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.

A214939 Number of squarefree words of length n in a 5-ary alphabet.

Original entry on oeis.org

5, 20, 80, 300, 1140, 4260, 15960, 59580, 222600, 830880, 3102120, 11578800, 43220940, 161324400, 602159940, 2247585300, 8389237320, 31313155560, 116877700500, 436250537520
Offset: 1

Views

Author

R. H. Hardin Jul 30 2012

Keywords

Comments

All terms are multiples of 5 by symmetry. Michael S. Branicky, May 20 2021

Examples

			Some solutions for n = 6:
..4....2....0....2....3....3....4....4....4....2....0....1....1....0....1....4
..2....1....4....4....1....2....0....3....0....4....2....0....3....3....0....3
..4....2....1....2....3....0....1....2....2....2....3....3....1....1....3....4
..0....4....2....3....4....1....4....1....3....3....0....2....2....2....4....2
..1....0....4....4....0....2....2....3....4....1....4....1....0....4....0....4
..0....1....1....2....1....1....0....4....3....3....3....0....3....1....4....1
		

Crossrefs

Column 4 of A214943.

Programs

  • Python
    from itertools import product
    def a(n):
      if n == 1: return 5
      squares = ["".join(u) + "".join(u)
        for r in range(1, n//2 + 1) for u in product("01234", repeat = r)]
      words = ("0"+"".join(w) for w in product("01234", repeat=n-1))
      return 5*sum(all(s not in w for s in squares) for w in words)
    print([a(n) for n in range(1, 10)]) # Michael S. Branicky, May 20 2021