A157348 Positive numbers y such that y^2 is of the form x^2+(x+281)^2 with integer x.
229, 281, 365, 1009, 1405, 1961, 5825, 8149, 11401, 33941, 47489, 66445, 197821, 276785, 387269, 1152985, 1613221, 2257169, 6720089, 9402541, 13155745, 39167549, 54802025, 76677301, 228285205, 319409609, 446908061, 1330543681
Offset: 1
Examples
(-60, a(1)) = (-60, 229) is a solution: (-60)^2+(-60+281)^2 = 3600+48841 = 52441 = 229^2. (A129626(1), a(2)) = (0, 281) is a solution: 0^2+(0+281)^2 = 78961 = 281^2. (A129626(3), a(4)) = (559, 1009) is a solution: 559^2+(559+281)^2 = 312481+705600 = 1018081 = 1009^2.
Links
- Index entries for linear recurrences with constant coefficients, signature (0,0,6,0,0,-1).
Crossrefs
Programs
-
PARI
{forstep(n=-60, 200000000, [3, 1], if(issquare(2*n^2+562*n+78961, &k), print1(k, ",")))}
Formula
a(n) = 6*a(n-3)-a(n-6) for n > 6; a(1)=229, a(2)=281, a(3)=365, a(4)=1009, a(5)=1405, a(6)=1961.
G.f.: x*(1-x)*(229+510*x+875*x^2+510*x^3+229*x^4) / (1-6*x^3+x^6).
a(3*k-1) = 281*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) = (297+68*sqrt(2))/281 for n mod 3 = {0, 2}.
Limit_{n -> oo} a(n)/a(n-1) = (130803+73738*sqrt(2))/281^2 for n mod 3 = 1.
Comments