|
| |
|
|
A197424
|
|
Number of subsets of {1, 2, ..., 4*n + 2} which do not contain two numbers whose difference is 4.
|
|
3
|
|
|
|
4, 36, 225, 1600, 10816, 74529, 509796, 3496900, 23961025, 164249856, 1125736704, 7716041281, 52886200900, 362488284900, 2484529385121, 17029223715904, 116720020119616, 800010960336225, 5483356589096100, 37583485459535236, 257601040852192129
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,1
|
|
|
COMMENTS
|
This sequence is an instance of a general result given in Math. Mag. Problem 1854 (see Links).
a(n) is the number of edge covers of a caterpillar graph with spine P_(4n+5), one pendant attached at vertex n+2 counting from the left end of the spine, a second one at 2n+3 and a third at 3n+4. The caterpillar graph for n=1 is as follows:
* * *
| | |
*--*--*--v1--*--v2--*--*--*
Each pendant edge must be included in an edge cover leaving only the middle six edges flexible. Every vertex except v1 and v2 is incident with at least one of the pendant edges. Therefore, if we label the middle six edges in the spine with numbers 3, 1, 5, 2, 6, 4 (starting from the left), the edges have to be chosen so that both 1,5 and 2,6 cannot be missing. This corresponds to choosing subsets of {1, 2, ..., 6} which do not contain two numbers whose difference is 4. (End)
|
|
|
REFERENCES
|
F. Alayont and E. Henning, Edge Covers of Caterpillars, Cycles with Pendants, and Spider Graphs; submitted.
|
|
|
LINKS
|
Marian Tetiva, Problem 1854, Mathematics Magazine, 84 (2011) 300.
|
|
|
FORMULA
|
a(n) = F(n+2)^2*F(n+3)^2 = A001654(n+2)^2, where F(n) denotes the n-th Fibonacci number A000045(n).
G.f.: ( -4-16*x+15*x^2+5*x^3-x^4 ) / ( (x-1)*(x^2+3*x+1)*(x^2-7*x+1) ). - R. J. Mathar, Oct 15 2011
For L=Lucas, a(n) = (1/25)*(L(2*(2*n+5)) - 2*(-1)^n*L(2*n+5) - 1), an instance of (F(n+p)*F(n+q))^2 = (1/25)*(L(2*(2*n+p+q)) - 2*(-1)^(n+q)*L(p-q)*L(2*n+p+q) + L(2*(p-q)) + 4*(-1)^(p-q)) which follows from squaring a specialization of identity 17b in the Vajda reference at A000045, F(n+p)*F(n+q) = (1/5)*(L(2*n+p+q) - (-1)*(n+q)*L(p-q)), then applying Vajda 17c, L(n)^2 = L(2*n) + 2*(-1)^n, to the expansion. - Ehren Metcalfe, Mar 26 2016
|
|
|
MATHEMATICA
|
Table[(1/25) (LucasL[2 (2 n + 5)] - 2 (-1)^n LucasL[2 n + 5] - 1), {n, 0, 20}] (* Michael De Vlieger, Mar 27 2016 *)
|
|
|
PROG
|
(PARI) Vec((4+16*x-15*x^2-5*x^3+x^4) / ((1-x)*(1-7*x+x^2)*(1+3*x+x^2)) + O(x^30)) \\ Colin Barker, Mar 26 2016
(PARI) a(n) = (fibonacci(n+2)*fibonacci(n+3))^2; \\ Altug Alkan, Mar 26 2016
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,easy
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
