|
Search: id:A121389
|
|
| |
|
| 0, 9, 9, 99, 999, 99999, 99999999, 9999999999999, 999999999999999999999, 9999999999999999999999999999999999, 9999999999999999999999999999999999999999999999999999999
(list; graph; listen)
|
|
|
OFFSET
|
0,2
|
|
|
COMMENT
|
Each a(n) has Fibonacci(n) (trailing) 9s. In general, if the same recurrence below is used with any a(0), a(1) >= 0, then, for all k >= 2, a(k) has the same number of trailing 9s as a(k-2) and a(k-1) have altogether. (See, for example, A121390).
|
|
FORMULA
|
a(n) = 10^Fibonacci(n) - 1 = 10^A000045(n) - 1 (= 9*A108047(n) for n>=1). a(0) = 0; a(1) = 9; a(n) = a(n-2)*a(n-1) + a(n-2) + a(n-1).
|
|
CROSSREFS
|
Cf. A000045, A063896, A108047, A121390.
Sequence in context: A165427 A050683 A092548 this_sequence A065242 A050720 A124116
Adjacent sequences: A121386 A121387 A121388 this_sequence A121390 A121391 A121392
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
Rick L. Shepherd (rshepherd2(AT)hotmail.com), Jul 26 2006
|
|
|
Search completed in 0.002 seconds
|