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 A077443 Numbers k such that (k^2 - 7)/2 is a square. 10
 3, 5, 13, 27, 75, 157, 437, 915, 2547, 5333, 14845, 31083, 86523, 181165, 504293, 1055907, 2939235, 6154277, 17131117, 35869755, 99847467, 209064253, 581953685, 1218515763, 3391874643, 7102030325, 19769294173, 41393666187 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS Lim_{n -> inf} a(n)/a(n-2) = 3 + 2*sqrt(2) = R1*R2. Lim_{k -> inf} a(2*k-1)/a(2*k) = (9 + 4*sqrt(2))/7 = R1 = A156649 (ratio #1). Lim_{k -> inf} a(2*k)/a(2*k-1) = (11 + 6*sqrt(2))/7 = R2 (ratio #2). Also gives solutions > 3 to the equation x^2-4 = floor(x*r*floor(x/r)) where r=sqrt(2). - Benoit Cloitre, Feb 14 2004 From Paul Curtz, Dec 15 2012: (Start) a(n-1) and A006452(n) are companions. Like A000129 and A001333. Reduced mod 10 this is a sequence of period 12: 3, 5, 3, 7, 5, 7, 7, 5, 7, 3, 5, 3. (End) The Pisano periods (periods of the sequence reducing a(n) modulo m) for m>=1 are 1,  1,  8,  4, 12,  8,  6,  4, 24, 12, 24,  8, 28, ... R. J. Mathar, Dec 15 2012 Positive values of x (or y) satisfying x^2 - 6xy + y^2 + 56 = 0. - Colin Barker, Feb 08 2014 From Wolfdieter Lang, Feb 05 2015: (Start) a(n+1) gives for n >= 0 all positive x solutions of the (generalized) Pell equation x^2 - 2*y^2 = +7. The corresponding y solutions are given in A077442(n), n >= 0. The, e.g., the Nagell reference for finding all solutions. Because the primitive Pythagorean triangle (3,4,5) is the only one with the sum of legs equal to 7 all positive solutions (x(n),y(n)) = (a(n+1),A077442(n)) of the Pell equation x^2 - 2*y^2 = +7 satisfy x(n) - y(n) < y(n) if n >= 1; only the first solution (x(0),y(0)) = (3,2) satisfies 3-1 > 1. Proof: Primitive Pythagorean triangles are characterized by the positive integer pairs [u,v] with  u+v odd, gcd(u,v) = 1 and u > v. See the Niven et al. reference, Theorem 5.5, p. 232. The leg sum is L = (u+v)^2 - 2*v^2. With L = 7, x = u+v and y = v, every solution (x(n),y(n)) with x(n)-y(n) = u(n) > v(n) = y(n) will correspond to a primitive Pythagorean triangle. Note that because of gcd(x,y) = 1 also gcd(u,v) = 1. But there is only one such triangle with L=7, namely the one with  [u(0),v(0)] = [2,1]. All other solutions with n >= 1 must therefore satisfy x(n)-y(n) < y(n). (End) REFERENCES A. H. Beiler, "The Pellian." Ch. 22 in Recreations in the Theory of Numbers: The Queen of Mathematics Entertains. Dover, New York, New York, pp. 248-268, 1966. L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. AMS Chelsea Publishing, Providence, Rhode Island, 1999, pp. 341-400. Peter G. L. Dirichlet, Lectures on Number Theory (History of Mathematics Source Series, V. 16); American Mathematical Society, Providence, Rhode Island, 1999, pp. 139-147. T. Nagell, Introduction to Number Theory, Chelsea Publishing Company, 1964, Theorem 109, pp. 207-208 with Theorem 104, pp. 197-198. Ivan Niven, Herbert S. Zuckerman and Hugh L. Montgomery, An Introduction to the Theory Of Numbers, Fifth Edition, John Wiley and Sons, Inc., NY 1991. LINKS Vincenzo Librandi, Table of n, a(n) for n = 1..1000 Jeremiah Bartz, Bruce Dearden, and Joel Iiams, Classes of Gap Balancing Numbers, arXiv:1810.07895 [math.NT], 2018. Jeremiah Bartz, Bruce Dearden, and Joel Iiams, Counting families of generalized balancing numbers, The Australasian Journal of Combinatorics (2020) Vol. 77, Part 3, 318-325. J. J. O'Connor and E. F. Robertson, History of Pell's Equation J. P. Robertson, Solving the Generalized Pell Equation Eric Weisstein's World of Mathematics, Pell Equation. Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1) FORMULA a(2n+1) = A038762(n). a(2n) = A101386(n-1). The same recurrences hold for the odd and the even indices: a(n+2) = 6*a(n) - a(n-2), a(n+1) = 3*a(n) + 2*(2*a(n)^2-14)^0.5 - Richard Choulet, Oct 11 2007 O.g.f.: -x*(x-1)*(3*x^2+8*x+3) / ( (x^2+2*x-1)*(x^2-2*x-1) ). - R. J. Mathar, Nov 23 2007 If n is even a(n) = (1/2)*(3+sqrt(2))*(3+2*sqrt(2))^-(1/2)*n) +(1/2)*(3-sqrt(2))*(3-2*sqrt(2))^-(1/2)*n); if n is odd a(n) = (1/2)*(3+sqrt(2))*(3+2*sqrt(2))^((1/2)n-1/2)) +(1/2)*(3-sqrt(2))*(3-2*sqrt(2))^((1/2)n-1/2)). - Antonio Alberto Olivares, Apr 20 2008 a(n) = A000129(n+1) + (-1)^n*A176981(n-1), n>1. - R. J. Mathar, Jul 03 2011 a(n) = A000129(n+1) -(-1)^n*A000129(n-2), rephrasing the formula above. - Paul Curtz, Dec 07 2012 a(n) = sqrt(8*A216134(n)^2 + 8*A216134(n) + 9) = 2*A124124(n) + 1. - Raphie Frank, May 24 2013 EXAMPLE a(3)^2 - 2*A077442(2)^2 = 13^2 - 2*9^2  = +7. - Wolfdieter Lang, Feb 05 2015 MATHEMATICA LinearRecurrence[{0, 6, 0, -1}, {3, 5, 13, 27}, 50] (* Sture Sjöstedt, Oct 09 2012 *) CROSSREFS Cf. A077442, A038762, A038761, A101386 A253811. Sequence in context: A035082 A005198 A160823 * A147196 A110225 A065047 Adjacent sequences:  A077440 A077441 A077442 * A077444 A077445 A077446 KEYWORD nonn,easy AUTHOR Gregory V. Richardson, Nov 06 2002 EXTENSIONS More terms from Richard Choulet, Oct 11 2007 Edited: replaced n by a(n) in the name. Moved Pell remarks to the comment section. Added cross references. - Wolfdieter Lang, Feb 05 2015 STATUS approved

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