login
Numbers k such that k+1 and 3*k+1 are perfect squares.
14

%I #81 Jul 14 2024 18:48:24

%S 0,8,120,1680,23408,326040,4541160,63250208,880961760,12270214440,

%T 170902040408,2380358351280,33154114877520,461777249934008,

%U 6431727384198600,89582406128846400,1247721958419651008,17378525011746267720,242051628206028097080

%N Numbers k such that k+1 and 3*k+1 are perfect squares.

%C Essentially the same as A051047.

%C It appears that a(n) = A046175(n)-A046174(n), that is, the triangular index of the n-th pentagonal triangular number minus its pentagonal index. - _Jonathan Vos Post_, Feb 28 2011

%C Sequence lists the nonnegative x solutions when (x + 1)*(3*x + 1) is a square. Positive x solutions when (x - 1)*(3*x - 1) is a square are in A011922. - _Bruno Berselli_, Feb 20 2018

%H G. C. Greubel, <a href="/A045899/b045899.txt">Table of n, a(n) for n = 1..870</a>

%H A. Baker and H. Davenport, <a href="http://dx.doi.org/10.1093/qmath/20.1.129">The Equations 3x^2-2=y^2 and 8x^2-7=z^2</a>, Quart. J. Math. Oxford 20 (1969).

%H A. Dujella and A. Pethoe, <a href="http://qjmath.oxfordjournals.org/content/49/3/291.extract">A generalization of a theorem of Baker and Davenport</a>, Quart. J. Math. Oxford Ser. (2) 49 (1998), 291-306.

%H A. Dujella, <a href="http://www.math.hr/~duje/ddbib.html">The Problem of Diophantus and Davenport, References</a>

%H A. Dujella, <a href="http://www.math.hr/~duje/papers.html">Publications of Andrej Dujella</a>

%H Z. Franusic, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL13/Franusic/franusic4.html">On the Extension of the Diophantine Pair {1,3} in Z[surd d]</a>, J. Int. Seq. 13 (2010) # 10.9.6

%H P. Gibbs, <a href="http://www.weburbia.com/pg/diophant.htm">1,3,8,120 ... A Diophantine Problem</a>

%H P. Gibbs, <a href="https://arxiv.org/abs/math/0107203">Diophantine quadruples and Cayley's hyperdeterminant</a>, arXiv:math/0107203 [math.NT], 2001.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (15,-15,1).

%F a(n) = A046184(n+1) - 1.

%F a(n) = 14*a(n-1) - a(n-2) + 8.

%F a(n) = ((2 + sqrt(3))*(7 + 4*sqrt(3))^n + (2 - sqrt(3))*(7 - 4*sqrt(3))^n - 4)/6. - Joseph Biberstine (jrbibers(AT)indiana.edu), Apr 23 2006

%F a(n) = 8*A076139(n-1) = 4*A217855(n-1) = 2*A123480(n-1) = 8/3*A076140(n-1). - _Peter Bala_, Dec 31 2012

%F From _Colin Barker_, Jul 30 2013: (Start)

%F G.f.: -8*x^2 / ((x - 1)*(x^2 - 14*x + 1)).

%F a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3). (End)

%F E.g.f.: (-4*exp(x) + (2 + sqrt(3))*exp((7-4*sqrt(3))*x) + (2 - sqrt(3))*exp((7+4*sqrt(3))*x))/6. - _Ilya Gutkovskiy_, Apr 28 2016

%t f[n_] := FullSimplify[((Sqrt[3] + 2)*(7 + 4*Sqrt[3])^n - (Sqrt[3] - 2) (7 - 4 Sqrt[3])^n - 4)/6]; Array[f, 18, 0] (* Joseph Biberstine (jrbibers(AT)indiana.edu), Apr 23 2006 *)

%t Rest[CoefficientList[Series[-8*x^2/((x - 1)*(x^2 - 14*x + 1)), {x,0,50}], x]] (* _G. C. Greubel_, Jun 07 2017 *)

%t LinearRecurrence[{15,-15,1},{0,8,120},20] (* _Harvey P. Dale_, Jul 14 2024 *)

%o (PARI) x='x+O('x^50); concat([0], Vec(-8*x^2/((x - 1)*(x^2 - 14*x + 1)))) \\ _G. C. Greubel_, Jun 07 2017

%Y Cf. A051047, A067900. A076139, A076140, A123480, A217855.

%Y Cf. A046184, A245031.

%K nonn,easy

%O 1,2

%A Andrej Dujella (duje(AT)math.hr)