OFFSET
1,2
COMMENTS
The case C=15 of finding k such that C*k+1 and (C+4)*k+2 are both perfect squares (A160682).
The 2 equations are equivalent to the Pell equation x^2 - 285*y^2 = 1, with x = (285*k+17)/2 and y = A*B/2.
LINKS
Andersen, K., Carbone, L. and Penta, D., Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
Index entries for linear recurrences with constant coefficients, signature (17, -1).
FORMULA
B(t+2) = 17*B(t+1) - B(t).
B(t) = ((285+19*w)*((17+w)/2)^(t-1)+(285-19*w)*((17-w)/2)^(t-1))/570 where w=sqrt(285).
G.f.: (1+x)*x/(1-17*x+x^2).
MAPLE
t:=0: for b from 1 to 1000000 do a:=sqrt((15*b^2+4)/19):
if (trunc(a)=a) then t:=t+1: n:=(b^2-1)/19: print(t, a, b, n): end if: end do:
MATHEMATICA
LinearRecurrence[{17, -1}, {1, 18}, 30] (* Harvey P. Dale, Jan 30 2024 *)
PROG
(Sage) [(lucas_number2(n, 17, 1)-lucas_number2(n-1, 17, 1))/15 for n in range(1, 20)] # Zerinvary Lajos, Nov 10 2009
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul Weisenhorn, Jun 14 2009
EXTENSIONS
Edited, extended by R. J. Mathar, Sep 02 2009
STATUS
approved