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
Examples
(-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.
Links
- Index entries for linear recurrences with constant coefficients, signature (0,0,6,0,0,-1).
Crossrefs
Programs
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Mathematica
LinearRecurrence[{0,0,6,0,0,-1},{221,241,265,1061,1205,1369},30] (* Harvey P. Dale, Nov 21 2011 *)
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PARI
{forstep(n=-24, 50000000, [3, 1], if(issquare(2*n^2+482*n+58081, &k), print1(k, ",")))}
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.
Comments