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.

A007040 Number of (marked) cyclic n-bit binary strings containing no runs of length > 2.

Original entry on oeis.org

2, 2, 6, 6, 10, 20, 28, 46, 78, 122, 198, 324, 520, 842, 1366, 2206, 3570, 5780, 9348, 15126, 24478, 39602, 64078, 103684, 167760, 271442, 439206, 710646, 1149850, 1860500, 3010348, 4870846, 7881198, 12752042, 20633238, 33385284, 54018520
Offset: 1

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For n >= 3, also the number of maximal independent vertex sets (and minimal vertex covers) in the n-prism graph. - Eric W. Weisstein, Mar 30 and Aug 07 2017
From Petros Hadjicostas, Jul 08 2018: (Start)
Let q and m be positive integers. We denote by f1(m,q,n) the number of (marked) cyclic q-ary strings of length n that contain no runs of lengths > m when no wrapping around is allowed, and by f2(m,q,n) when wrapping around is allowed.
It is clear that f1(m,q,n) = f2(m,q,n) for n > m, but f1(m,q,n) = q^n and f2(m,q,n) = q^n - q when 1 <= n <= m.
Burstein and Wilf (1997) and Edlin and Zeilberger (2000) considered f1(m,q,n) while Hadjicostas and Zhang considered f2(m,q,n).
Let g(m, q, x) = (m+1-m*q*x)/(1-q*x+(q-1)*x^(m+1)) - (m+1)/(1-x^(m+1)).
By generalizing Moser (1991, 1993), Burstein and Wilf (1997) proved that the g.f. of the numbers f1(m,q,n) is F1(m,q,x) = ((1-x^m)/(1-x))*(q*x + (q-1)*x* g(m, q, x)).
Using the above formula by Burstein and Wilf (1997), Hadjicostas and Zhang (2018) proved that the g.f. of the numbers f2(m,q,n) is F2(m,q,x) = ((q-1)*x*(1-x^m)/(1-x))*g(m, q, x).
A necklace is an unmarked cyclic string. If f3(m,q,n) is the number of q-ary necklaces of length n with no runs of length > m (and wrapping around is allowed), then f3(m,q,n) = (1/n)*Sum_{d|n} phi(n/d)*f2(m,q,d), where phi(.) is Euler's totient function. Using this formula and F2(m,q,x), Hadjicostas and Zhang (2018) proved that the g.f. of the numbers f3(m,q,n) is given by F3(m,q,x) = -(q-1)*x*(1-x^m)/((1-x)*(1-x^(m+1))) - Sum_{s>=1} (phi(s)/s)*log(1 - (q-1)*(x^s - x^(s*(m+1)))/(1-x^s)).
For the current sequence, we have q = 2 and m = 2. We have a(n) = f1(m=2, q=2, n) = f2(m=2, q=2, n) for n >= 3, but for a(1) and a(2) it is unclear what approach the author of the sequence is following. He has a(1) = q^1 = 2, but a(2) = q^2 - q = 2^2 - 2 = 2. (Note that, for q = m = 2, we have f1(m=2, q=2, 1) = 2, f1(m=2, q=2, 2) = 4, f2(m=2, q=2, 1) = 0, and f2(m=2, q=2, 2) = 2.)
If A(x) is the g.f. of the current sequence, we have A(x) = F1(m=2,q=2, x) - 2*x^2 = F2(m=2, q=2, x) + 2*x.
When m = 1 and q = 3, we have f1(m=1, q=3, n) = number of marked cyclic words on three letters with no two consecutive like letters. We have f1(m=1, q=3, n) = A092297(n) for n >= 2. This was first stated in the comments of that sequence by G. Critzer.
When m = 1 and q = 4, we have f1(m=1, q=4, n) = number of marked cyclic words on four letters with no two consecutive like letters. We have f1(m=1, q=4, n) = A218034(n) for n >= 1. This was first stated in the comments of that sequence by J. Arndt.
A generalization of the above formula by Burstein and Wilf (1997) was given by Taylor (2014) in Section 5 of his paper. (End)

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Mathematica
    Join[{2}, LinearRecurrence[{0, 1, 2, 1}, {2, 6, 6, 10}, 40]] (* Harvey P. Dale, Nov 09 2011 *)
    Join[{2}, Table[n Sum[Binomial[2 k, n - 2 k]/k, {k, n}], {n, 2, 40}]] (* Harvey P. Dale, Nov 09 2011 *)
    Table[LucasL[n] + 2 Cos[2 n Pi/3], {n, 20}] (* Eric W. Weisstein, Mar 30 2017 *)
  • PARI
    a(n)=if(n<3,2,([0,1,0,0; 0,0,1,0; 0,0,0,1; 1,2,1,0]^(n-2)*[2;6;6;10])[1,1]) \\ Charles R Greathouse IV, Jun 15 2015

Formula

a(n) = a(n-2) + 2*a(n-3) + a(n-4), n >= 7. - David W. Wilson
a(n) = n*Sum_{k=1..n} binomial(2*k, n-2*k)/k for n > 1 with a(0) = 0 and a(1) = 2. - Vladimir Kruchinin, Oct 12 2011
G.f.: 2*x*(1 + x + 2*x^2 - x^4)/((1 - x - x^2)*(1 + x + x^2)). - Colin Barker, Mar 15 2012
a(n) = A000032(n) + 2*cos(2*Pi*n/3) for n > 1. - Eric W. Weisstein, Mar 30 2017
a(n) = 2*A100886(n-1), n > 1. - R. J. Mathar, Jan 20 2018
a(n) = A000032(n) - A061347(n) for n > 1. - Wojciech Florek, Feb 18 2018

Extensions

Name clarified by Petros Hadjicostas, Jul 08 2018