|
Search: id:A081018
|
|
|
| A081018 |
|
(Lucas(4n+1)-1)/5, or Fibonacci(2n)*Fibonacci(2n+1), or A081017/5. |
|
+0
2
|
|
| 0, 2, 15, 104, 714, 4895, 33552, 229970, 1576239, 10803704, 74049690, 507544127, 3478759200, 23843770274, 163427632719, 1120149658760, 7677619978602, 52623190191455, 360684711361584, 2472169789339634, 16944503814015855
(list; graph; listen)
|
|
|
OFFSET
|
0,2
|
|
|
COMMENT
|
Another interpretation of this sequence is: nonnegative integers k such that (k + 1)^2 + (2k)^2 is a perfect square. So apart from a(0) = 0, a(n) + 1 and 2a(n) form the legs of a pythagorean triple. - Nick Hobson, Jan 13 2007
|
|
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. 28.
|
|
FORMULA
|
a(n) = 8a(n-1)-8a(n-2)+a(n-3)
F(3) + F(7) + F(11) +...+ F(4n+3).
|
|
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+1)-1)/5) od:
|
|
CROSSREFS
|
Cf. A000045 (Fibonacci numbers), A000032 (Lucas numbers), A081017.
Partial sums of A033891. Bisection of A001654 and A059840.
Equals A089508 + 1.
Sequence in context: A037753 A037641 A027080 this_sequence A006675 A037524 A037733
Adjacent sequences: A081015 A081016 A081017 this_sequence A081019 A081020 A081021
|
|
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 0.003 seconds
|
