

A032091


Number of reversible strings with n1 beads of 2 colors. 4 beads are black. String is not palindromic.


6



2, 6, 16, 32, 60, 100, 160, 240, 350, 490, 672, 896, 1176, 1512, 1920, 2400, 2970, 3630, 4400, 5280, 6292, 7436, 8736, 10192, 11830, 13650, 15680, 17920, 20400, 23120, 26112, 29376, 32946, 36822, 41040, 45600, 50540, 55860, 61600, 67760, 74382, 81466, 89056
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OFFSET

6,1


COMMENTS

Also, number of 4element subsets of the set {1,...,n1} whose elements sum up to an odd integer, i.e., 4th column of the triangle A159916, cf. there.  M. F. Hasler, May 01 2009
Also, if the offset is changed to 3, so that a(3)=2, a(n) = number of nonequivalent (mod D_3) ways to place 2 indistinguishable points on a triangular grid of side n so that they are not adjacent.  Heinrich Ludwig, Mar 23 2014


LINKS

Colin Barker, Table of n, a(n) for n = 6..1000
C. G. Bower, Transforms (2)
Elizabeth Wilmer, Notes on Stephan's conjectures 72, 73 and 74
Index entries for linear recurrences with constant coefficients, signature (3,1,5,5,1,3,1)


FORMULA

"BHK[ 5 ]" (reversible, identity, unlabeled, 5 parts) transform of 1, 1, 1, 1...
From M. F. Hasler, May 01 2009: (Start)
G.f.: 2*x^6 / ((x1)^5*(x+1)^2).  Corrected by Colin Barker, Mar 07 2015
a(n) = [(n5)(n3)(n1)^2 + (6n15) X[2Z](n)]/48, where X[2Z] is the characteristic function of 2Z.
(End)
From Colin Barker, Mar 07 2015: (Start)
a(n) = (n^410*n^3+32*n^232*n)/48 if n is even.
a(n) = (n^410*n^3+32*n^238*n+15)/48 if n is odd.
(End)


PROG

(PARI) A032091(n)=polcoeff(2/(1x)^5/(1+x)^2+O(x^(n5)), n6)
A032091(n)=((n5)*(n3)*(n1)^2+if(n%2==0, 6*n15))/48 \\ M. F. Hasler, May 01 2009


CROSSREFS

a(n+6) = 2*A002624(n), A239572.
Sequence in context: A192735 A218212 A171218 * A182994 A235792 A192706
Adjacent sequences: A032088 A032089 A032090 * A032092 A032093 A032094


KEYWORD

nonn,easy


AUTHOR

Christian G. Bower


STATUS

approved



