A276647 Number of squares after the n-th generation in a symmetric (with 45-degree angles) non-overlapping Pythagoras tree.
1, 3, 7, 15, 31, 59, 107, 183, 303, 483, 755, 1151, 1735, 2571, 3787, 5511, 7999, 11507, 16547, 23631, 33783, 48027, 68411, 96983, 137839, 195075, 276883, 391455, 555175, 784427, 1111979, 1570599, 2225823, 3143187, 4453763, 6288623, 8909911, 12579771
Offset: 0
Links
- Colin Barker, Table of n, a(n) for n = 0..1000
- Ernst van de Kerkhof, Illustration of a(6)
- "QuantumKiwi", A Year in the Life of a Pythagoras Tree, YouTube, (2008).
- Wikipedia, Pythagoras tree (fractal)
- Index entries for linear recurrences with constant coefficients, signature (3,-1,-5,6,-2).
Crossrefs
Partial sums of A276677.
Programs
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Mathematica
TableForm[Table[{n, 20 * 2^Floor[n/2] + 28*2^Floor[(n-1)/2] - (2n^2 + 10n + 33)}, {n, 0, 100, 1}], TableSpacing -> {1, 5}] LinearRecurrence[{3,-1,-5,6,-2},{1,3,7,15,31},50] (* Harvey P. Dale, May 07 2019 *) -
PARI
Vec((1+x)^2*(1-2*x+2*x^2)/((1-x)^3*(1-2*x^2)) + O(x^50)) \\ Colin Barker, Sep 20 2016
Formula
Theorem: a(n) = 20*2^floor(n/2) + 28*2^floor((n-1)/2) - (2*n^2+10*n+33).
From Colin Barker, Sep 20 2016: (Start)
G.f.: (1+x)^2*(1-2*x+2*x^2) / ((1-x)^3*(1-2*x^2)).
a(n) = 3*a(n-1)-a(n-2)-5*a(n-3)+6*a(n-4)-2*a(n-5) for n>4.
a(n) = (-25+2^((n-1)/2)*(24-24*(-1)^n+17*sqrt(2)+17*(-1)^n*sqrt(2))-4*(1+n)-2*(1+n)*(2+n)). Therefore:
a(n) = 17*2^(n/2+1)-2*n^2-10*n-33 for n even.
a(n) = 3*2^((n+7)/2)-2*n^2-10*n-33 for n odd. (End)
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