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A276677
Number of squares added at the n-th generation of a symmetric (with 45-degree angles), non-overlapping Pythagoras tree.
2
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
OFFSET
0,2
COMMENTS
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.
FORMULA
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.
From Colin Barker, Sep 20 2016: (Start)
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)
MATHEMATICA
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 *)
PROG
(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
CROSSREFS
With an offset of 4, auxiliary sequence C(n) is equal to A077866: C(n+4) = A077866(n).
Partial sums give A276647.
Sequence in context: A260881 A089055 A305122 * A112128 A208933 A227036
KEYWORD
nonn,easy
AUTHOR
STATUS
approved