A032098 "BHK" (reversible, identity, unlabeled) transform of 3,3,3,3,...

3, 6, 21, 87, 363, 1491, 6051, 24387, 97923, 392451, 1571331, 6288387, 25159683, 100651011, 402628611, 1610563587, 6442352643, 25769607171, 103078821891, 412316073987, 1649265868803, 6597066620931

Offset: 1

Author

Christian G. Bower

Keywords

• nonn

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) = 3 for all n >= 1, and thus, C(x) = 3*x/(1-x). Substituting into the above formula for A(x), and doing the algebra, we get A(x) = 3*x*(7*x^2 - 5*x + 1)/((1-x)*(1-2*x)*(1-4*x)), which is the formula conjectured by Colin Barker below.

Using partial fraction decomposition, we get A(x) = -21/8 + 3/(1-x) - 3/(4*(1-2*x)) + 3/(8*(1-4*x)), from which we get a(n) = 3 * (2^(2*n-3) - 2^(n-2) + 1), which is Ralf Stephan's conjecture 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) = 3 for all n >= 1, each box can have one of three colors, say A, B or C. 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) = 3 possible arrays are 1_A, 1_B, and 1_C. For n=2, the a(2) = 6 possible arrays for the boxes are 2_A, 2_B, 2_C, 1_A 1_B, 1_A 1_C, 1_B 1_C.
For n=3, the a(3) = 21 possible arrays for the boxes are:
  3_A, 3_B, 3_C (one box on the line);
  1_A 2_A, 1_A 2_B, 1_A 2_C, 1_B 2_A, 1_B 2_B, 1_B 2_C, 1_C 2_A, 1_C 2_B, 1_C 2_C    (two boxes on the line);
  1_A 1_A 1_B, 1_A 1_A 1_C, 1_A 1_B 1_B, 1_A 1_B 1_C, 1_A 1_C 1_B, 1_A 1_C 1_C,
    1_B 1_B 1_B, 1_B 1_B 1_C, 1_B 1_C 1_C (three boxes on the line).
(End)
		

Links

• C. G. Bower, Transforms (2)

Formula

Conjecture: a(n) = 3 * (2^(2*n-3) - 2^(n-2) + 1). - Ralf Stephan, Sep 11 2003

From Colin Barker, Sep 22 2012: (Start)

Conjecture: a(n) = 7*a(n-1) - 14*a(n-2) + 8*a(n-3).

G.f.: 3*x*(1-5*x+7*x^2)/((1-x)*(1-2*x)*(1-4*x)). (End)