A321257 Start with an equilateral triangle, and repeatedly append along the triangles of the previous step equilateral triangles with half their side length that do not overlap with any prior triangle; a(n) gives the number of triangles appended at n-th step.
1, 6, 21, 60, 147, 330, 705, 1464, 2991, 6054, 12189, 24468, 49035, 98178, 196473, 393072, 786279, 1572702, 3145557, 6291276, 12582723, 25165626, 50331441, 100663080, 201326367, 402652950, 805306125, 1610612484, 3221225211, 6442450674, 12884901609, 25769803488
Offset: 1
Keywords
Links
- Rémy Sigrist, Illustration of the construction after 7 steps
- Index entries for linear recurrences with constant coefficients, signature (4,-5,2).
Crossrefs
Cf. A321237.
Programs
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GAP
Concatenation([1],List([2..35],n->3*(2^(n-1)+3*(2^(n-1)-n)))); # Muniru A Asiru, Nov 02 2018
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Magma
[1] cat [3*(2^(n-1) + 3*(2^(n-1)-n)): n in [2..35]]; // Vincenzo Librandi, Nov 02 2018
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Maple
1,seq(3*(2^(n-1)+3*(2^(n-1)-n)),n=2..35); # Muniru A Asiru, Nov 02 2018
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Mathematica
CoefficientList[Series[(1 + 2 x + 2 x^2 + 4 x^3) / ((1 - 2 x) (1 - x)^2), {x, 0, 40}], x] (* Vincenzo Librandi, Nov 02 2018 *) -
PARI
a(n) = if (n==1, 1, 3*(2^(n-1) + 3*(2^(n-1)-n)))
Formula
a(n) = 3*(2^(n-1) + 3*(2^(n-1)-n)) for any n > 1.
Sum_{n > 0} a(n) / 4^(n-1) = 6.
G.f.: x*(1 + 2*x + 2*x^2 + 4*x^3)/((1-2*x)*(1-x)^2). - Vincenzo Librandi, Nov 02 2018
a(n) - 4*a(n-1) + 5*a(n-2) - 2*a(n-3) = 0, with n>1. - Vincenzo Librandi, Nov 02 2018
Comments