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A077446 Numbers n such that 2*n^2 + 14 is a square. 7

%I #56 Aug 15 2023 18:03:19

%S 1,5,11,31,65,181,379,1055,2209,6149,12875,35839,75041,208885,437371,

%T 1217471,2549185,7095941,14857739,41358175,86597249,241053109,

%U 504725755,1404960479,2941757281,8188709765,17145817931,47727298111

%N Numbers n such that 2*n^2 + 14 is a square.

%C The equation "2*n^2 + 14 is a square" is a version of the generalized Pell Equation x^2 - D*y^2 = C where x^2 - 2*y^2 = 14.

%C Numbers n such that (ceiling(sqrt(n*n/2)))^2 = (7+n^2)/2. - _Ctibor O. Zizka_, Nov 09 2009

%C From _Wolfdieter Lang_, Feb 26 2015: (Start)

%C This sequence gives all positive solutions x = a(n+1), n >= 0, of the Pell equation x^2 - 2*y^2 = -7. For the corresponding y-solutions see y(n) = 2*A006452(n+2) = A077447(n+1)/2. This implies that X^2 - 2*Y^2 = 14 has the general solutions (X(n),Y(n)) = (2*y(n), x(n)). See the first comment above.

%C For the positive first class solutions see (A054490(n), 2*A038723(n)) and for the second class solutions (A255236(n), 2*A038725(n+1)). (End)

%C For n > 0, a(n) is the n-th almost Lucas-balancing number of second type (see Tekcan and Erdem). - _Stefano Spezia_, Nov 26 2022

%D 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.

%D L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. AMS Chelsea Publishing, Providence, Rhode Island, 1999, pp. 341-400.

%D 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.

%H Michael De Vlieger, <a href="/A077446/b077446.txt">Table of n, a(n) for n = 1..2612</a>

%H Jeremiah Bartz, Bruce Dearden, and Joel Iiams, <a href="https://arxiv.org/abs/1810.07895">Classes of Gap Balancing Numbers</a>, arXiv:1810.07895 [math.NT], 2018.

%H Jeremiah Bartz, Bruce Dearden, and Joel Iiams, <a href="https://ajc.maths.uq.edu.au/pdf/77/ajc_v77_p318.pdf">Counting families of generalized balancing numbers</a>, The Australasian Journal of Combinatorics (2020) Vol. 77, Part 3, 318-325.

%H J. J. O'Connor and E. F. Robertson, <a href="https://mathshistory.st-andrews.ac.uk/HistTopics/Pell/">Pell's Equation</a>

%H Ahmet Tekcan and Alper Erdem, <a href="https://arxiv.org/abs/2211.08907">General Terms of All Almost Balancing Numbers of First and Second Type</a>, arXiv:2211.08907 [math.NT], 2022.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PellEquation.html">Pell Equation</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,6,0,-1).

%F 2*(a(n))^2 + 14 = (A077447(n))^2.

%F Lim. n-> Inf. a(n)/a(n-2) = 5.8284271247461... = 3 + 2*sqrt(2) = A156035 = RG (Great Ratio).

%F Lim. k-> Inf. a(2*k+1)/a(2*k) = 2.09383632135605... = (9 + 4*sqrt(2))/7 = A156649 = R1 (Ratio 1).

%F Lim. k -> Inf. a(2*k)/a(2*k-1) = 2.78361162489122432754 = (11 + 6*sqrt(2))/7 = R2 (Ratio 2); RG = R1*R2.

%F a(2*k-1) = [ 2*[(3+2*Sqrt(2))^n - (3-2*Sqrt(2))^n] - [(3+2*Sqrt(2))^(n-1) - (3-2*Sqrt(2))^(n-1)] + [(3+2*Sqrt(2))^(n-2) - (3-2*Sqrt(2))^(n-2)] ] / (4*Sqrt(2)) a(2*k) = [ 5*[(3+2*Sqrt(2))^n - (3-2*Sqrt(2))^n] + [(3+2*Sqrt(2))^(n-1) - (3-2*Sqrt(2))^(n-1)] ] / (4*Sqrt(2)).

%F a(n) = 6*a(n-2) - a(n-4).

%F G.f.: x*(1+x)*(x^2+4*x+1) / ( (x^2+2*x-1)*(x^2-2*x-1) ). - _R. J. Mathar_, Jul 03 2011

%F a(n) = 6*a(n-2) - a(n-4) with a(1)=1, a(2)=5, a(3)=11, a(4)=31. - _Sture Sjöstedt_, Oct 08 2012

%F Bisection: a(2*k+1) = S(k, 6) + 5*S(k-1, 6), a(2*k) = 5*S(k-1, 6) + S(k-2, 6), with the Chebyshev polynomials S(n, x) (A049310) with S(-2, x) = -1, S(-1, x) = 0, evaluated at x = 6. S(n, 6) = A001109(n+1). See A054490 and A255236, and the given g.f.s. - _Wolfdieter Lang_, Feb 26 2015

%F E.g.f.: 1 - cosh(sqrt(2)*x)*(cosh(x) - 3*sinh(x)) - sqrt(2)*(cosh(x) - 2*sinh(x))*sinh(sqrt(2)*x). - _Stefano Spezia_, Nov 26 2022

%e n = 3: (A077447(3))^2 - 2*a(3)^2 = 16^2 - 2*11^2 = 14;

%e a(3)^2 - 2*(2*A006452(3+1))^2 = 11^2 - 2*(2*4)^2 = -7. - _Wolfdieter Lang_, Feb 26 2015

%t LinearRecurrence[{0,6,0,-1},{1,5,11,31},50] (* _Sture Sjöstedt_, Oct 08 2012 *)

%Y Cf. A001109, A006452, A038723, A038725, A049310, A054490, A077447, A155765, A156035, A156649, A255236.

%K nonn,easy

%O 1,2

%A _Gregory V. Richardson_, Nov 09 2002

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