Ernst van de Kerkhof - cp's OEIS frontend

User: Ernst van de Kerkhof

Ernst van de Kerkhof has authored 2 sequences.

A276677 Number of squares added at the n-th generation of a symmetric (with 45-degree angles), non-overlapping Pythagoras tree.

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

Ernst van de Kerkhof, Sep 13 2016

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.

Colin Barker, Table of n, a(n) for n = 0..1000
Ernst van de Kerkhof, Illustration of a(7)
Index entries for linear recurrences with constant coefficients, signature (2,1,-4,2).

With an offset of 4, auxiliary sequence C(n) is equal to A077866: C(n+4) = A077866(n).

Partial sums give A276647.

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 *)

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

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)

A276647 Number of squares after the n-th generation in a symmetric (with 45-degree angles) non-overlapping Pythagoras tree.

Original entry on oeis.org

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

Ernst van de Kerkhof, Sep 13 2016

Non-overlapping is to be understood as: any two different squares in the tree can never share more than one side, disallowing area overlap. In branches where an area overlap is about to occur, growth is terminated.

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).

Partial sums of A276677.

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 *)

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

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)

cp's OEIS Frontend

User: Ernst van de Kerkhof

A276677 Number of squares added at the n-th generation of a symmetric (with 45-degree angles), non-overlapping Pythagoras tree.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula