A075870 Numbers k such that 2*k^2 - 4 is a square.

2, 10, 58, 338, 1970, 11482, 66922, 390050, 2273378, 13250218, 77227930, 450117362, 2623476242, 15290740090, 89120964298, 519435045698, 3027489309890, 17645500813642, 102845515571962, 599427592618130, 3493720040136818, 20362892648202778, 118683635849079850

Offset: 1

Author

Gregory V. Richardson, Oct 16 2002

Keywords

• nonn
• easy

Comments

Lim_{n->infinity} a(n)/a(n-1) = 3 + 2*sqrt(2).

Also gives solutions to the equation x^2-2 = floor(x*r*floor(x/r)) where r=sqrt(2). - Benoit Cloitre, Feb 14 2004

The upper intermediate convergents to 2^(1/2) beginning with 10/7, 58/41, 338/239, 1970/1393 form a strictly decreasing sequence; essentially, numerators = A075870, denominators = A002315. - Clark Kimberling, Aug 27 2008

Numbers n such that sqrt(floor(n^2/2 - 1)) is an integer. The integer square roots are given by A002315. - Richard R. Forberg, Aug 01 2013

a(n) are the integer square roots of m^2 + (m+2)^2. The values of m are given by A065113 (except for m = 0). The values of this expression are given by A165518. - Richard R. Forberg, Aug 15 2013

Values of x (or y) in the solutions to x^2 - 6*x*y + y^2 + 16 = 0. - Colin Barker, Feb 04 2014

Also integers k such that k^2 is equal to the sum of four consecutive triangular numbers. - Colin Barker, Dec 20 2014

Equivalently, numbers x such that (x-1)*x/2 + x*(x+1)/2 = (y-1)^2 + (y+1)^2. y-values are listed in A002315. Example: for x=58 and y=41, 57*58/2 + 58*59/2 = 40^2 + 42^2. - Bruno Berselli, Mar 19 2018

Examples

			From _Muniru A Asiru_, Mar 19 2018: (Start)
For k=2, 2*2^2 - 4 = 8 - 4 = 4 = 2^2.
For k=10, 2*10^2 - 4 = 200 - 4 =  196 = 14^2.
For k=58, 2*58^2 - 4 = 6728 - 4 =  6724 = 82^2.
... (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.
• P.-F. Teilhet, Reply to Query 2094, L'Intermédiaire des Mathématiciens, 10 (1903), 235-238. - N. J. A. Sloane, Mar 03 2022

Links

• Colin Barker, Table of n, a(n) for n = 1..1000
• Tanya Khovanova, Recursive Sequences
• J. J. O'Connor and E. F. Robertson, Pell's Equation
• Eric Weisstein's World of Mathematics, Pell Equation.
• Index entries for linear recurrences with constant coefficients, signature (6,-1).

Crossrefs

Cf. A000129, A000217, A000290, A002315.

Twice A001653.

Programs

GAP

a:=[2,10];; for n in [3..25] do a[n]:=6*a[n-1]-a[n-2]; od; a; # Muniru A Asiru, Mar 19 2018

Maple

a:= proc(n) option remember: if n = 1 then 2 elif n = 2 then 10 elif  n >= 3 then 6*procname(n-1) - procname(n-2) fi; end: seq(a(n), n = 0..25); # Muniru A Asiru, Mar 19 2018

Mathematica

LinearRecurrence[{6,-1},{2,10},30] (* Harvey P. Dale, Sep 27 2018 *)

PARI

Vec(2*x*(1-x)/(1-6*x+x^2) + O(x^100)) \\ Colin Barker, Dec 20 2014

Formula

a(n) = 2 * A001653(n).

a(n) = (1/sqrt(2))*((1+sqrt(2))^(2*n-1) - (1-sqrt(2))^(2*n-1)) = 6*a(n-1) - a(n-2).

G.f.: 2*x*(1-x)/(1-6*x+x^2). - Philippe Deléham, Nov 17 2008

a(n) = round(((2+sqrt(2))*(3+2*sqrt(2))^(n-1))/2). - Paul Weisenhorn, Jun 11 2020

From Peter Bala, Aug 21 2022: (Start)

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

1/a(n) - 1/a(n+1) = 1/(Pell(2*n) + 1/Pell(2*n)), where Pell(n) = A000129(n). (End)

Extensions

More terms from Colin Barker, Dec 20 2014