|
Search: id:A120694
|
|
|
| A120694 |
|
Sequence demonstrating the Pythagorean theorem. |
|
+0 1
|
|
| 1, 25, 1201, 58825, 2882401, 141237625, 6920643601, 339111536425
(list; graph; listen)
|
|
|
OFFSET
|
0,2
|
|
|
COMMENT
|
Each of the two terms in M^n * [1,0] are the hypotenuse (left term); and longest leg in a Pythagorean triple: e.g. M^3 * [1,0] = [58825, 58824]; such that 58825^2 - 58824^2 = 7^3; generally: sqrt:((a(n)^2 - (a(n)-1)^2)) = 7^n. Characteristic polynomial of M = x^2 - 50x + 49.
|
|
FORMULA
|
a(n) = 50*a(n-1) - 49*a(n-2), n>1. Let M = the 2 X 2 matrix [25, 24; 24, 25]. Then, a(n) = left term in M^n * [1,0].
a(n)=(1/2)*(1+*49^n), with n>=0 [From Paolo P. Lava (ppl(AT)spl.at), Aug 28 2008]
|
|
EXAMPLE
|
58825 = a(3) = 50*a(2) - 49*a(1) = 50*1201 - 49*25.
58825 = a(3) = left term in M^3 * [1,0] = [58825, 58824].
|
|
CROSSREFS
|
Adjacent sequences: A120691 A120692 A120693 this_sequence A120695 A120696 A120697
Sequence in context: A012508 A112102 A012799 this_sequence A012809 A014769 A012851
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
Gary W. Adamson (qntmpkt(AT)yahoo.com), Jun 28 2006
|
|
|
Search completed in 0.002 seconds
|