

A032094


Number of reversible strings with n1 beads of 2 colors. 7 beads are black. String is not palindromic.


7



4, 16, 60, 160, 396, 848, 1716, 3200, 5720, 9696, 15912, 25152, 38760, 58080, 85272, 122496, 173052, 240240, 328900, 443872, 592020, 780208, 1017900, 1314560, 1682928, 2135744, 2689808, 3361920, 4173840, 5147328, 6310128, 7689984, 9321780, 11240400
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OFFSET

9,1


COMMENTS

If the offset is changed to 3, this is the 2nd Witt transform of A000292 [Moree].  R. J. Mathar, Nov 08 2008
Also 7th column of A159916, i.e., number of 7element subsets of {1,...,n1} whose elements add up to an odd integer.  M. F. Hasler, May 02 2009
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 = 8, and (according to C. G. Bower) a(n) = a_{k=8}(n) is the number of reversible nonpalindromic compositions of n with 8 positive parts. If n = b_1 + b_2 + b_3 + b_4 + b_5 + b_6 + b_7 + b_8 is such a composition of n (with b_i >= 1), then it is equivalent to the composition n = b_8 + 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.
The fact that we are finding the BHK[8] 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 k1 = 7 black balls and nk = n8 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_8 = b_8 + b_7 + b_6 + b_5 + b_4 + b_3 + b_2 + b_1, gives two strings of length n1 with 7 black balls and n8 white balls that are mirror images of each other.
Hence, for n >= 2, a(n) = a_{k=8}(n) is also the number of reversible nonpalindromic strings of length n1 that have k1 = 7 black balls and nk = n8 white balls. (Clearly, a(n) = a_{k=8}(n) > 0 only for n >= 9.)
(End)


LINKS

Colin Barker, Table of n, a(n) for n = 9..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 (4,2,12,17,8,28,8,17,12,2,4,1).


FORMULA

"BHK[ 8 ]" (reversible, identity, unlabeled, 8 parts) transform of 1, 1, 1, 1, ...
From R. J. Mathar, Nov 08 2008: (Start)
G.f.: 4*x^9*(1+x^2)/((1x)^8*(1+x)^4).
a(n) = 4*A031164(n9). (End)
From Colin Barker, Mar 07 2015: (Start)
a(n) = (n^728*n^6+322*n^51960*n^4+6664*n^311872*n^2+8448*n)/10080 if n is even.
a(n) = (n^728*n^6+322*n^51960*n^4+6769*n^313132*n^2+13068*n5040)/10080 if n is odd.
(End)
From Petros Hadjicostas, May 19 2018: (Start)
a(n) = (1/2)*(binomial(n1, n8)  binomial((n/2)1, (n8)/2)) if n is even.
a(n) = (1/2)*binomial(n1, n8) if n is odd.
G.f.: (1/2)*((x/(1x))^8  (x^2/(1x^2))^4).
These formulae agree with the above formulae by R. J. Mathar and Colin Barker. Clearly, the first two formulae (those about a(n)) can be combined into the one given by M. F. Hasler below in the PROG section.
(End)


MATHEMATICA

f[n_] := Binomial[n  1, n  8]/2  If[ OddQ@ n, 0, Binomial[(n/2)  1, (n  8)/2]/2]; Array[f, 36, 9] (* or *)
CoefficientList[ Series[ 4x^9 (x^2 + 1)/((x  1)^8 (x + 1)^4), {x, 0, 40}], x] (* or *)LinearRecurrence[{4, 2, 12, 17, 8, 28, 8, 17, 12, 2, 4, 1}, {4, 16, 60, 160, 396, 848, 1716, 3200, 5720, 9696, 15912, 25152}, 34] (* Robert G. Wilson v, May 20 2018 *)


PROG

(PARI) A032094(n)=(binomial(n, 7)if(n%2, binomial(n\2, 3)))\2 \\ M. F. Hasler, May 02 2009
(PARI) Vec(4*x^9*(1+x^2)/((1x)^8*(1+x)^4) + O(x^100)) \\ Colin Barker, Mar 07 2015


CROSSREFS

Cf. A005995, A018210, A032091, A032092, A032093, A159916.  M. F. Hasler, May 02 2009 and Petros Hadjicostas, May 19 2018
Sequence in context: A047123 A297096 A330791 * A282083 A261563 A265955
Adjacent sequences: A032091 A032092 A032093 * A032095 A032096 A032097


KEYWORD

nonn,easy


AUTHOR

Christian G. Bower


STATUS

approved



