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A097741
Pell equation solutions (10*a(n))^2 - 101*b(n)^2 = -1 with b(n) = A097742(n), n >= 0.
4
1, 403, 162005, 65125607, 26180332009, 10524428342011, 4230794013156413, 1700768668860536015, 683704774087922321617, 274847618414675912754019, 110488058897925629004794021, 44415924829347688184014442423, 17855091293338872724344801060025
OFFSET
0,2
FORMULA
G.f.: (1 + x)/(1 - 2*201*x + x^2).
a(n) = S(n, 2*201) + S(n-1, 2*201) = S(2*n, 2*sqrt(101)), with Chebyshev polynomials of the 2nd kind. See A049310 for the triangle of S(n, x)= U(n, x/2) coefficients. S(-1, x) := 0 =: U(-1, x).
a(n) = ((-1)^n)*T(2*n+1, 10*i)/(10*i) with the imaginary unit i and Chebyshev polynomials of the first kind. See the T-triangle A053120.
a(n) = 402*a(n-1) - a(n-2) for n > 1, a(0)=1, a(1)=403. - Philippe Deléham, Nov 18 2008
a(n) = (1/10)*sinh((2*n + 1)*arcsinh(10)). - Bruno Berselli, Apr 03 2018
Let h = (10 + sqrt(101))^(2*n+1) then a(n) = (h-1/h)/20 and a(n) = floor(h/20). - Peter Luschny, Apr 05 2018
EXAMPLE
(x,y) = (10*1=10;1), (4030=10*403;401), (1620050=10*162005;161201), ... give the positive integer solutions to x^2 - 101*y^2 =-1.
MATHEMATICA
LinearRecurrence[{402, -1}, {1, 403}, 20] (* or *) CoefficientList[Series[(1 + x)/(1 - 402 x + x^2), {x, 0, 20}], x] (* Harvey P. Dale, Apr 20 2011 *)
a[n_] := Floor[(10 + Sqrt[101])^(2 n + 1)]/20;
Table[a[n], {n, 0, 11}] (* Peter Luschny, Apr 05 2018 *)
PROG
(PARI) x='x+O('x^99); Vec((1+x)/(1-2*201*x+x^2)) \\ Altug Alkan, Apr 05 2018
CROSSREFS
Cf. A097740 for S(n, 2*201).
Cf. similar sequences of the type (1/k)*sinh((2*n+1)*arcsinh(k)) listed in A097775.
Sequence in context: A261857 A165808 A283662 * A117836 A185640 A185637
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Aug 31 2004
STATUS
approved