A032096 "BHK" (reversible, identity, unlabeled) transform of 2,2,2,2,...

2, 3, 8, 23, 74, 227, 704, 2135, 6482, 19523, 58808, 176663, 530714, 1592867, 4780784, 14344535, 43040162, 129127043, 387400808, 1162222103, 3486725354, 10460235107, 31380882464, 94142824535, 282429005042

Offset: 1

Author

Christian G. Bower

Keywords

• nonn
• easy

Comments

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(n) = 2 for all n >= 1, and thus, C(x) = 2*x/(1-x). Substituting into the above formula for A(x), and doing the algebra, we get A(x) = x*(2-5*x-4*x^2+15*x^3)/((1-x)*(1-3*x)*(1-3*x^2)), which is the formula conjectured by Colin Barker below.

Using partial fraction decomposition, we get A(x) = (1/3) + 2*x/(1-x) + (1/3)/(1-3*x) - (1/3 + 1/(2*sqrt(3)))/(1-sqrt(3)*x) + (-1/3 + 1/(2*sqrt(3)))/(1 + sqrt(3)*x). From this, we get a(n) = 2 + 3^{n-1} - 2*3^{(n/2) -1} when n is even and a(n) = 2 + 3^{n-1} -3^{(n-1)/2} when n is odd, which verify Ralf Stephan's formula 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(n) = 2 for all n >= 1, each box can have one of two colors, say A and B. Then a(n) = BIK(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) = 3 possible arrays for the boxes are 1_A 1_B, 2_A, and 2_B. (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) = 8 possible arrays for the boxes are:
  3_A, 3_B (one box on the line);
  1_A 2_B, 1_B 2_A, 1_A 2_A, 1_B 2_B (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) = 23 possible arrays for the boxes are:
  4_A, 4_B (one box on the line);
  2_A 2_B, 3_A 1_A, 3_A 1_B, 3_B 1_A, 3_B 1_B (two boxes on the line);
  1_A 1_A 2_A, 1_A 1_A 2_B, 1_A 1_B 2_A, 1_A 1_B 2_B, 1_B 1_A 2_A, 1_B 1_A 2_B,
    1_B 1_B 2_A, 1_B 1_B 2_B, 1_A 2_A 1_B, 1_A 2_B 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_B 1_A 1_B 1_A (for boxes on the line).
(End)
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..200
• C. G. Bower, Transforms (2)
• Index entries for linear recurrences with constant coefficients, signature (4,0,-12,9).

Programs

Magma

[(1/6)*((-1)^n-5)*3^Floor(n/2) + 3^(n-1) + 2: n in [1..30]]; // Vincenzo Librandi, Oct 19 2013

Mathematica

Table[((1/6) ((-1)^n - 5) 3^(Floor[n/2]) + 3^(n - 1) + 2), {n, 1, 40}] (* Vincenzo Librandi, Oct 19 2013 *)

Formula

a(n) = (1/6)*((-1)^n-5)*3^floor(n/2) + 3^(n-1) + 2. - Ralf Stephan, May 11 2004

From Colin Barker, Sep 22 2012: (Start)

Conjecture: a(n) = 4*a(n-1) - 12*a(n-3) + 9*a(n-4).

G.f.: x*(2-5*x-4*x^2+15*x^3)/((1-x)*(1-3*x)*(1-3*x^2)). (End)

3*a(n) = 6+3^n-A038754(n+1). - R. J. Mathar, Mar 24 2023