cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Previous Showing 11-13 of 13 results.

A174196 Those positive integers that are both in sequence A166934 and in sequence A166935.

Original entry on oeis.org

23, 29, 40, 58, 79, 104, 116, 121, 176, 223, 232, 242, 251, 287, 288, 303, 335, 351, 485, 489, 497, 501, 502, 672, 736, 800, 815, 831, 847, 864, 971, 979, 994, 996, 1002, 1004, 1011, 1087, 1119, 1200, 1216, 1232, 1247, 1311, 1376, 1471, 1696, 1760, 1824, 1839
Offset: 1

Views

Author

Ray Chandler, Mar 11 2010

Keywords

Comments

Each integer in this sequence, when written in binary, contains at least two palindromic substrings that are tied for the longest such substring, one with identical digits and one with different digits.

Crossrefs

A193159 Numbers having in binary representation longest palindromic subwords not longer than 3.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 20, 23, 24, 26, 28, 29, 40, 52, 56, 58, 104, 116, 232
Offset: 1

Views

Author

Reinhard Zumkeller, Jul 16 2011

Keywords

Comments

A050430(a(n)) <= 3, 1 <= n <= 26.

Programs

  • Haskell
    a193159 n = a193159_list !! (n-1)
    a193159_list = map (+ 1) $ findIndices (<= 3) a050430_list

A215256 Longest palindromic factor of (n base 2); in case of tie choose largest; if it begins with 0 complement it.

Original entry on oeis.org

1, 1, 1, 11, 11, 101, 11, 111, 111, 1001, 101, 101, 11, 101, 111, 1111, 1111, 10001, 1001, 1001, 101, 10101, 1001, 111, 111, 1001, 101, 11011, 111, 111, 1111, 11111, 11111, 100001, 10001, 10001, 11011, 1001, 1001, 1001, 101, 1001, 10101, 10101, 1001, 101101
Offset: 0

Views

Author

N. J. A. Sloane, Aug 15 2012

Keywords

Comments

The "if it begins with 0 complement it" clause is required because nonzero terms in the OEIS may not begin with 0.

Examples

			n=10 = 1010, longest palindromic factor is 101.
n=12 = 1100, there are two palindromic factors of length 2, namely 11 and 00, and we choose the larger, 11.
n=24 = 11000, longest palindromic factor is 000, complementing gives 111.
		

Crossrefs

Extensions

More terms from Lars Blomberg, Jun 29 2014
Previous Showing 11-13 of 13 results.