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.

Showing 1-3 of 3 results.

A230127 Number of binary strings of length n avoiding "squares" (that is, repeated blocks of the form xx) with |x| > 1.

Original entry on oeis.org

1, 2, 4, 8, 12, 20, 26, 38, 42, 52, 56, 56, 48, 42, 32, 22, 10, 4, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
Offset: 0

Views

Author

Nathaniel Johnston, Oct 10 2013

Keywords

Comments

Entringer et al. showed that a(n) = 0 for all n >= 19.

Examples

			a(4) = 12 because there are 16 binary strings of length 4, but 4 of these strings (namely 0000, 0101, 1010, and 1111) repeat a substring of length 2. Thus a(4) = 16 - 4 = 12.
a(18) = 2 because there are 2 strings of length 18 not containing any "squares" of length greater than 1: 010011000111001101 and 101100111000110010.
		

Crossrefs

Programs

  • Mathematica
    a[n_] := a[n] = Select[PadLeft[#, n]& /@ IntegerDigits[Range[0, 2^n-1], 2], {} == SequencePosition[#, {b__, b__} /; Length[{b}]>1, 1]&] // Length;
    Table[Print[n, " ", a[n]]; a[n], {n, 0, 20}] (* Jean-François Alcover, Nov 10 2021 *)

A229614 Number of binary strings of length n avoiding "squares" (that is, repeated blocks of the form xx) with |x| > 2.

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 56, 104, 178, 314, 536, 930, 1558, 2666, 4482, 7574, 12686, 21360, 35812, 60152, 100812, 169122, 283498, 475356, 796292, 1334558, 2235888, 3746534, 6276048, 10515080, 17614726, 29510362, 49434792, 82815016, 138729368, 232399846, 389306052
Offset: 0

Views

Author

Jeffrey Shallit, Sep 26 2013

Keywords

Comments

Entringer et al. showed that this sequence is always nonzero (in contrast with A230127, which is zero for all n >= 19). - Nathaniel Johnston, Oct 10 2013
For n >= 1, terms are even by symmetry. - Michael S. Branicky, Nov 11 2021

Examples

			For n = 6 there are 8 strings omitted, namely 000000, 001001, ..., 111111, so a(6) = 64 - 8 = 56.
		

Crossrefs

Programs

  • Mathematica
    a[n_] := a[n] = Select[PadLeft[#, n]& /@ IntegerDigits[Range[0, 2^n-1], 2], {} == SequencePosition[#, {b__, b__} /; Length[{b}] >= 3, 1]&] // Length;
    Table[Print[n, " ", a[n]]; a[n], {n, 0, 22}] (* Jean-François Alcover, Nov 11 2021 *)
  • Python
    # see link for a faster program based on binary operations
    def isf(s): # incrementally squarefree (check factors ending in last letter)
        for l in range(3, len(s)//2 + 1):
            if s[-2*l:-l] == s[-l:]: return False
        return True
    def aupton(nn, verbose=False):
        alst, sfs = [1], set("0")
        for n in range(1, nn+1):
            an = 2*len(sfs) # by symmetry
            sfsnew = set(s+i for s in sfs for i in "01" if isf(s+i))
            alst, sfs = alst+[an], sfsnew
            if verbose: print(n, an)
        return alst
    print(aupton(24)) # Michael S. Branicky, Nov 11 2021

Extensions

a(23)-a(36) from Nathaniel Johnston, Oct 10 2013

A230216 Number of binary strings of length n avoiding "squares" (that is, repeated blocks of the form xx) with |x| = 3.

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 56, 104, 192, 352, 648, 1192, 2192, 4032, 7416, 13640, 25088, 46144, 84872, 156104, 287120, 528096, 971320, 1786536, 3285952, 6043808, 11116296, 20446056, 37606160, 69168512, 127220728, 233995400, 430384640, 791600768, 1455980808
Offset: 0

Views

Author

Nathaniel Johnston, Oct 11 2013

Keywords

Examples

			For n = 6 there are 8 strings omitted, namely 000000, 001001, ..., 111111, so a(6) = 64-8 = 56.
		

Crossrefs

Programs

  • PARI
    Vec((1 + x + x^2 + x^3 + 2*x^4 + 4*x^5) / (1 - x - x^2 - x^3) + O(x^40)) \\ Colin Barker, Aug 09 2019

Formula

a(n) = 8*A000073(n) for n >= 3.
From Colin Barker, Aug 09 2019: (Start)
G.f.: (1 + x + x^2 + x^3 + 2*x^4 + 4*x^5) / (1 - x - x^2 - x^3).
a(n) = a(n-1) + a(n-2) + a(n-3) for n>5.
(End)
Showing 1-3 of 3 results.