A179999 Length of the n-th term in the modified Look and Say sequence A110393.
1, 2, 2, 4, 6, 8, 10, 14, 18, 24, 30, 40, 50, 66, 82, 108, 134, 176, 218, 286, 354, 464, 574, 752, 930, 1218, 1506, 1972, 2438, 3192, 3946, 5166, 6386, 8360, 10334, 13528, 16722, 21890, 27058, 35420, 43782, 57312, 70842, 92734, 114626, 150048
Offset: 1
Examples
The 6th term in A110393 is 21112211, so a(6) = 8.
Links
- Colin Barker, Table of n, a(n) for n = 1..1000
- N. Johnston, Further Variants of the “Look-and-Say” Sequence
- Index entries for linear recurrences with constant coefficients, signature (1,1,-1,1,-1).
Programs
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Mathematica
CoefficientList[Series[((1+x) (-1-x+x^2) (1-x+x^2))/((1-x) (-1+x^2+x^4)),{x,0,99}],x] (* Peter J. C. Moses, Jun 23 2013 *)
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PARI
Vec(x*(1 + x)*(1 + x - x^2)*(1 - x + x^2) / ((1 - x)*(1 - x^2 - x^4)) + O(x^50)) \\ Colin Barker, Aug 10 2019
Formula
a(n) = length(A110393(n)).
From Colin Barker, Aug 10 2019: (Start)
G.f.: x*(1 + x)*(1 + x - x^2)*(1 - x + x^2) / ((1 - x)*(1 - x^2 - x^4)).
a(n) = a(n-1) + a(n-2) - a(n-3) + a(n-4) - a(n-5) for n>6. (End)
From A.H.M. Smeets, Aug 10 2019 (Start)
Limit_{n->oo} a(n+1)/a(n) = (1+phi)/2 = (3+sqrt(5))/4 = A239798 for odd n.
Limit_{n->oo} a(n+1)/a(n) = 2/phi = 4/(1+sqrt(5)) = A134972 for even n.
Limit_{n->oo} a(n+2)/a(n) = (1+phi)/phi = phi = A001622. (End)
For odd n > 1, a(n) = 4*Fibonacci((n + 1)/2) - 2. For even n, a(n) = 2*Fibonacci(n/2 + 2) - 2. - Ehren Metcalfe, Aug 10 2019
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