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A065455
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Number of (binary) bit strings of length n in which no even block of 0's is followed by an odd block of 1's.
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8
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1, 2, 4, 7, 14, 25, 49, 89, 172, 316, 605, 1120, 2131, 3965, 7513, 14026, 26504, 49591, 93538, 175277, 330205, 619369, 1165892, 2188312, 4117045, 7730828, 14539447, 27309529, 51349169, 96468034, 181357036, 340753271, 640539142, 1203616849
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OFFSET
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0,2
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COMMENTS
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The limit of the ratio of successive terms as n increases can be shown to be 2*cos(Pi/9). In the opposite direction, as n—->-oo (see A052545), a(n+1)/a(n) approaches 2*cos(5*Pi/9). For example, a(-6)/a(-7) = -92/265, which is close to 2*cos(5*Pi/9). - Richard Locke Peterson, Apr 22 2019
Let P(n, j, m) = Sum_{r=1..m} (2^n*(1-(-1)^r)*cos(Pi*r/(m+1))^n*cot(Pi*r/(2*(m+1)))* sin(j*Pi*r/(m+1)))/(m+1) denote the number of paths of length n starting at the j-th node on the path graph P_m. We have a(n) = P(n, 3, 8). - Herbert Kociemba, Sep 17 2020
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LINKS
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FORMULA
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G.f.: (1+x)^2/(1-3*x^2-x^3).
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EXAMPLE
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a(5) = 32-7 = 25 because 00111, 00101, 00100, 10010, 01001, 11001, 00001 are forbidden.
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MATHEMATICA
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LinearRecurrence[{0, 3, 1}, {1, 2, 4}, 40] (* G. C. Greubel, May 31 2019 *)
a[n_, j_, m_]:=Sum[(2^(n+1)Cos[Pi r/(m+1)]^n Cot[Pi r/(2(m+1))] Sin[j Pi r/(m+1)])/(m+1), {r, 1, m, 2}]
CoefficientList[Series[(1+x)^2/(1-3x^2-x^3), {x, 0, 50}], x] (* Harvey P. Dale, Jul 16 2021 *)
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PROG
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(Magma) I:=[1, 2, 4]; [n le 3 select I[n] else 3*Self(n-2) +Self(n-3): n in [1..40]]; // G. C. Greubel, May 31 2019
(Sage) ((1+x)^2/(1-3*x^2-x^3)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, May 31 2019
(GAP) a:=[1, 2, 4];; for n in [4..40] do a[n]:=3*a[n-2]+a[n-3]; od; a; # G. C. Greubel, May 31 2019
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CROSSREFS
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Cf. A052545 (this is what we get if n takes negative values).
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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