|
| |
|
|
A075843
|
|
Numbers k such that 99*k^2 + 1 is a square.
|
|
8
|
|
|
|
0, 1, 20, 399, 7960, 158801, 3168060, 63202399, 1260879920, 25154396001, 501827040100, 10011386405999, 199725901079880, 3984506635191601, 79490406802752140, 1585823629419851199, 31636982181594271840
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,3
|
|
|
COMMENTS
|
From Wolfdieter Lang, Nov 08 2002: (Start)
Chebyshev's polynomials U(n,x) evaluated at x=10.
The a(n) give all (unsigned, integer) solutions of Pell equation b(n)^2 - 99*a(n)^2 = +1 with b(n)= A001085(n). (End)
For n>=2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 20's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imagianry unit). - John M. Campbell, Jul 08 2011
For n>=1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,19}. Milan Janjic, Jan 25 2015
|
|
|
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.
|
|
|
LINKS
|
Vincenzo Librandi, Table of n, a(n) for n = 0..700
Tanya Khovanova, Recursive Sequences
J. J. O'Connor and E. F. Robertson, Pell's Equation
Eric Weisstein's World of Mathematics, Pell Equation.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (20,-1).
|
|
|
FORMULA
|
a(n) = ((10+3*sqrt(11))^n - (10-3*sqrt(11))^n) / (6*sqrt(11)).
a(n) = 20*a(n-1) - a(n-2), n>=1, a(0)=0, a(1)=1.
a(n) = S(n-1, 20), with S(n, x) := U(n, x/2), Chebyshev's polynomials of the second kind. S(-1, x) := 0. See A049310.
G.f.: x/(1 - 20*x + x^2).
a(n) = sqrt((A001085(n)^2 - 1)/99).
Lim_{n->inf.} a(n)/a(n-1) = 10 + 3*sqrt(11).
a(n+1) = sum_{k=0..n} A101950(n,k)*19^k. - Philippe Deléham, Feb 10 2012
Product_{n>=1} (1 + 1/a(n)) = 1/3*(3 + sqrt(11)). - Peter Bala, Dec 23 2012
Product_{n>=2} (1 - 1/a(n)) = 3/20*(3 + sqrt(11)). - Peter Bala, Dec 23 2012
|
|
|
MATHEMATICA
|
lst={}; Do[AppendTo[lst, GegenbauerC[n, 1, 10]], {n, 0, 8^2}]; lst (* Vladimir Joseph Stephan Orlovsky, Sep 11 2008 *)
CoefficientList[Series[x/(1 - 20 x + x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Dec 24 2012 *)
|
|
|
PROG
|
sage: [lucas_number1(n, 20, 1) for n in xrange(0, 20)] # Zerinvary Lajos, Jun 25 2008
(MAGMA) I:=[0, 1, 20]; [n le 3 select I[n] else 20*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Dec 24 2012
|
|
|
CROSSREFS
|
Cf. A001084.
Sequence in context: A218722 A158534 A171325 * A208072 A208122 A207372
Adjacent sequences: A075840 A075841 A075842 * A075844 A075845 A075846
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
Gregory V. Richardson, Oct 14 2002
|
|
|
STATUS
|
approved
|
| |
|
|
