|
%I #48 Jul 01 2018 11:09:55
%S 3,9,28,60,126,226,396,636,1001,1491,2184,3080,4284,5796,7752,10152,
%T 13167,16797,21252,26532,32890,40326,49140,59332,71253,84903,100688,
%U 118608,139128,162248,188496,217872,250971,287793,329004,374604,425334,481194,543004
%N Number of reversible strings with n-1 beads of 2 colors. 5 beads are black. String is not palindromic.
%C If the offset is changed to 3, this is the 2nd Witt transform of A000217 [Moree]. - _R. J. Mathar_, Nov 08 2008
%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 even and c(n) = 1 for all n >= 1, we get C(x) = x/(1-x) and A_k(x) = (1/2)*((x/(1-x))^k - (x^2/(1-x^2))^{k/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((n/2)-1, (n-k)/2)) for even n >= k+1 and a_k(n) = (1/2)*binomial(n-1, n-k) for odd n >= k+1. (Clearly, a_k(1) = ... = a_k(k) = 0.)
%C In this sequence, k = 6, and (according to C. G. Bower) a(n) = a_{k=6}(n) is the number of reversible non-palindromic compositions of n with 6 positive parts. If n = b_1 + b_2 + b_3 + b_4 + b_5 + b_6 is such a composition of n (with b_i >= 1), then it is equivalent to the composition n = 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[6] 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 5 black balls and n-6 white balls. This process, applied to the equivalent compositions n = b_1 + b_2 + b_3 + b_4 + b_5 + b_6 = b_6 + b_5 + b_4 + b_3 + b_2 + b_1, gives two strings of length n-1 with 5 black balls and n-6 white balls that are mirror images of each other.
%C Hence, for n >= 2, a(n) = a_{k=6}(n) is also the number of reversible non-palindromic strings of length n-1 that have k-1 = 5 black balls and n-k = n-6 white balls. (Clearly, a(n) = a_{k=6}(n) > 0 only for n >= 7.)
%C (End)
%H Colin Barker, <a href="/A032092/b032092.txt">Table of n, a(n) for n = 7..1000</a>
%H C. G. Bower, <a href="/transforms2.html">Transforms (2)</a>
%H Pieter Moree, <a href="http://dx.doi.org/10.1016/j.disc.2005.03.004">The formal series Witt transform</a>, Discr. Math. no. 295 vol. 1-3 (2005) 143-160. - _R. J. Mathar_, Nov 08 2008
%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (3,0,-8,6,6,-8,0,3,-1).
%F "BHK[ 6 ]" (reversible, identity, unlabeled, 6 parts) transform of 1, 1, 1, 1, ...
%F G.f.: x^7*(3+x^2)/((1-x)^6*(1+x)^3). - _R. J. Mathar_, Nov 08 2008
%F From _Colin Barker_, Mar 07 2015: (Start)
%F a(n) = (2*n^5-30*n^4+170*n^3-480*n^2+728*n-480)/480 if n is even.
%F a(n) = (2*n^5-30*n^4+170*n^3-450*n^2+548*n-240)/480 if n is odd.
%F (End)
%F From _Petros Hadjicostas_, May 19 2018: (Start)
%F a(n) = (1/2)*(binomial(n-1, n-6) - binomial((n/2)-1, (n-6)/2)) if n is even.
%F a(n) = (1/2)*binomial(n-1, n-6) if n is odd.
%F G.f.: (1/2)*((x/(1-x))^6 - (x^2/(1-x^2))^3).
%F These formulae agree with the above formulae by R. J. Mathar and C. Barker.
%F (End)
%e From _Petros Hadjicostas_, May 19 2018: (Start)
%e For n=7, we have the following 3 reversible non-palindromic compositions with 6 parts of n: 1+1+1+1+1+2 (= 2+1+1+1+1+1), 1+1+1+1+2+1 (= 1+2+1+1+1+1), and 1+1+1+2+1+1 (= 1+1+2+1+1+1). Using the process described in the comments, we get the following reversible non-palindromic strings with 5 black balls and n-6 = 1 white balls: BBBBBW (= WBBBBB), BBBBWB (= BWBBBB), and BBBWBB (= BBWBBB).
%e For n=8, we get the following 9 compositions and 9 corresponding strings:
%e 1+1+1+1+1+3 <-> BBBBBWW
%e 1+1+1+1+3+1 <-> BBBBWWB
%e 1+1+1+3+1+1 <-> BBBWWBB
%e 1+1+1+1+2+2 <-> BBBBWBW
%e 1+1+1+2+1+2 <-> BBBWBBW
%e 1+1+2+1+1+2 <-> BBWBBBW
%e 1+2+1+1+1+2 <-> BWBBBBW
%e 1+1+1+2+2+1 <-> BBBWBWB
%e 1+1+2+1+2+1 <-> BBWBBWB
%e (End)
%t LinearRecurrence[{3,0,-8,6,6,-8,0,3,-1},{3,9,28,60,126,226,396,636,1001},50] (* _Harvey P. Dale_, Mar 19 2017 *)
%t f[n_] := Binomial[n - 1, n - 6]/2 - If[ OddQ@ n, 0, Binomial[(n/2) - 1, (n - 6)/2]/2]; Array[a, 40, 7] (* or *)
%t CoefficientList[ Series[(x^7 (x^2 + 3))/((x - 1)^6 (x + 1)^3), {x, 0, 46}], x] (* _Robert G. Wilson v_, May 20 2018 *)
%o (PARI) Vec(x^7*(3+x^2)/((1-x)^6*(1+x)^3) + O(x^100)) \\ _Colin Barker_, Mar 07 2015
%Y Cf. A002620, A006584, A032091, A032093, A032094, A282011.
%K nonn,easy
%O 7,1
%A _Christian G. Bower_
|
