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A276677
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Number of squares added at the n-th generation of a symmetric (with 45-degree angles), non-overlapping Pythagoras tree.
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2
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1, 2, 4, 8, 16, 28, 48, 76, 120, 180, 272, 396, 584, 836, 1216, 1724, 2488, 3508, 5040, 7084, 10152, 14244, 20384, 28572, 40856, 57236, 81808, 114572, 163720, 229252, 327552, 458620, 655224, 917364, 1310576, 1834860, 2621288, 3669860, 5242720, 7339868
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OFFSET
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0,2
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COMMENTS
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The auxiliary sequence C(n), which appears in the recurrence relation for a(n), is defined as the number of collisions (squares touching each other, halting tree growth at that point) in generation n.
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LINKS
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FORMULA
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a(0) = 1, a(n) = 2*a(n-1) - 4*C(n-1), where:
C(0) = 0; for n >= 1, C(n) = C(n-1) + 2^(floor(n/2)-1) - 1. Also:
C(0) = 0; for n >= 1, C(n) = 2^floor(n/2) + 2^floor((n-1)/2) - (n+1).
a(0) = 1; for n >= 1, a(n) = 6*2^floor(n/2) + 8*2^floor((n-1)/2) - (4*n+8).
All formulas are proved.
G.f.: (1 + x)^2*(1 - 2*x + 2*x^2) / ((1 - x)^2*(1 - 2*x^2)).
a(n) = 2*a(n-1) + a(n-2) - 4*a(n-3) + 2*a(n-4) for n>4.
a(n) = -4+2^((n-1)/2)*(7-7*(-1)^n+5*sqrt(2)+5*(-1)^n*sqrt(2))-4*(1+n) for n>0. Therefore:
a(n) = 5*2^(n/2+1)-8-4*n for n>0 and even;
a(n) = 7*2^((n+1)/2)-8-4*n for n>0 and odd. (End)
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MATHEMATICA
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TableForm[Table[{n, 6*2^Floor[n/2] + 8*2^Floor[(n-1)/2] - (4n + 8)}, {n, 1, 100, 1}], TableSpacing -> {1, 5}]
LinearRecurrence[{2, 1, -4, 2}, {1, 2, 4, 8, 16}, 70] (* Harvey P. Dale, Jan 21 2019 *)
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PROG
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(PARI) Vec((1+x)^2*(1-2*x+2*x^2)/((1-x)^2*(1-2*x^2)) + O(x^50)) \\ Colin Barker, Sep 20 2016
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CROSSREFS
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With an offset of 4, auxiliary sequence C(n) is equal to A077866: C(n+4) = A077866(n).
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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