A032097 "BHK" (reversible, identity, unlabeled) transform of 2,1,1,1,...

2, 2, 5, 14, 39, 107, 289, 772, 2047, 5402, 14213, 37325, 97905, 256622, 672337, 1760998, 4611643, 12075527, 31617521, 82781216, 216732891, 567428402, 1485570025, 3889310329, 10182407329, 26657986682, 69791674109, 182717232062, 478360339887

Offset: 1

Author

Christian G. Bower

Keywords

• nonn
• easy

Comments

First five terms match the total dominating number of the n-Sierpinski sieve graph. - Eric W. Weisstein, Apr 18 2018

From Petros Hadjicostas, May 20 2018: (Start)

Using the formulae in C. B. Bower's web link below about transforms, it can be proved that, for k >= 2, the BHK[k] transform of sequence (c(n): n >= 1), which has g.f. C(x) = Sum_{n >= 1} c(n)*x^n, has generating function B_k(x) = (1/2)*(C(x)^k - C(x^2)^{k/2}) if k is even, and B_k(x) = C(x)*B_{k-1}(x) = (C(x)/2)*(C(x)^{k-1} - C(x^2)^{(k-1)/2}) if k is odd. For k=1, Bower assumes that the BHK[k=1] transform of (c(n): n >= 1) is itself, which means that the g.f. of the output sequence is C(x). (This assumption is not accepted by all mathematicians because a sequence of length 1 is not only reversible but palindromic as well.)

Since a(m) = BHK(c(n): n >= 1)(m) = Sum_{k=1..m} BHK[k](c(n): n >= 1)(m) for m = 1,2,3,..., it can be easily proved (using sums of infinite geometric series) that the g.f. of BHK(c(n): n >= 1) is A(x) = (C(x)^2 - C(x^2))/(2*(1-C(x))*(1-C(x^2))) + C(x). (The extra C(x) is due of course to the special assumption made for the BHK[k=1] transform.)

Here, BHK(c(n): n >= 1)(m) indicates the m-th element of the output sequence when the transform is BHK and the input sequence is (c(n): n >= 1). Similarly, BHK[k](c(n): n >= 1)(m) indicates the m-th element of the output sequence when the transform is BHK[k] (i.e., with k boxes) and the input sequence is (c(n): n >= 1).

For the current sequence, c(1) = 2 and c(n) = 1 for all n >= 2, and thus, C(x) = x + x/(1-x). Substituting into the above formula for A(x), and doing the algebra, we get A(x) = x*(x^5-4*x^4+x^3+9*x^2-8*x+2)/((x-1)*(x^2-3*x+1)*(x^2+x-1)), which is the formula given by Colin Barker below.

(End)

Examples

			From _Petros Hadjicostas_, May 20 2018: (Start)
According to C. G. Bower, in his website above, we have boxes of different colors and sizes (the size of the box is determined by the number of balls it can hold). Since c(1) = 2, each box of size 1 can have one of two colors, say A and B. On the other hand, since c(n) = 1 for n >= 2, each box of size >= 2 can be of one color only (and there is no need to specify it). Then a(n) = BHK(c(n): n >= 1)(n) = number of ways we can have boxes on a line such that the total number of balls is n and the array of boxes is reversible but not palindromic (with the exception when having only one box on the line).
Hence, for n=1, the a(1) = 2 possible arrays are 1_A and 1_B. For n=2, the a(2) = 2 possible arrays for the boxes are 1_A 1_B and 2. (Note that 1_A 1_B is not palindromic because the boxes have different colors even though each one has only 1 ball.)
For n=3, the a(3) = 5 possible arrays for the boxes are:
  3 (one box on the line);
  1_A 2, 1_B 2 (two boxes on the line);
  1_A 1_B 1_B, 1_A 1_A 1_B (three boxes on the line).
For n=4, the a(4) = 14 possible arrays for the boxes are:
  4 (one box on the line);
  1_A 3, 1_B 3 (two boxes on the line);
  1_A 1_A 2, 1_A 1_B 2, 1_B 1_A 2, 1_B 1_B 2, 1_A 2 1_B (three boxes on the line);
  1_A 1_A 1_A 1_B, 1_A 1_A 1_B 1_A, 1_A 1_A 1_B 1_B, 1_A 1_B 1_B 1_B,
    1_B 1_A 1_B 1_B, 1_A 1_B 1_A 1_B (four boxes on the line).
(End)
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• C. G. Bower, Transforms (2)
• Index entries for linear recurrences with constant coefficients, signature (5,-7,1,3,-1).

Programs

Magma

[2] cat [1/2*(Fibonacci(2*n+1) - Fibonacci(n+2) + 2): n in [2..30]]; // Vincenzo Librandi, Oct 19 2013

Mathematica

CoefficientList[Series[(x^5 - 4 x^4 + x^3 + 9 x^2 - 8 x + 2)/((x - 1) (x^2 - 3 x + 1) (x^2 + x - 1)), {x, 0, 40}], x] (* Vincenzo Librandi, Oct 19 2013 *)
Join[{2}, Table[(Fibonacci[2 n + 1] - Fibonacci[n + 2])/2 + 1, {n, 2, 20}]] (* Eric W. Weisstein, Apr 18 2018 *)
Join[{2}, LinearRecurrence[{5, -7, 1, 3, -1}, {2, 5, 14, 39, 107}, 20]] (* Eric W. Weisstein, Apr 18 2018 *)

Formula

For n > 1, a(n) = (1/2)*(F(2n+1) - F(n+2) + 2), where F(n) = A000045(n). - Ralf Stephan, May 04 2004

G.f.: x*(x^5-4*x^4+x^3+9*x^2-8*x+2)/((x-1)*(x^2-3*x+1)*(x^2+x-1)). - Colin Barker, Sep 22 2012

Extensions

More terms from Vincenzo Librandi, Oct 19 2013