|
Search: id:A081016
|
|
|
| A081016 |
|
(Lucas(4n+3)+1)/5, or Fibonacci(2n+1)*Fibonacci(2n+2), or A081015/5. |
|
+0 6
|
|
| 1, 6, 40, 273, 1870, 12816, 87841, 602070, 4126648, 28284465, 193864606, 1328767776, 9107509825, 62423800998, 427859097160, 2932589879121, 20100270056686, 137769300517680, 944284833567073, 6472224534451830
(list; graph; listen)
|
|
|
OFFSET
|
0,2
|
|
|
COMMENT
|
a(n-1) is, together with b(n) := A089508(n), n>=1, the solution to a binomial problem - see A089508.
Numbers k such that 1 - 2 x + 5 x^2 is a square [From Artur Jasinski (grafix(AT)csl.pl), Oct 26 2008]
|
|
REFERENCES
|
Hugh C. Williams, Edouard Lucas and Primality Testing, John Wiley and Sons, 1998, p. 75
A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 26.
|
|
FORMULA
|
a(n) = 8a(n-1)-8a(n-2)+a(n-3)
G.f.: (1-2*x)/((1-x)*(1-7*x+x^2)).
F(1) + F(5) + F(9) +...+ F(4n+1) = F(2n)*F(2n+3) + 1.
a(n)=(1/5)+(2/5)*{[(7/2)-(3/2)*sqrt(5)]^n+[(7/2)+(3/2)*sqrt(5)]^n+(1/5)*sqrt(5)*{[(7/2)+(3/2) *sqrt(5)]^n-[(7/2)-(3/2)*sqrt(5)]^n}, with n>=0 [From Paolo P. Lava (ppl(AT)spl.at), Dec 01 2008]
|
|
MAPLE
|
luc := proc(n) option remember: if n=0 then RETURN(2) fi: if n=1 then RETURN(1) fi: luc(n-1)+luc(n-2): end: for n from 0 to 25 do printf(`%d, `, (luc(4*n+3)+1)/5) od:
|
|
CROSSREFS
|
Cf. A000045 (Fibonacci numbers), A000032 (Lucas numbers), A081015.
Partial sums of A033889. Bisection of A001654. Equals A003482 + 1.
A145995 [From Artur Jasinski (grafix(AT)csl.pl), Oct 26 2008]
Sequence in context: A135032 A122074 A123357 this_sequence A083426 A122471 A090041
Adjacent sequences: A081013 A081014 A081015 this_sequence A081017 A081018 A081019
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
R. K. Guy (rkg(AT)cpsc.ucalgary.ca), Mar 01, 2003
|
|
EXTENSIONS
|
More terms and Maple code from James A. Sellers (sellersj(AT)math.psu.edu), Mar 03, 2003
|
|
|
Search completed in 0.002 seconds
|