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URL: https://oeis.org/A098247

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A098247
First differences of Chebyshev polynomials S(n,227) = A098245(n) with Diophantine property.
5
1, 226, 51301, 11645101, 2643386626, 600037119001, 136205782626601, 30918112619119426, 7018275358757483101, 1593117588325329544501, 361630674274491049118626, 82088569942721142820383601, 18633743746323424929177958801, 4229777741845474737780576264226
OFFSET
0,2
COMMENTS
(15*b(n))^2 - 229*a(n)^2 = -4 with b(n) = A098246(n) give all positive solutions of this Pell equation.
FORMULA
a(n) = S(n, 227) - S(n-1, 227) = T(2*n+1, sqrt(229)/2)/(sqrt(229)/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 second kind, A053120.
a(n) = ((-1)^n)*S(2*n, 15*i) with the imaginary unit i and the S(n, x) = U(n, x/2) Chebyshev polynomials.
G.f.: (1-x)/(1-227*x+x^2).
a(n) = 227*a(n-1) - a(n-2), n > 1; a(0)=1, a(1)=226. - Philippe Deléham, Nov 18 2008
Sum_{n>=0} 1/(a(n)+1) = sqrt(229)/30. - Amiram Eldar, Jan 01 2026
EXAMPLE
All positive solutions of Pell equation x^2 - 229*y^2 = -4 are (15 = 15*1, 1), (3420 = 15*228, 226), (776325 = 15*51755, 51301), (176222355 = 15*11748157, 11645101), ...
MATHEMATICA
LinearRecurrence[{227, -1}, {1, 226}, 20] (* G. C. Greubel, Aug 01 2019 *)
PROG
(PARI) my(x='x+O('x^20)); Vec((1-x)/(1-227*x+x^2)) \\ G. C. Greubel, Aug 01 2019
(Magma) I:=[1, 226]; [n le 2 select I[n] else 227*Self(n-1) - Self(n-2): n in [1..20]]; // G. C. Greubel, Aug 01 2019
(SageMath) ((1-x)/(1-227*x+x^2)).series(x, 20).coefficients(x, sparse=False) # G. C. Greubel, Aug 01 2019
(GAP) a:=[1, 226];; for n in [3..20] do a[n]:=227*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Aug 01 2019
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Sep 10 2004
STATUS
approved