|
| |
|
|
A015448
|
|
a(0)=1, a(1)=1, and a(n) = 4*a(n-1) + a(n-2) for n>=2.
|
|
32
| |
|
|
1, 1, 5, 21, 89, 377, 1597, 6765, 28657, 121393, 514229, 2178309, 9227465, 39088169, 165580141, 701408733, 2971215073, 12586269025, 53316291173, 225851433717, 956722026041, 4052739537881, 17167680177565
(list; graph; refs; listen; history; internal format)
|
|
|
|
OFFSET
| 0,3
|
|
|
COMMENTS
| a(n) = A167808(3*n-1) for n>0. [From Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Nov 12 2009]
If one deletes the leading 0 in A084326, takes the inverse binomial transform, and adds a(0)=1 in front, one obtains this sequence here. [From Al Hakanson (hawkuu(AT)gmail.com), May 02 2009]
|
|
|
LINKS
| T. D. Noe, Table of n, a(n) for n=0..200
Joerg Arndt, Fxtbook, pp.313-315
Tanya Khovanova, Recursive Sequences
Index entries for sequences related to linear recurrences with constant coefficients
|
|
|
FORMULA
| a(n) = Fibonacci(3n-1) = ( (1+sqrt(5))*(2-sqrt(5))^n - (1-sqrt(5))*(2+sqrt(5))^n )/ (2*sqrt(5)).
O.g.f.: (1-3*x)/(1-4*x-x^2). - Len Smiley (smiley(AT)math.uaa.alaska.edu), Dec 09 2001
a(n)=Sum_{k, 0<=k<=n}3^k*A055830(n,k) . - Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Oct 18 2006
a(n) = upper left term in the 2 X 2 matrix [1,2; 2,3]^n - Gary W. Adamson (qntmpkt(AT)yahoo.com), Mar 02 2008
[a(n), A001076(n)] = [1,4; 1,3]^n * [1,0] - Gary W. Adamson (qntmpkt(AT)yahoo.com), Mar 21 2008
a(n)= Fibonacci(3n+1) mod Fibonacci(3n), n>0
a(n)= (A000032(3*n)-Fibonacci(3*n))/2 = (A014448(n)-A014445(n)/2.
|
|
|
MAPLE
| with(combinat): a:=n->fibonacci(n, 4)-3*fibonacci(n-1, 4): seq(a(n), n=1..23); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Apr 04 2008
|
|
|
MATHEMATICA
| Fibonacci/@(3*Range[30]-1) [From Vladimir Orlovsky (4vladimir(AT)gmail.com), Mar 01 2010]
|
|
|
CROSSREFS
| Cf. A001076.
Sequence in context: A010925 A019992 A010917 * A099843 A035011 A113987
Adjacent sequences: A015445 A015446 A015447 * A015449 A015450 A015451
|
|
|
KEYWORD
| nonn,easy
|
|
|
AUTHOR
| Olivier Gerard (olivier.gerard(AT)gmail.com)
|
| |
|
|
