

A032092


Number of reversible strings with n1 beads of 2 colors. 5 beads are black. String is not palindromic.


7



3, 9, 28, 60, 126, 226, 396, 636, 1001, 1491, 2184, 3080, 4284, 5796, 7752, 10152, 13167, 16797, 21252, 26532, 32890, 40326, 49140, 59332, 71253, 84903, 100688, 118608, 139128, 162248, 188496, 217872, 250971, 287793, 329004, 374604, 425334, 481194, 543004
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OFFSET

7,1


COMMENTS

If the offset is changed to 3, this is the 2nd Witt transform of A000217 [Moree].  R. J. Mathar, Nov 08 2008
From Petros Hadjicostas, May 19 2018: (Start)
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)^{k1}  C(x^2)^{(k1)/2}) if k is odd. This follows easily from the formulae in C. G. Bower's web link below about transforms.
When k is even and c(n) = 1 for all n >= 1, we get C(x) = x/(1x) and A_k(x) = (1/2)*((x/(1x))^k  (x^2/(1x^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(n1, nk)  binomial((n/2)1, (nk)/2)) for even n >= k+1 and a_k(n) = (1/2)*binomial(n1, nk) for odd n >= k+1. (Clearly, a_k(1) = ... = a_k(k) = 0.)
In this sequence, k = 6, and (according to C. G. Bower) a(n) = a_{k=6}(n) is the number of reversible nonpalindromic 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.
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).
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 nonpalindromic string of length n1 that has 5 black balls and n6 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 n1 with 5 black balls and n6 white balls that are mirror images of each other.
Hence, for n >= 2, a(n) = a_{k=6}(n) is also the number of reversible nonpalindromic strings of length n1 that have k1 = 5 black balls and nk = n6 white balls. (Clearly, a(n) = a_{k=6}(n) > 0 only for n >= 7.)
(End)


LINKS

Colin Barker, Table of n, a(n) for n = 7..1000
C. G. Bower, Transforms (2)
Pieter Moree, The formal series Witt transform, Discr. Math. no. 295 vol. 13 (2005) 143160.  R. J. Mathar, Nov 08 2008
Index entries for linear recurrences with constant coefficients, signature (3,0,8,6,6,8,0,3,1).


FORMULA

"BHK[ 6 ]" (reversible, identity, unlabeled, 6 parts) transform of 1, 1, 1, 1, ...
G.f.: x^7*(3+x^2)/((1x)^6*(1+x)^3).  R. J. Mathar, Nov 08 2008
From Colin Barker, Mar 07 2015: (Start)
a(n) = (2*n^530*n^4+170*n^3480*n^2+728*n480)/480 if n is even.
a(n) = (2*n^530*n^4+170*n^3450*n^2+548*n240)/480 if n is odd.
(End)
From Petros Hadjicostas, May 19 2018: (Start)
a(n) = (1/2)*(binomial(n1, n6)  binomial((n/2)1, (n6)/2)) if n is even.
a(n) = (1/2)*binomial(n1, n6) if n is odd.
G.f.: (1/2)*((x/(1x))^6  (x^2/(1x^2))^3).
These formulae agree with the above formulae by R. J. Mathar and C. Barker.
(End)


EXAMPLE

From Petros Hadjicostas, May 19 2018: (Start)
For n=7, we have the following 3 reversible nonpalindromic 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 nonpalindromic strings with 5 black balls and n6 = 1 white balls: BBBBBW (= WBBBBB), BBBBWB (= BWBBBB), and BBBWBB (= BBWBBB).
For n=8, we get the following 9 compositions and 9 corresponding strings:
1+1+1+1+1+3 <> BBBBBWW
1+1+1+1+3+1 <> BBBBWWB
1+1+1+3+1+1 <> BBBWWBB
1+1+1+1+2+2 <> BBBBWBW
1+1+1+2+1+2 <> BBBWBBW
1+1+2+1+1+2 <> BBWBBBW
1+2+1+1+1+2 <> BWBBBBW
1+1+1+2+2+1 <> BBBWBWB
1+1+2+1+2+1 <> BBWBBWB
(End)


MATHEMATICA

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 *)
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 *)
CoefficientList[ Series[(x^7 (x^2 + 3))/((x  1)^6 (x + 1)^3), {x, 0, 46}], x] (* Robert G. Wilson v, May 20 2018 *)


PROG

(PARI) Vec(x^7*(3+x^2)/((1x)^6*(1+x)^3) + O(x^100)) \\ Colin Barker, Mar 07 2015


CROSSREFS

Cf. A002620, A006584, A032091, A032093, A032094, A282011.
Sequence in context: A102558 A022767 A015638 * A026524 A282081 A022631
Adjacent sequences: A032089 A032090 A032091 * A032093 A032094 A032095


KEYWORD

nonn,easy


AUTHOR

Christian G. Bower


STATUS

approved



