A290745 Maximum number of distinct Lyndon factors that can appear in words of length n over an alphabet of size 10.
10, 11, 13, 16, 20, 25, 31, 38, 46, 55, 64, 74, 85, 97, 110, 124, 139, 155, 172, 190, 208, 227, 247, 268, 290, 313, 337, 362, 388, 415, 442, 470, 499, 529, 560, 592, 625, 659, 694, 730, 766, 803, 841, 880, 920, 961, 1003, 1046, 1090, 1135
Offset: 1
Keywords
Links
- Vincenzo Librandi, Table of n, a(n) for n = 1..1000
- 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).
Programs
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Magma
[Binomial(n+1,2)-(10-(n-10*Floor(n/10)))*Binomial(Floor(n/10)+1,2)-(n-10*Floor(n/10))*Binomial(Floor(n/10)+2,2)+ 10: n in [1..50]]; // Vincenzo Librandi, Oct 04 2017
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Mathematica
Table[(Binomial[n+1, 2] - (10 - (n-10 Floor[n/10])) Binomial[Floor[n/10]+1, 2]- (n-10 Floor[n/10]) Binomial[Floor[n/10]+2, 2]+10), {n, 60}] (* Vincenzo Librandi, Oct 04 2017 *)
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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)(10); \\ Andrew Howroyd, Aug 14 2017
Formula
a(n) = binomial(n+1,2) - (s-p)*binomial(m+1,2) - p*binomial(m+2,2) + s where s=10, m=floor(n/s), p=n-m*s. - Andrew Howroyd, Aug 14 2017
G.f.: x*(10 - 9*x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 + x^9 - 10*x^10 + 10*x^11) / ((1 - x)^3*(1 + x)*(1 - x + x^2 - x^3 + x^4)*(1 + x + x^2 + x^3 + x^4)) (conjectured). - Colin Barker, Oct 03 2017
Extensions
a(11)-a(50) from Andrew Howroyd, Aug 14 2017