A361496 Inventory of positions as an irregular table; row 0 contains 0, subsequent rows contain the 0-based positions (mod 2) of 0's, followed by the positions (mod 2) of 1's in prior rows flattened.

0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1

Offset

0

Comments

For n >= 2, the n-th row contains 2^(n - 1) terms and their sum equals to 2^(n - 2).

Links

Table of n, a(n) for n=0..100.

Example

Table begins:
   0,
   0,
   0, 1,
   0, 1, 0, 1,
   0, 1, 0, 0, 0, 1, 1, 1,
   ...
For n = 4:
- the terms in rows 0..3 are: 0, 0, 0, 1, 0, 1, 0, 1,
- we have 0's at positions 0 mod 2 = 0, 1 mod 2 = 1, 2 mod 2 = 0, 4 mod 2 = 0, 6 mod 2 = 0,
- we have 1's at positions 3 mod 2 = 1, 5 mod 2 = 1, 7 mod 2 = 1,
- so row 4 is: 0, 1, 0, 0, 0, 1, 1, 1.

Mathematica

row[0] = {0}; row[n_] := row[n] = Module[{t = Flatten@ Table[row[i], {i, 0, n-1}], p}, Join@@ (Mod[-1 + If[(p = Position[t, #]) == {}, {}, Flatten[p]], 2]& /@ {0, 1})]; Table[row[i], {i, 0, 7}] // Flatten (* Amiram Eldar, Mar 15 2023 *)

Crossrefs

Cf. A342585, A356784, A358066.

Sequence in context: A064911 A174898 A099618 * A360125 A324883 A106002

Adjacent sequences: A361493 A361494 A361495 * A361497 A361498 A361499

Keyword

nonn,tabf

Author

Ctibor O. Zizka, Mar 14 2023

Status

approved