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WARNING: It is unclear whether this sequence should start at offset 1 (as written) or offset 0 (in analogy to many similar sequences, which seems to be assumed in many of the given formulas).

Inverse binomial transform of A057079. - Paul Barry, May 15 2003

The unsigned version, with g.f. (1 + x + 2*x^2)/(1 - x^3), has a(n) = 4/3 -cos(2*Pi*n/3)/3 - sqrt(3)*sin(2*Pi*n/3)/3 = gcd(Fib(n+4), Fib(n+1)). - Paul Barry, Apr 02 2004

a(n) = L(n-2,-1), where L is defined as in A108299; see also A010892 for L(n,+1). - Reinhard Zumkeller, Jun 01 2005

From the Taylor expansion of log(1 + x + x^2) at x = 1, Sum_{k > 0} a(k)/k = log(3) = A002391. This is case n = 3 of the general expression Sum_{k > 0} (1-n*!(k mod n))/k = log(n). - Jaume Oliver Lafont, Oct 16 2009

If used with offset zero, a non-simple continued fraction representation of 2+sqrt(2). - R. J. Mathar, Mar 08 2012

Periodic sequences of this type can be also calculated by a(n) = c + floor(q/(p^m-1)*p^n) mod p, where c is a constant, q is the number representing the periodic digit pattern and m is the period length. c, p and q can be calculated as follows: Let D be the array representing the number pattern to be repeated, m = size of D, max = maximum value of elements in D, min = minimum value of elements in D. Than c := min, p := max - min + 1 and q := p^m*Sum_{i=1..m} (D(i)-min)/p^i. Example: D = (1, 1, -2), c = -2, p = 4 and q = 60 for this sequence. - Hieronymus Fischer, Jan 04 2013

This is the Dirichlet inverse of A117997. - Petros Hadjicostas, Jul 25 2020

John W. Layman observes that the second differences give the sequence shifted to the right.

This sequence was investigated in cooperation with Paul Barry.

Generating floretion: - 0.5'i - 0.5'k - 0.5j' - 0.5'ii' + 0.5'jj' - 0.5'kk' + 0.5'ik' - 0.5'ki' ("tes").

From Joshua P. Bowman, Sep 28 2023: (Start)

a(n) is equal to the number of circular binary sequences of length n+1 with an even number of 0's and no consecutive 1's. A circular binary sequence is a finite sequence of 0's and 1's for which the first and last digits are considered to be adjacent. Rotations are distinguished from each other.

a(n) is also equal to the number of matchings in the cycle graph C_{n+1} for which the number of edges plus the number of unmatched vertices is even.

a(n) is also equal to the number of circular compositions of n+1 into an even number of 1's and 2's. (End)

A generalized tribonacci (A001590) sequence.

For n > 2, 6*a(n) is the number of quaternary sequences of length n in which all triples (q(i),q(i+1),q(i+2)) contain two (arbitrarily chosen) digits, e.g., 0 and 3.

Similarly, recurrences a(n) = a(n-1) + a(n-2) + k*a(n-3) are related to binary (k=0, the Fibonacci sequence A000045), ternary (k=1, the tribonacci sequence A001590), quinary (k=3) and so on sequences with all triples (t(i),t(i+1),t(i+2)) containing two (arbitrarily chosen) digits (usually 0 and k+1).

For n > 0, a(n) is the number of ways to tile a strip of length n with squares, dominoes, and two colors of trominoes, with the restriction that the first tile cannot be a tromino. - Greg Dresden and Bora Bursalı, Aug 31 2023

For n > 1, a(n) is the number of ways to tile a strip of length n-2 with squares, dominoes, and two colors of trominoes, where the strip begins with an upper level of two cells. For example, when n=7 we have this strip of length 5:

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|||_____

|||_|||. - Guanji Chen and Greg Dresden, Jun 17 2024

This is the "unmarked" version of sequence A007040. An unmarked cyclic string is a necklace. Notice that we define a(1) = 0 and a(2) = 1 because wrapping around the circle is allowed here (otherwise we would have to let a(1) = 2 and a(2) = 3).

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 (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).

If f3(m,q,n) is the number of q-ary necklaces (= unmarked cyclic strings) 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)).

If A(x) is the g.f. of the current sequence (a(n): n >= 1), we have A(x) = F3(m=2, q=2, x). Also, a(n) = f3(m=2, q=2, n) = (1/n)*Sum_{d|n} phi(n/d)*f2(m=2, q=2, d). Note that f2(m=2, q=2, n=1) = 0 and f2(m=2, q=2, n) = A007040(n) for n >= 2.

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)).

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 = 3. We have a(n) = f1(m=3, q=2, n) = f2(m=3, q=2, n) for n >= 4, but we have f1(m=3, q=2, n) = 2^n and f2(m=3, q=2, n) = 2^n - 2 for n = 1,2,3.

If A(x) is the g.f. of the current sequence, we have A(x) = F1(m=3,q=2, x) = F2(m=3, q=2, x) + 2*(x+x^2+x^3).

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.

When m=2 and q=2, we have f1(m=2, q=2, n) = number of marked cyclic words on two letters containing no runs of length > 2. We have f1(m=2, q=2, n) = A007040(n) for n >= 3.

A generalization of the above formula by Burstein and Wilf (1997) was given by Taylor (2014) in Section 5 of his paper.