A077444 Numbers k such that (k^2 + 4)/2 is a square.

2, 14, 82, 478, 2786, 16238, 94642, 551614, 3215042, 18738638, 109216786, 636562078, 3710155682, 21624372014, 126036076402, 734592086398, 4281516441986, 24954506565518, 145445522951122, 847718631141214, 4940866263896162, 28797478952235758, 167844007449518386

Offset

1,1

Comments

The equation "(k^2 + 4)/2 is a square" is a version of the generalized Pell Equation x^2 - D*y^2 = C where x^2 - 2*y^2 = -4.

Sequence of all positive integers k such that continued fraction [k,k,k,k,k,k,...] belongs to Q(sqrt(2)). - Thomas Baruchel, Sep 15 2003

Equivalently, 2*n^2 + 8 is a square.

Numbers n such that (ceiling(sqrt(n*n/2)))^2 = 2 + n^2/2. - Ctibor O. Zizka, Nov 09 2009

The continued fraction [a(n);a(n),a(n),...] = (1 + sqrt(2))^(2*n-1). - Thomas Ordowski, Jun 07 2013

a((p+1)/2) == 2 (mod p) where p is an odd prime. - Altug Alkan, Mar 17 2016

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.

Links

Amiram Eldar, Table of n, a(n) for n = 1..1306

S. Falcon, Relationships between Some k-Fibonacci Sequences, Applied Mathematics, 2014, 5, 2226-2234.

Tanya Khovanova, Recursive Sequences

J. J. O'Connor and E. F. Robertson, Pell's Equation

Eric Weisstein's World of Mathematics, Pell Equation

Eric Weisstein's World of Mathematics, NSW Number

Index entries for linear recurrences with constant coefficients, signature (6,-1).

Formula

a(n) = (((3 + 2*sqrt(2))^n - (3 - 2*sqrt(2))^n) + ((3 + 2*sqrt(2))^(n-1) - 3 - 2*sqrt(2))^(n-1))) / (2*sqrt(2)).

a(n) = 2*A002315(n-1).

Recurrence: a(n) = 6*a(n-1) - a(n-2), starting 2, 14.

Offset 0, with a=3+2*sqrt(2), b=3-2*sqrt(2): a(n) = a^((2n+1)/2) - b^((2n+1)/2). a(n) = 2*(A001109(n+1) + A001109(n)) = (A003499(n+1) - A003499(n))/2 = 2*sqrt(A001108(2n+1)) = sqrt(A003499(2n+1)-2). - Mario Catalani (mario.catalani(AT)unito.it), Mar 31 2003

Limit_{n->oo} a(n)/a(n-1) = 5.82842712474619009760... = 3 + 2*sqrt(2). See A156035.

From R. J. Mathar, Nov 16 2007: (Start)

G.f.: 2*x*(1+x)/(1-6*x+x^2).

a(n) = 2*(7*A001109(n) - A001109(n+1)). (End)

a(n) = (1+sqrt(2))^(2*n-1) - (1+sqrt(2))^(1-2*n). - Gerson Washiski Barbosa, Sep 19 2010

a(n) = floor((1 + sqrt(2))^(2*n-1)). - Thomas Ordowski, Jun 07 2013

a(n) = sqrt(2*A075870(n)^2-4). - Derek Orr, Jun 18 2015

a(n) = 2*sqrt((2*A001653(n)^2)-1). - César Aguilera, Jul 13 2023

Mathematica

LinearRecurrence[{6, -1}, {2, 14}, 30] (* Harvey P. Dale, Jul 25 2018 *)

Prog

(PARI) for(n=1, 20, q=(1+sqrt(2))^(2*n-1); print1(contfrac(q)[1], ", ")) \\ Derek Orr, Jun 18 2015
(PARI) Vec(2*x*(1+x)/(1-6*x+x^2) + O(x^100)) \\ Altug Alkan, Mar 17 2016
(Magma) [n: n in [0..10^8] | IsSquare((n^2 + 4) div 2)]; // Vincenzo Librandi, Jun 20 2015

Crossrefs

(A077445(n))^2 - 2*a(n) = 8.

First differences of A001541.

Pairwise sums of A001542.

Bisection of A002203 and A080039.

Cf. A001653.

Sequence in context: A361813 A102401 A077461 * A335693 A138126 A268881

Adjacent sequences: A077441 A077442 A077443 * A077445 A077446 A077447

Keyword

nonn,easy

Author

Gregory V. Richardson, Nov 09 2002

Status

approved