A097841 - OEIS

%I #48 Jan 01 2026 02:57:39

%S 1,82,6805,564733,46866034,3889316089,322766369353,26785719340210,

%T 2222891938868077,184473245206710181,15309056460218076946,

%U 1270467212952893676337,105433469618629957059025,8749707511133333542222738,726120289954448054047428229,60259234358708055152394320269

%N First differences of Chebyshev polynomials S(n,83) = A097839(n) with Diophantine property.

%C (9*b(n))^2 - 85*a(n)^2 = -4 with b(n) = A097840(n) give all positive solutions of this Pell equation.

%C For n > 0, a(n) is the hypotenuse of the Pythagorean triple (x(n), y(n), a(n)) that is primitive for n == 0, 2 (mod 3) where (x(n)) and (y(n)) are recurrences of the form (82,82,-1) with the initial values x(0) = 1, x(1) = 80, x(2) = 6643; y(0) = 0, y(1) = 18, y(2) = 1476. - _Klaus Purath_, Jul 19 2025

%H Indranil Ghosh, <a href="/A097841/b097841.txt">Table of n, a(n) for n = 0..520</a>

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

%H Giovanni Lucca, <a href="https://web.archive.org/web/20200713023901/http://forumgeom.fau.edu/FG2019volume19/FG201902index.html">Integer Sequences and Circle Chains Inside a Hyperbola</a>, Forum Geometricorum, Vol. 19 (2019), 11-16.

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (83,-1).

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

%F a(n) = ((-1)^n)*S(2*n, 9*i) with the imaginary unit i and the S(n, x) = U(n, x/2) Chebyshev polynomials.

%F G.f.: (1-x)/(1 - 83*x + x^2).

%F a(n) = S(n, 83) - S(n-1, 83) = T(2*n+1, sqrt(85)/2)/(sqrt(85)/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) = 83*a(n-1) - a(n-2) for n > 1, a(0)=1, a(1)=82. - _Philippe Deléham_, Nov 18 2008

%F From _Klaus Purath_, Jul 19 2025: (Start)

%F a(n) = A099371(2n+1) = A099371(n)^2 + A099371(n+1)^2.

%F a(n) = (t(i+2*n+1) + t(i))/(t(i+n+1) + t(i+n)) as long as t(i+n+1) + t(i+n) != 0 for any integer i and n >= 1 where (t) is a sequence satisfying t(i+3) = 82*t(i+2) + 82*t(i+1) - t(i) or t(i+2) = 83*t(i+1) - t(i) regardless of initial values and including this sequence itself. (End)

%F Sum_{n>=0} 1/(a(n)+1) = sqrt(85)/18 = A010536 / 18. - _Amiram Eldar_, Jan 01 2026

%e All positive solutions of Pell equation x^2 - 85*y^2 = -4 are (9 = 9*1, 1), (756 = 9*84, 82), (62739 = 9*6971, 6805), (5206581 = 9*578509, 564733), ...

%t CoefficientList[Series[(1-x)/(1-83x+x^2), {x, 0, 20}], x] (* _Michael De Vlieger_, Feb 08 2017 *)

%t LinearRecurrence[{83,-1}, {1,82}, 20] (* _G. C. Greubel_, Jan 13 2019 *)

%o (PARI) my(x='x+O('x^20)); Vec((1-x)/(1-83*x+x^2)) \\ _G. C. Greubel_, Jan 13 2019

%o (Magma) m:=20; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!( (1-x)/(1-83*x+x^2) )); // _G. C. Greubel_, Jan 13 2019

%o (SageMath) ((1-x)/(1-83*x+x^2)).series(x, 20).coefficients(x, sparse=False) # _G. C. Greubel_, Jan 13 2019

%o (GAP) a:=[1,82];; for n in [3..20] do a[n]:=83*a[n-1]-a[n-2]; od; a; # _G. C. Greubel_, Jan 13 2019

%Y Cf. A097839, A097840.

%Y Cf. A049310, A053120.

%Y Cf. A010536, A099371.

%K nonn,easy

%O 0,2

%A _Wolfdieter Lang_, Sep 10 2004