OFFSET
0,2
COMMENTS
Numbers n such that 8*n^2 + 41 is a square.
(x, y) = (a(n), a(n+1)) are solutions to x^2 + y^2 - 6*x*y = 41. - John O. Oladokun, Mar 17 2021
REFERENCES
A. H. Beiler, Recreations in the Theory of Numbers, Dover, N. Y., 1964, pp. 122-125, 194-196.
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..1000
I. Adler, Three Diophantine equations - Part II, Fib. Quart., 7 (1969), pp. 181-193.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (6,-1).
FORMULA
a(n) = 6*a(n-1) - a(n-2), a(0)=1, a(1)=10.
a(n) = (10*((3+2*sqrt(2))^n - (3-2*sqrt(2))^n) - ((3+2*sqrt(2))^(n-1) - (3-2*sqrt(2))^(n-1)))/(4*sqrt(2)).
From G. C. Greubel, Jan 19 2020: (Start)
a(n) = ChebyshevU(n,3) + 4*ChebyshevU(n-1,3).
a(n) = (Pell(2*n+2) + 4*Pell(2*n))/2 = (Pell-Lucas(2*n+1) + 3*Pell(2*n))/2.
E.g.f.: exp(3*x)*( cosh(2*sqrt(2)*x) + 7*sinh(2*sqrt(2)*x)/(2*sqrt(2)) ). (End)
MAPLE
a[0]:=1: a[1]:=10: for n from 2 to 26 do a[n]:=6*a[n-1]-a[n-2] od: seq(a[n], n=0..19); # Zerinvary Lajos, Jul 26 2006
MATHEMATICA
Table[(LucasL[2*n+1, 2] + 3*Fibonacci[2*n, 2])/2, {n, 0, 30}] (* G. C. Greubel, Jan 19 2020 *)
LinearRecurrence[{6, -1}, {1, 10}, 20] (* Harvey P. Dale, Jun 11 2024 *)
PROG
(PARI) vector(31, n, polchebyshev(n-1, 2, 3) +4*polchebyshev(n-2, 2, 3) ) \\ G. C. Greubel, Jan 19 2020
(Magma) I:=[1, 10]; [n le 2 select I[n] else 6*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 19 2020
(Sage) [chebyshev_U(n, 3) +4*chebyshev_U(n-1, 3) for n in (0..30)] # G. C. Greubel, Jan 19 2020
(GAP) a:=[1, 10];; for n in [3..30] do a[n]:=6*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 19 2020
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Barry E. Williams, May 04 2000
EXTENSIONS
More terms from James A. Sellers, May 05 2000
STATUS
approved