OFFSET
3,2
COMMENTS
Restate the formula and divide both sides by 4, then y^2 - (-7)*x^2 = 2^n and (y/2)^2 - (-7)*(x/2)^2 = 2^(n-2). Let y = V_n, x = U_n, -7 = D, and 2^(n-2) = Q^n. We then have this sequence as the absolute values for V_n = A002249(n)(excluding a(0) = 2) in relation to the Lucas sequence identity: (V_n/2)^2 - D*(U_n/2)^2 = Q^n with V_n = (a^n + b^n), U_n = (a^n - b^n)/(a - b), D = (a - b)^2 = P^2 - 4*Q = -7, Q = (a*b) = 2; a = (1 + sqrt(-7))/2, b = (1 - sqrt(-7))/2, P = 1. By the Ramanujan-Nagell Theorem, iff y is in +- {1, 3, 5, 11, 181} = +-A038198, then |x| = 1 and we are left with 2^n = 7 + y^2. See A060728 and note that a(A060728(n) - 3) = A038198(n). - Raphie Frank, Dec 05 2015
REFERENCES
A. Engel, Problem-Solving Strategies, p. 126.
LINKS
T. D. Noe, Table of n, a(n) for n=3..500
Paul Barry, On a Central Transform of Integer Sequences, arXiv:2004.04577 [math.CO], 2020.
Eric Weisstein's World of Mathematics, Diophantine Equations 2nd Powers
FORMULA
a(n+2) = abs(A002249(n)). - Artur Jasinski, Oct 05 2008 [With correction by Jianing Song, Nov 21 2018]
MATHEMATICA
a = {}; Do[k = Expand[((1 + I Sqrt[7])/2)^n + ((1 - I Sqrt[7])/2)^n]; AppendTo[a, Abs[k]], {n, 1, 50}]; a (* Artur Jasinski, Oct 05 2008 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Ed Pegg Jr, Oct 17 2002
STATUS
approved