|
Search: id:A082574
|
|
|
| A082574 |
|
a(1)=1, a(n)=ceiling(r(3)*a(n-1)) where r(3)= (1/2) *(3+sqrt(13)) is the positive root of X^2=3*X+1. |
|
+0
2
|
|
| 1, 4, 14, 47, 156, 516, 1705, 5632, 18602, 61439, 202920, 670200, 2213521, 7310764, 24145814, 79748207, 263390436, 869919516, 2873148985, 9489366472, 31341248402, 103513111679, 341880583440, 1129154862000, 3729345169441
(list; graph; listen)
|
|
|
OFFSET
|
1,2
|
|
|
COMMENT
|
More generally the sequence a(1)=1, a(n)=ceiling(r(z)*a(n-1)) where r(z)= (1/2) *(z+sqrt(z^2+4)) is the positive root of X^2=z*X+1 satisfies the linear recurrence : n>3 a(n)=(z+1)*a(n-1)-(z-1)*a(n-2)-a(n-3) and the closed form formula : a(n)=floor(t(z)*r(z)^n) where t(z)=1/(2*z)*(1+(z+2)/sqrt(z^2+4)) is the positive root of z*(z^2+4)*X^2=(z^2+4)*X+1.
|
|
FORMULA
|
a(1)=1, a(2)=4, a(3)=14, a(n)=4a(n-1)-2a(n-2)-a(n-3); a(n)=floor(t(3)*r(3)^n) where t(3)=1/6*(1+5/sqrt(13)) is the positive root of 39*X^2=13*X+1.
G.f.: 1/((1-x)(1-3x-x^2)). Partial sums of A006190. - Paul Barry (pbarry(AT)wit.ie), Jul 10 2004
|
|
MAPLE
|
a:=n->sum(fibonacci(i, 3), i=0..n): seq(a(n), n=1..25); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Mar 20 2008
|
|
CROSSREFS
|
Cf. A000071, A048739, A049652.
Sequence in context: A046718 A104487 A094789 this_sequence A137284 A121095 A000908
Adjacent sequences: A082571 A082572 A082573 this_sequence A082575 A082576 A082577
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
Benoit Cloitre (benoit7848c(AT)orange.fr), May 06 2003
|
|
|
Search completed in 0.002 seconds
|
