|
|
A075835
|
|
Numbers n such that 13*n^2 + 4 is a square.
|
|
1
|
|
|
0, 3, 33, 360, 3927, 42837, 467280, 5097243, 55602393, 606529080, 6616217487, 72171863277, 787274278560, 8587845200883, 93679022931153, 1021881407041800, 11147016454528647, 121595299592773317
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
Lim_{n->infinity} a(n)/a(n-1) = (11 + sqrt(13))/2.
|
|
REFERENCES
|
A. H. Beiler, "The Pellian", ch. 22 in Recreations in the Theory of Numbers: The Queen of Mathematics Entertains. Dover, New York, New York, pp. 248-268, 1966.
L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. AMS Chelsea Publishing, Providence, Rhode Island, 1999, pp. 341-400.
Peter G. L. Dirichlet, Lectures on Number Theory (History of Mathematics Source Series, V. 16); American Mathematical Society, Providence, Rhode Island, 1999, pp. 139-147.
S. Falcon, Relationships between Some k-Fibonacci Sequences, Applied Mathematics, 2014, 5, 2226-2234; http://www.scirp.org/journal/am; http://dx.doi.org/10.4236/am.2014.515216
|
|
LINKS
|
|
|
FORMULA
|
a(n) = ((11 + 3*sqrt(13))^n - (11 - 3*sqrt(13))^n) / ((2^n) * sqrt(13)).
a(n) = 11*a(n-1) - a(n-2) with a(1)=0 and a(2)=3.
G.f.: 3x^2/(1-11x+x^2). (End)
|
|
MATHEMATICA
|
LinearRecurrence[{11, -1}, {0, 3}, 20] (* Harvey P. Dale, Dec 27 2011 *)
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|