A080204 Number of fixed points under n-fold inflation for the substitution rule a->abc, b->ab, c->b that underlies the Kolakoski (3,1) sequence.
1, 3, 10, 23, 51, 114, 253, 559, 1234, 2723, 6007, 13250, 29225, 64459, 142170, 313567, 691595, 1525362, 3364293, 7420183, 16365730, 36095755, 79611695, 175589122, 387274001, 854159699, 1883908522, 4155091047, 9164341795, 20212592114, 44580275277, 98324892351, 216862376818
Offset: 1
Links
- Metin Sariyar, Table of n, a(n) for n = 1..300
- J. E. Anderson and I. F. Putnam, Topological invariants for substitution tilings and their associated C*-algebras, Ergod. Theory and Dyn. Systems 18 (1998) 509-537. See also, Technical report DMS-720-IR, University of Victoria, 1995.
- Michael Baake and B. Sing, Kolakoski-(3,1) is a (deformed) model set, arXiv:math/0206098 [math.MG], 2002-2003.
- Greg Dresden and Michael Tulskikh, Tilings of 2 X n boards with dominos and L-shaped trominos, Washington & Lee University (2021).
- Index entries for linear recurrences with constant coefficients, signature (3,-2,1,-1).
Crossrefs
Equals one less than A332647.
Programs
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Magma
I:=[1,3,10,23]; [n le 4 select I[n] else 3*Self(n-1)-2*Self(n-2)+Self(n-3)-Self(n-4): n in [1..40]]; // Vincenzo Librandi, Mar 12 2015
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Mathematica
CoefficientList[Series[(1 + 3 x^2 - 2 x^3) / ((1 - x) (1 - 2 x - x^3)), {x, 0, 40}], x] (* Vincenzo Librandi, Mar 12 2015 *)
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Maxima
a(n):=(n*sum(sum(binomial(m,i)*binomial(n-2*i-1,m-1),i,0,(n-m)/2)/m,m,1,n)); /* Vladimir Kruchinin, Mar 10 2015 */
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PARI
x='x+O('x^66); Vec(x*(1+3*x^2-2*x^3)/((1-x)*(1-2*x-x^3))) \\ Joerg Arndt, Jun 15 2013
Formula
Obtained from dynamical zeta function Z(x)=(1-x)/(1-2x-x^3), so that x Z'(x)/Z(x) is the ordinary power series generating function.
From Colin Barker, Jul 02 2012: (Start)
a(n) = 3*a(n-1) - 2*a(n-2) + a(n-3) - a(n-4).
G.f.: x*(1+3*x^2-2*x^3)/((1-x)*(1-2*x-x^3)). (End)
a(n) = n*Sum_{m=1..n} Sum_{i=0..(n-m)/2} binomial(m,i)*binomial(n-2*i-1,m-1)/m. - Vladimir Kruchinin, Mar 10 2015
a(n) = 2*a(n-1) + a(n-3) + 2. - Greg Dresden, Feb 08 2020
a(n) = w1^n + w2^n + w3^n - 1, for w1, w2, w3 roots of x^3-2x^2-1=0. - Greg Dresden, Feb 18 2020
Comments