|
| |
|
|
A032091
|
|
Number of reversible strings with n-1 beads of 2 colors. 4 beads are black. String is not palindromic.
|
|
3
|
|
|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
6,1
|
|
|
COMMENTS
|
Also, number of 4-element subsets of the set {1,...,n-1} whose elements sum up to an odd integer, i.e. 4-th column of the triangle A158816, cf. there. [From M. F. Hasler, May 01 2009]
|
|
|
LINKS
|
Table of n, a(n) for n=6..43.
C. G. Bower, Transforms (2)
Elizabeth Wilmer, Notes on Stephan's conjectures 72, 73 and 74
|
|
|
FORMULA
|
"BHK[ 5 ]" (reversible, identity, unlabeled, 5 parts) transform of 1, 1, 1, 1...
Contribution from M. F. Hasler, May 01 2009: (Start)
G.f.: 2 x^5 (1-x)^-5 (1+x)^-2.
a(n) = [(n-5)(n-3)(n-1)^2 + (6n-15) X[2Z](n)]/48, where X[2Z] is the characteristic function of 2Z. (End)
|
|
|
PROG
|
Contribution from M. F. Hasler, May 01 2009: (Start)
(PARI) A032091(n)=polcoeff(2/(1-x)^5/(1+x)^2+O(x^(n-5)), n-6)
A032091(n)=((n-5)*(n-3)*(n-1)^2+if(n%2==0, 6*n-15))/48 \\\\ (End)
|
|
|
CROSSREFS
|
a(n+6) = 2*A002624(n).
Sequence in context: A192735 A218212 A171218 * A182994 A192706 A199477
Adjacent sequences: A032088 A032089 A032090 * A032092 A032093 A032094
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
Christian G. Bower
|
|
|
STATUS
|
approved
|
| |
|
|