a href="/index/X#XOR-triangles">Index entries for sequences related to XOR-triangles</a>
%e For n = 151, we have rotationally symmetrical T(151) as below, replacing 0 with "." for clarity:
%e 1 . . 1 . 1 1 1
%e 1 . 1 1 1 . .
%e 1 1 . . 1 .
%e . 1 . 1 1
%e 1 1 1 .
%e . . 1
%e . 1
%e 1
%e At the center of the figure we see a CZT with s = 2, ringed by 1s, with k = 2. Since 151 is the first term of A334769, a(1) = 2.
%e For n = 599, we have a rotationally symmetrical T(599) with s = 4 and k = 2.
%e 1 . . 1 . 1 . 1 1 1
%e 1 . 1 1 1 1 1 . .
%e 1 1 . . . . 1 .
%e . 1 . . . 1 1
%e 1 1 . . 1 .
%e . 1 . 1 1
%e 1 1 1 .
%e . . 1
%e . 1
%e 1
%e Since A334769(4) = 599, a(4) = 4.
%t Block[{f, s = Rest[Import["https://oeis.org/A334556/b334556.txt", "Data"][[All, -1]] ]}, f[n_] := NestWhileList[Map[BitXor @@ # &, Partition[#, 2, 1]] &, IntegerDigits[n, 2], Length@ # > 1 &]; Array[Block[{n = s[[#]]}, If[# == 0, Nothing, #] &@ FirstCase[MapIndexed[If[2 #2 > #3 + 1, Nothing, #1[[#2 ;; -#2]]] & @@ {#1, First[#2], Length@ #1} &, f[n][[1 ;; Ceiling[IntegerLength[#, 2]/(2 Sqrt[3])] + 3]] ], r_List /; FreeQ[r, 1] :> Length@ r] /. k_ /; MissingQ@ k -> 0] &, Lengths - 1, 2] ]
%Y Cf. A038554, A070939, A334556, A334769, A334771, A334796.
%K nonn
%O 1,1
%A _Michael De Vlieger_, May 10 2020