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First differences of Chebyshev polynomials S(n,51) = A097836(n) with Diophantine property.
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%I #32 Sep 08 2022 08:45:14

%S 1,50,2549,129949,6624850,337737401,17217982601,877779375250,

%T 44749530155149,2281348258537349,116304011655249650,

%U 5929223246159194801,302274081542463685201,15410048935419488750450,785610221624851462587749

%N First differences of Chebyshev polynomials S(n,51) = A097836(n) with Diophantine property.

%C (7*b(n))^2 - 53*a(n)^2 = -4 with b(n)=A097837(n) give all positive solutions of this Pell equation.

%H Indranil Ghosh, <a href="/A097838/b097838.txt">Table of n, a(n) for n = 0..584</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/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (51, -1).

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

%F a(n) = ((-1)^n)*S(2*n, 7*i) with the imaginary unit i and the S(n, x) = U(n, x/2) Chebyshev polynomials.

%F G.f.: (1-x)/(1 - 51*x + x^2).

%F a(n) = S(n, 51) - S(n-1, 51) = T(2*n+1, sqrt(53)/2)/(sqrt(53)/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) = 51*a(n-1) - a(n-2), a(0)=1, a(1)=50. - _Philippe Deléham_, Nov 18 2008

%e All positive solutions of Pell equation x^2 - 53*y^2 = -4 are (7=7*1,1), (364=7*52,50), (18557=7*2651,2549), (946043=7*135149,129949), ...

%t LinearRecurrence[{51,-1}, {1,50}, 20] (* _G. C. Greubel_, Jan 13 2019 *)

%o (PARI) my(x='x+O('x^20)); Vec((1-x)/(1-51*x+x^2)) \\ _G. C. Greubel_, Jan 13 2019

%o (Magma) m:=20; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!( (1-x)/(1-51*x+x^2) )); // _G. C. Greubel_, Jan 13 2019

%o (Sage) ((1-x)/(1-51*x+x^2)).series(x, 20).coefficients(x, sparse=False) # _G. C. Greubel_, Jan 13 2019

%o (GAP) a:=[1,50];; for n in [3..20] do a[n]:=51*a[n-1]-a[n-2]; od; a; # _G. C. Greubel_, Jan 13 2019

%K nonn,easy

%O 0,2

%A _Wolfdieter Lang_, Sep 10 2004