

A032096


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


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
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OFFSET

1,1


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_{k1}(x) = (C(x)/2)*(C(x)^{k1}  C(x^2)^{(k1)/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 not only is 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*(1C(x))*(1C(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 mth 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 mth 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/(1x). Substituting into the above formula for A(x), and doing the algebra, we get A(x) = x*(25*x4*x^2+15*x^3)/((1x)*(13*x)*(13*x^2)), which is the formula conjectured by Colin Barker below.
Using partial fraction decomposition, we get A(x) = (1/3) + 2*x/(1x) + (1/3)/(13*x)  (1/3 + 1/(2*sqrt(3)))/(1sqrt(3)*x) + (1/3 + 1/(2*sqrt(3)))/(1 + sqrt(3)*x). From this, we get a(n) = 2 + 3^{n1}  2*3^{(n/2) 1} when n is even and a(n) = 2 + 3^{n1} 3^{(n1)/2} when n is odd, which verify Ralf Stephan's formula below.
(End)


LINKS

Vincenzo Librandi, Table of n, a(n) for n = 1..200
C. G. Bower, Transforms (2)


FORMULA

a(n) = (1/6)*((1)^n5)*3^floor(n/2) + 3^(n1) + 2.  Ralf Stephan, May 11 2004
From Colin Barker, Sep 22 2012: (Start)
Conjecture: a(n) = 4*a(n1)  12*a(n3) + 9*a(n4).
G.f.: x*(25*x4*x^2+15*x^3)/((1x)*(13*x)*(13*x^2)). (End)


EXAMPLE

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)


MATHEMATICA

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


PROG

(MAGMA) [(1/6)*((1)^n5)*3^Floor(n/2) + 3^(n1) + 2: n in [1..30]]; // Vincenzo Librandi, Oct 19 2013


CROSSREFS

Sequence in context: A263459 A261061 A086628 * A301462 A120763 A120708
Adjacent sequences: A032093 A032094 A032095 * A032097 A032098 A032099


KEYWORD

nonn


AUTHOR

Christian G. Bower


STATUS

approved



