A159565 Positive numbers y such that y^2 is of the form x^2+(x+241)^2 with integer x.

221, 241, 265, 1061, 1205, 1369, 6145, 6989, 7949, 35809, 40729, 46325, 208709, 237385, 270001, 1216445, 1383581, 1573681, 7089961, 8064101, 9172085, 41323321, 47001025, 53458829, 240849965, 273942049, 311580889, 1403776469, 1596651269

Offset

1,1

Comments

(-21,a(1)) and (A129991(n), a(n+1)) are solutions (x, y) to the Diophantine equation x^2+(x+241)^2 = y^2.

Links

Table of n, a(n) for n=1..29.

Index entries for linear recurrences with constant coefficients, signature (0,0,6,0,0,-1).

Formula

a(n) = 6*a(n-3)-a(n-6) for n > 6; a(1)=221, a(2)=241, a(3)=265, a(4)=1061, a(5)=1205, a(6)=1369.

G.f.: x*(1-x)*(221+462*x+727*x^2+462*x^3+221*x^4) / (1-6*x^3+x^6).

a(3*k-1) = 241*A001653(k) for k >= 1.

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

Limit_{n -> oo} a(n)/a(n-1) = (243+22*sqrt(2))/241 for n mod 3 = {0, 2}.

Limit_{n -> oo} a(n)/a(n-1) = (137283+87958*sqrt(2))/241^2 for n mod 3 = 1.

Example

(-21, a(1)) = (-21, 221) is a solution: (-21)^2+(-21+241)^2 = 441+48400 = 48841 = 221^2.
(A129993(1), a(2)) = (0, 241) is a solution: 0^2+(0+241)^2 = 58081= 241^2.
(A129993(3), a(4)) = (620, 1061) is a solution: 620^2+(620+241)^2 = 384400+741321 = 1125721 = 1061^2.

Mathematica

LinearRecurrence[{0, 0, 6, 0, 0, -1}, {221, 241, 265, 1061, 1205, 1369}, 30] (* Harvey P. Dale, Nov 21 2011 *)

Prog

(PARI) {forstep(n=-24, 50000000, [3, 1], if(issquare(2*n^2+482*n+58081, &k), print1(k, ", ")))}

Crossrefs

Cf. A129991, A001653, A156035 (decimal expansion of 3+2*sqrt(2)), A159566 (decimal expansion of (243+22*sqrt(2))/241), A159567 (decimal expansion of (137283+87958*sqrt(2))/241^2).

Sequence in context: A339680 A266237 A371899 * A330281 A345512 A048931

Adjacent sequences: A159562 A159563 A159564 * A159566 A159567 A159568

Keyword

nonn,easy

Author

Klaus Brockhaus, Apr 16 2009

Status

approved