A237502 Number of binary strings of length 2n that contain the ones' complements and the reversals of each of their two halves.

2, 6, 8, 16, 26, 46, 72, 116, 184, 288, 434, 646, 980, 1444, 2122, 3066, 4500, 6452, 9362, 13294, 19256, 27280, 39242, 55306, 79546, 111990, 160514, 225466, 323108, 453520, 648970, 909942, 1301932, 1825044, 2609142, 3655406, 5225912, 7320748, 10461942, 14651858

Offset

1,1

Links

Table of n, a(n) for n=1..40.

Michael S. Branicky, Python program

Example

The two halves of 1000111000 are 10001 and 11000. Their reversals are 10001 and 00011, and their complements are 01110 and 00111 and all are substrings of 1000111000. Since there are 25 other strings of length 2*5 with this property, a(5) = 26.

Mathematica

sQ[L_, {s__}] := MatchQ[L, {___, s, ___}]; a[n_] := Length@ Select[ Tuples[{0, 1}, 2*n], sQ[#, Reverse[Take[#, n]]] && sQ[#, Reverse[Take[#, -n ]]] && sQ[#, 1 - Take[#, n]] && sQ[#, 1 - Take[#, -n]] &]; Array[a, 8]

Prog

(Python) # see link for faster version
from itertools import product as prod
def comp(s): z, o = ord('0'), ord('1'); return s.translate({z:o, o:z})
def ok(s):
  a, b = s[len(s)//2:], s[:len(s)//2]
  return comp(a) in s and comp(b) in s and a[::-1] in s and b[::-1] in s
def a(n): return 2*sum(ok("0"+"".join(p)) for p in prod("01", repeat=2*n-1))
print([a(n) for n in range(1, 12)]) # Michael S. Branicky, Feb 06 2021

Crossrefs

Cf. A237500, A237501.

Sequence in context: A266074 A191822 A238549 * A279726 A337881 A331974

Adjacent sequences: A237499 A237500 A237501 * A237503 A237504 A237505

Keyword

nonn

Author

Giovanni Resta, Feb 08 2014

Extensions

a(21)-a(40) from Max Alekseyev, Feb 01 2024

Status

approved