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A290743
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Maximum number of distinct Lyndon factors that can appear in words of length n over an alphabet of size 2.
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7
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2, 3, 4, 6, 8, 11, 14, 18, 22, 27, 32, 38, 44, 51, 58, 66, 74, 83, 92, 102, 112, 123, 134, 146, 158, 171, 184, 198, 212, 227, 242, 258, 274, 291, 308, 326, 344, 363, 382, 402, 422, 443, 464, 486, 508, 531, 554, 578, 602, 627, 652, 678, 704, 731, 758
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OFFSET
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1,1
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COMMENTS
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See theorem 1 of reference for formula.
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LINKS
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Amy Glen, Jamie Simpson, and W. F. Smyth, Counting Lyndon Factors, Electronic Journal of Combinatorics 24(3) (2017), #P3.28.
Ryo Hirakawa, Yuto Nakashima, Shunsuke Inenaga, and Masayuki Takeda, Counting Lyndon Subsequences, arXiv:2106.01190 [math.CO], 2021. See MDF(n, s).
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FORMULA
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a(n) = binomial(n+1,2) - (s-p)*binomial(m+1,2) - p*binomial(m+2,2) + s where s=2, m=floor(n/s), p=n-m*s. - Andrew Howroyd, Aug 14 2017
G.f.: x*(2 - x - 2*x^2 + 2*x^3) / ((1 - x)^3*(1 + x)).
a(n) = (2*n^2 + 16) / 8 for n even.
a(n) = (2*n^2 + 14) / 8 for n odd.
a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4) for n > 4. (End)
E.g.f.: ((8 + x + x^2)*cosh(x) + (7 + x + x^2)*sinh(x) - 8)/4. - Stefano Spezia, Jul 06 2021
Sum_{n>=1} 1/a(n) = coth(sqrt(2)*Pi)*Pi/(2*sqrt(2)) + tanh(sqrt(7)*Pi/2)*Pi/sqrt(7) - 1/4. - Amiram Eldar, Sep 16 2022
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MATHEMATICA
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Table[(Binomial[n+1, 2] - (2-(n - 2 Floor[n/2])) Binomial[Floor[n/2]+1, 2] - (n-2 Floor[n/2]) Binomial[Floor[n/2]+2, 2] + 2), {n, 60}] (* Vincenzo Librandi, Oct 04 2017 *)
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PROG
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(PARI) a(n)=(s->my(m=n\s, p=n%s); binomial(n+1, 2)-(s-p)*binomial(m+1, 2)-p*binomial(m+2, 2)+s)(2); \\ Andrew Howroyd, Aug 14 2017
(Magma) [Binomial(n+1, 2)-(2-(n-2*Floor(n/2)))*Binomial(Floor(n/2)+1, 2)-(n-2*Floor(n/2))*Binomial(Floor(n/2)+2, 2)+2: n in [1..60]]; // Vincenzo Librandi, Oct 04 2017
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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