|
Search: id:A131435
|
|
|
| A131435 |
|
Recursive sequence generated from a Petersen graph. |
|
+0 1
|
|
| 1, 6, 35, 198, 1124, 6373, 36142, 204959, 1162306, 6591376, 37379241, 211975382, 1202098747
(list; graph; listen)
|
|
|
OFFSET
|
1,2
|
|
|
COMMENT
|
Characteristic polynomial of M = x^5 - 3x^4 - 15x^3 - 3x^2 + 13x - 4. a(n)/a(n-1) tends to 5.6709364838...the largest root of the polynomial and an eigenvalue of the matrix.
|
|
REFERENCES
|
Stephan G. Wagner, "The Fibonacci Number of Generalized Petersen Graphs", Fibonacci Quarterly, Vol. 44, Number 4, November 2006, p. 366.
|
|
FORMULA
|
Let M = the 5x5 adjacency matrix of a Petersen graph, [Wagner]: [2,1,1,1,0; 1,1,0,1,0; 8,5,0,3,0; 3,2,0,0,1; 5,3,0,3,0]. Then a(n) = M^n (2,1); = second term from the left of M^n * [1,0,0,0,0]. For n>5, a(n) = 3*a(n-1) + 15*a(n-2) + 3*a(n-3) - 13*a(n-4) + 4*a(n-5).
G.f.: x(1+x)(1+2x)/(1-3x-15x^2-3x^3+13x^4-4x^5). [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Oct 30 2008]
|
|
EXAMPLE
|
a(8) = 204959 = 3*36142 + 15*6373 + 3*1124 - 13*198 + 4*35, = 3*a(7) + 15*a(6) + 3*a(5) - 13*a(4) + 4*a(3).
a(5) = 1124 = second term from the left of M^5 * [1,0,0,0,0] = [2669, 1124, 6148, 2580, 4324].
|
|
CROSSREFS
|
Sequence in context: A009572 A027202 A026934 this_sequence A079027 A081105 A161727
Adjacent sequences: A131432 A131433 A131434 this_sequence A131436 A131437 A131438
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
Gary W. Adamson (qntmpkt(AT)yahoo.com), Jul 11 2007
|
|
|
Search completed in 0.002 seconds
|