A137397 Number of distinct palindromic subwords in the binary representation of n.

2, 2, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8

Offset

0,1

Comments

Equals A070941 from a(1) to a(202) and continues a(203)=8, a(204)=a(205)=9.

Omitting "distinct" in the definition, we get 1, 2, 4, 4, 7, 7, 7, 7, 11, 11, 11, 11, 11, 11, 11, 11, 16, 16,... which apparently is built by repeating entries of A000124 in blocks of length 2,4,8,16,32..

Links

Michael S. Branicky, Table of n, a(n) for n = 0..10000

A. Glen, J. Justin, S. Widmer, and L. Q. Zamboni, Palindromic Richness, arXiv:0801.1656 [math.CO], 2008.

Example

For n=10 the binary representation is A007088(10)=1010, which contains the a(10)=5 palindromic substrings {}, {0}, {1}, {101}, {010}. The empty subword is always included in the count.

Prog

(Python)
def ispal(s): return s == s[::-1]
def a(n):
  s = bin(n)[2:]
  return 1 + len(set(s[i:j] for i in range(len(s))
    for j in range(i+1, len(s)+1) if ispal(s[i:j])))
print([a(n) for n in range(105)]) # Michael S. Branicky, Feb 02 2021

Crossrefs

Cf. A070941.

Sequence in context: A328003 A262558 A156876 * A239684 A062571 A277903

Adjacent sequences: A137394 A137395 A137396 * A137398 A137399 A137400

Keyword

nonn,base

Author

R. J. Mathar, Apr 11 2008

Status

approved