A051041 Number of squarefree quaternary words of length n.
1, 4, 12, 36, 96, 264, 696, 1848, 4848, 12768, 33480, 87936, 230520, 604608, 1585128, 4156392, 10895952, 28566216, 74887056, 196322976, 514662960, 1349208600, 3536962584, 9272217936, 24307198464, 63721617888, 167046745992, 437914664688, 1147996820376, 3009483583056, 7889385389784, 20682088837608, 54218261608896
Offset: 0
Keywords
Examples
There are 96 quaternary squarefree words of length 4: each of the words 0102, 0120, 0121, 0123 has 4!=24 images under the permutations of the set {0,1,2,3}. - _Arseny Shur_, Apr 26 2015 G.f. = 1 + 4*x + 12*x^2 + 36*x^3 + 96*x^4 + 264*x^5 + 696*x^6 + 1848*x^7 + ....
Links
- A. M. Shur, Growth properties of power-free languages, Computer Science Review, Vol. 6 (2012), 187-208.
- A. M. Shur, Numerical values of the growth rates of power-free languages, arXiv:1009.4415 [cs.FL], 2010.
- Eric Weisstein's World of Mathematics, Squarefree Word.
Crossrefs
Cf. A006156.
Third column of A215075, multiplied by 24 (except for the first three terms). - Arseny Shur, Apr 26 2015
Programs
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Python
def isf(s): # incrementally squarefree (check factors ending in last letter) for l in range(1, len(s)//2 + 1): if s[-2*l:-l] == s[-l:]: return False return True def aupton(nn, verbose=False): alst, sfs = [1], set("1") for n in range(1, nn+1): an = 4*len(sfs) sfsnew = set(s+i for s in sfs for i in "0123" if isf(s+i)) alst, sfs = alst+[an], sfsnew if verbose: print(n, an) return alst print(aupton(14)) # Michael S. Branicky, Jun 20 2022
Formula
Let L be the limit lim a(n)^{1/n}, which exists because a(n) is a submultiplicative sequence. Then L=2.6215080... (Shur 2010). See (Shur 2012) for the methods of estimating such limits. - Arseny Shur, Apr 26 2015
Extensions
More terms from David Wasserman, Feb 27 2002
a(13)-a(15) from John W. Layman, Mar 03 2004
a(16)-a(25) from Max Alekseyev, Jul 03 2006
a(26)-a(30) from Giovanni Resta, Mar 20 2020
Comments