|
%I #35 Jun 30 2018 20:54:26
%S 3,12,40,100,226,452,848,1484,2485,3976,6160,9240,13524,19320,27072,
%T 37224,50391,67188,88440,114972,147862,188188,237328,296660,367913,
%U 452816,553504,672112,811240,973488,1161984,1379856
%N Number of reversible strings with n-1 beads of 2 colors. 6 beads are black. Strings are not palindromic.
%C From _Petros Hadjicostas_, May 19 2018: (Start)
%C Let k be an integer >= 2. The g.f. of the BHK[k] transform of the sequence (c(n): n>=1), with g.f. C(x) = Sum_{n>=1} c(n)*x^n, is A_k(x) = (C(x)^k - C(x^2)^(k/2))/2 if k is even, and A_k(x) = (C(x)/2)*(C(x)^{k-1} - C(x^2)^{(k-1)/2}) if k is odd. This follows easily from the formulae in C. G. Bower's web link below about transforms.
%C When k is odd and c(n) = 1 for all n>=1, we get C(x) = x/(1-x) and A_k(x) = (1/2)*(x/(1-x))*((x/(1-x))^{k-1} - (x^2/(1-x^2))^{(k-1)/2}). If (a_k(n): n>=1) is the output sequence (with g.f. A_k(x)), then it can be proved (using Taylor expansions) that a_k(n) = (1/2)*(binomial(n-1, n-k) - binomial(floor((n-1)/2), floor((n-k)/2))) for n >= k+1. (Clearly, a_k(1) = ... = a_k(k) = 0.)
%C In this sequence, k = 7, and (according to C. G. Bower) a(n) = a_{k=7}(n) is the number of reversible non-palindromic compositions of n with 7 positive parts. If n = b_1 + b_2 + b_3 + b_4 + b_5 + b_6 + b_7 is such a composition of n (with b_i >=1), then it is equivalent to the composition n = b_7 + b_6 + b_5 + b_4 + b_3 + b_2 + b_1, and each equivalent class has two elements because here linear palindromes are not allowed as compositions of n.
%C The fact that we are finding the BHK[7] transform of 1, 1, 1, ... means that each part of each composition of n can have exactly one color (see Bower's link below about transforms).
%C In each such composition replace each b_i with one black (B) ball followed by b_i - 1 white (W) balls. Then drop the first black (B) ball. We then get a reversible non-palindromic string of length n-1 that has 6 black balls and n-7 white balls. This process, applied to the equivalent compositions n = b_1 + b_2 + b_3 + b_4 + b_5 + b_6 + b_7 = b_7 + b_6 + b_5 + b_4 + b_3 + b_2 + b_1, gives two strings of length n-1 with 6 black balls and n-7 white balls that are mirror images of each other.
%C Hence, for n>=2, a(n) = a_{k=7}(n) is also the number of reversible non-palindromic strings of length n-1 that have k-1 = 6 black balls and n-k = n-7 white balls. (Clearly, a(n) = a_{k=7}(n) > 0 only for n >= 8. For n=7, the composition 1+1+1+1+1+1+1, which corresponds to string BBBBBB, is discarded because it is palindromic.)
%C (End)
%H C. G. Bower, <a href="/transforms2.html">Transforms (2)</a>
%F "BHK[ 7 ]" (reversible, identity, unlabeled, 7 parts) transform of 1, 1, 1, 1, ...
%F Empirical G.f.: -x^8*(x^2+3)/((x-1)^7*(x+1)^3). - _Colin Barker_, Nov 24 2012
%F From _Petros Hadjicostas_, May 19 2018: (Start)
%F a(n) = (1/2)*(binomial(n-1, n-7) - binomial(floor((n-1)/2), floor((n-7)/2))) for n >= 8.
%F G.f.: (1/2)*(x/(1-x))*((x/(1-x))^6 - (x^2/(1-x^2))^3), which is the same as the g.f. given by _Colin Barker_ above.
%F (End)
%e From _Petros Hadjicostas_, May 19 2018: (Start)
%e For n=8, we have the following 3 reversible non-palindromic compositions with 7 parts of n: 1+1+1+1+1+1+2 (= 2+1+1+1+1+1+1), 1+1+1+1+1+2+1 (= 1+2+1+1+1+1+1), and 1+1+1+1+2+1+1 (= 1+1+2+1+1+1+1). Using the process described in the comments, we get the following reversible non-palindromic strings with 6 black balls and n-7=1 white balls: BBBBBBW (= WBBBBBB), BBBBBWB (= BWBBBBB), and BBBBWBB (= BBWBBBB).
%e For n=9, we get the following 12 compositions and 12 corresponding strings:
%e 1+1+1+1+1+1+3 <-> BBBBBBWW
%e 1+1+1+1+1+3+1 <-> BBBBBWWB
%e 1+1+1+1+3+1+1 <-> BBBBWWBB
%e 1+1+1+1+1+2+2 <-> BBBBBWBW
%e 1+1+1+1+2+1+2 <-> BBBBWBBW
%e 1+1+1+2+1+1+2 <-> BBBWBBBW
%e 1+1+2+1+1+1+2 <-> BBWBBBBW
%e 1+2+1+1+1+1+2 <-> BWBBBBBW
%e 1+1+1+1+2+2+1 <-> BBBBWBWB
%e 1+1+1+2+1+2+1 <-> BBBWBBWB
%e 1+1+2+1+1+2+1 <-> BBWBBBWB
%e 1+1+1+2+2+1+1 <-> BBBWBWBB
%e (End)
%Y Cf. A002620, A006584, A032091, A032092, A032094, A239572, A282011.
%K nonn
%O 8,1
%A _Christian G. Bower_
%E Definition changed slightly by _Harvey P. Dale_, Oct 02 2017
| 