cp's OEIS Frontend

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

%I A032098 #37 May 25 2024 19:14:08
%S A032098 3,6,21,87,363,1491,6051,24387,97923,392451,1571331,6288387,25159683,
%T A032098 100651011,402628611,1610563587,6442352643,25769607171,103078821891,
%U A032098 412316073987,1649265868803,6597066620931
%N A032098 "BHK" (reversible, identity, unlabeled) transform of 3,3,3,3,...
%C A032098 From _Petros Hadjicostas_, May 20 2018: (Start)
%C A032098 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.)
%C A032098 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.)
%C A032098 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).
%C A032098 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.
%C A032098 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.
%C A032098 (End)
%H A032098 C. G. Bower, <a href="/transforms2.html">Transforms (2)</a>
%F A032098 Conjecture: a(n) = 3 * (2^(2*n-3) - 2^(n-2) + 1). - _Ralf Stephan_, Sep 11 2003
%F A032098 From _Colin Barker_, Sep 22 2012: (Start)
%F A032098 Conjecture: a(n) = 7*a(n-1) - 14*a(n-2) + 8*a(n-3).
%F A032098 G.f.: 3*x*(1-5*x+7*x^2)/((1-x)*(1-2*x)*(1-4*x)). (End)
%e A032098 From _Petros Hadjicostas_, May 20 2018: (Start)
%e A032098 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).
%e A032098 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.
%e A032098 For n=3, the a(3) = 21 possible arrays for the boxes are:
%e A032098   3_A, 3_B, 3_C (one box on the line);
%e A032098   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);
%e A032098   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,
%e A032098     1_B 1_B 1_B, 1_B 1_B 1_C, 1_B 1_C 1_C (three boxes on the line).
%e A032098 (End)
%K A032098 nonn
%O A032098 1,1
%A A032098 _Christian G. Bower_