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A097834
Chebyshev polynomials S(n,27) + S(n-1,27) with Diophantine property.
5
1, 28, 755, 20357, 548884, 14799511, 399037913, 10759224140, 290100013867, 7821941150269, 210902311043396, 5686540457021423, 153325690028535025, 4134107090313424252, 111467565748433919779, 3005490168117402409781
OFFSET
0,2
COMMENTS
(5*a(n))^2 - 29*b(n)^2 = -4 with b(n) = A097835(n) give all positive solutions of this Pell equation.
From Klaus Purath, Sep 24 2024: (Start)
a(n) = (t(i+2n+1) - t(i))/(t(i+n+1) - t(i+n)), where (t) is any sequence satisfying t(i) = 28t(i-1) - 28t(i-2) + t(i-3) or t(i) = 27t(i-1) - t(i-2) without regard to initial values and including this sequence itself, as long as t(i+n+1) - t(i+n) != 0 for integer i and n >= 0.
a(n) = (Sum_{i .. i+2n} t(i))/t(i+n), where (t) is any recurrence of the form (27,-1) without regard to initial values and including this sequence itself, as long as t(i+n) != 0 for integer i and n >= 0.
a(n) = t(n) - t(n-1) = (t(n+1) - t(n-2))/28, where (t) is any third order recurrence with constant coefficients (28,-28,1) and initial values t(0) = x, t(1) = x + 1, t(2) = x + 29 for any integer x.
a(n) = t(n-1) + t(n) = (t(n-2) + t(n+1))/26, where (t) is any third order recurrence with constant coefficients (26,26,-1) and initial values t(0) = x, t(1) = 1 - x, t(2) = x + 27 for any integer x.
a(n) = (t(i+4n+2) - t(i))/(t(i+2n+2) - t(i+2n)), where (t) is any recurrence of the form (5,1) without regard to initial values, as long as t(i+2n+2) - t(i+2n) != 0 for nonnegative integer i and n. (End)
LINKS
Indranil Ghosh, Table of n, a(n) for n = 0..697 (terms 0..200 from Vincenzo Librandi)
Tanya Khovanova, Recursive Sequences
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Hyperbola, Forum Geometricorum (2019) Vol. 19, 11-16.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
FORMULA
a(n) = S(n, 27) + S(n-1, 27) = S(2*n, sqrt(29)), with S(n, x)=U(n, x/2) Chebyshev's polynomials of the second kind, A049310. S(-1, x) = 0 = U(-1, x). S(n, 27)=A097781(n).
a(n) = (-2/5)*i*((-1)^n)*T(2*n+1, 5*i/2) with the imaginary unit i and Chebyshev's polynomials of the first kind. See the T-triangle A053120.
G.f.: (1+x)/(1-27*x+x^2).
a(n) = - a(-1-n) for all n in Z. - Michael Somos, Nov 01 2008
From Peter Bala, Aug 26 2022: (Start)
a(n) = (2/5)*(5/2 o 5/2 o ... o 5/2) (2*n+1 terms), where the binary operation o is defined on real numbers by x o y = x*sqrt(1 + y^2) + y*sqrt(1 + x^2). The operation o is commutative and associative with identity 0.
The aerated sequence (b(n))n>=1 = [1, 0, 28, 0, 755, 0, 20357, 0, ...], with o.g.f. x*(1 + x^2)/(1 - 27*x^2 + x^4), is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -25, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials.
b(n) = (1/2)*( (-1)^n - 1 )*F(n,5) + (1/5)*( 1 + (-1)^(n+1) )*F(n+1,5), where F(n,x) is the n-th Fibonacci polynomial - see A168561 (but with row indexing starting at n = 1).
Exp( Sum_{n >= 1} 10*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 10*A052918(n)*x^n.
Exp( Sum_{n >= 1} (-10)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 10*A052918(n)*(-x)^n.
(End)
a(n) = (a(n-1)*a(n-2) - 783)/a(n-3) for n >= 3. - Klaus Purath, Sep 24 2024
EXAMPLE
All positive solutions of Pell equation x^2 - 29*y^2 = -4 are
(5=5*1,1), (140=5*28,26), (3775=5*755,701), (101785=5*20357,18901), ...
MATHEMATICA
a[n_] := -2/5*I*(-1)^n*ChebyshevT[2*n + 1, 5*I/2]; Table[a[n], {n, 0, 15}] (* Jean-François Alcover, Jun 21 2013, from 2nd formula *)
PROG
(PARI) {a(n) = (-1)^n * subst(2 * I / 5 * poltchebi(2*n + 1), 'x, -5/2 * I)}; /* Michael Somos, Nov 04 2008 */
CROSSREFS
A087130(2*n + 1) = 5 * a(n). - Michael Somos, Nov 01 2008
Sequence in context: A226991 A277060 A229463 * A162830 A163187 A163548
KEYWORD
nonn,easy,changed
AUTHOR
Wolfdieter Lang, Sep 10 2004
STATUS
approved