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First differences of Chebyshev polynomials S(n,531)=A098257(n) with Diophantine property.
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%I #30 Sep 08 2022 08:45:14

%S 1,530,281429,149438269,79351439410,42135464888441,22373852504322761,

%T 11880473544330497650,6308509078186989929389,

%U 3349806440043747322007909,1778740911154151640996270290,944508074016414477621697516081

%N First differences of Chebyshev polynomials S(n,531)=A098257(n) with Diophantine property.

%C (23*b(n))^2 - 533*a(n)^2 = -4 with b(n)=A098258(n) give all positive solutions of this Pell equation.

%H G. C. Greubel, <a href="/A098259/b098259.txt">Table of n, a(n) for n = 0..365</a>

%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>

%H Giovanni Lucca, <a href="http://forumgeom.fau.edu/FG2019volume19/FG201902index.html">Integer Sequences and Circle Chains Inside a Hyperbola</a>, Forum Geometricorum (2019) Vol. 19, 11-16.

%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (531,-1).

%F a(n) = ((-1)^n)*S(2*n, 23*i) with the imaginary unit i and the S(n, x) = U(n, x/2) Chebyshev polynomials.

%F G.f.: (1-x)/(1-531*x+x^2).

%F a(n) = S(n, 531) - S(n-1, 531) = T(2*n+1, sqrt(533)/2)/(sqrt(533)/2), with S(n, x) = U(n, x/2) Chebyshev's polynomials of the second kind, A049310. S(-1, x)= 0 = U(-1, x) and T(n, x) Chebyshev's polynomials of the first kind, A053120.

%F a(n) = 531*a(n-2) - a(n-2), n>1; a(0)=1, a(1)=530. - _Philippe Deléham_, Nov 18 2008

%e All positive solutions of Pell equation x^2 - 533*y^2 = -4 are (23=23*1,1), (12236=23*532,530), (6497293=23*282491,281429), (3450050347=23*150002189,149438269), ...

%t LinearRecurrence[{531,-1}, {1,530}, 20] (* _G. C. Greubel_, Aug 01 2019 *)

%o (PARI) my(x='x+O('x^20)); Vec((1-x)/(1-531*x+x^2)) \\ _G. C. Greubel_, Aug 01 2019

%o (Magma) I:=[1,530]; [n le 2 select I[n] else 531*Self(n-1) - Self(n-2): n in [1..20]]; // _G. C. Greubel_, Aug 01 2019

%o (Sage) ((1-x)/(1-531*x+x^2)).series(x, 20).coefficients(x, sparse=False) # _G. C. Greubel_, Aug 01 2019

%o (GAP) a:=[1,530];; for n in [3..20] do a[n]:=531*a[n-1]-a[n-2]; od; a; # _G. C. Greubel_, Aug 01 2019

%K nonn,easy

%O 0,2

%A _Wolfdieter Lang_, Sep 10 2004