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

Author

R. J. Mathar, Apr 11 2008

Keywords

• nonn
• base

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..

Examples

			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.
		

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.

Crossrefs

Cf. A070941.

Programs

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