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%I #29 Jul 23 2020 19:20:42
%S 1,3,5,1,11,9,13,31,5,57,67,47,181,87,275,449,101,999,797,1201,2795,
%T 393,5197,5983,4411,16377,7555,25199,40309,10089,90707,70529,110885,
%U 251943,30173,473713,534059,413367,1481485,654751,2308219,3617721
%N a(n) is the unique odd positive solution y of 2^n = 7x^2 + y^2.
%C 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
%D A. Engel, Problem-Solving Strategies, p. 126.
%H T. D. Noe, <a href="/A077021/b077021.txt">Table of n, a(n) for n=3..500</a>
%H Paul Barry, <a href="https://arxiv.org/abs/2004.04577">On a Central Transform of Integer Sequences</a>, arXiv:2004.04577 [math.CO], 2020.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/DiophantineEquation2ndPowers.html">Diophantine Equations 2nd Powers</a>
%F a(n+2) = abs(A002249(n)). - _Artur Jasinski_, Oct 05 2008 [With correction by _Jianing Song_, Nov 21 2018]
%t 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 *)
%Y Cf. A077020 (x).
%Y Cf. A002249, A038198, A060728.
%K nonn
%O 3,2
%A _Ed Pegg Jr_, Oct 17 2002
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