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 A077021 a(n) is the unique odd positive solution y of 2^n = 7x^2 + y^2. 6
 1, 3, 5, 1, 11, 9, 13, 31, 5, 57, 67, 47, 181, 87, 275, 449, 101, 999, 797, 1201, 2795, 393, 5197, 5983, 4411, 16377, 7555, 25199, 40309, 10089, 90707, 70529, 110885, 251943, 30173, 473713, 534059, 413367, 1481485, 654751, 2308219, 3617721 (list; graph; refs; listen; history; text; internal format)
 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 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)/2)^n + ((1 - I Sqrt)/2)^n]; AppendTo[a, Abs[k]], {n, 1, 50}]; a (* Artur Jasinski, Oct 05 2008 *) CROSSREFS Cf. A077020 (x). Cf. A002249, A038198, A060728. Sequence in context: A026253 A259182 A138259 * A143250 A221494 A174883 Adjacent sequences:  A077018 A077019 A077020 * A077022 A077023 A077024 KEYWORD nonn AUTHOR Ed Pegg Jr, Oct 17 2002 STATUS approved

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Last modified May 31 19:11 EDT 2020. Contains 334748 sequences. (Running on oeis4.)