1,1

(-40, a(1)) and (A130645(n), a(n+1)) are solutions (x, y) to the Diophantine equation x^2+(x+439)^2 = y^2.

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

lim_{n -> infinity} a(n)/a(n-1) = (443+42*sqrt(2))/439 for n mod 3 = {0, 2}.

lim_{n -> infinity} a(n)/a(n-1) = (450483+287918*sqrt(2))/439^2 for n mod 3 = 1.

For the generic case x^2+(x+p)^2=y^2 with p=m^2-2 a prime number in A028871, m>=5, the x values are given by the sequence defined by: a(n)=6*a(n-3)-a(n-6)+2p with a(1)=0, a(2)=2m+2, a(3)=3m^2-10m+8, a(4)=3p, a(5)=3m^2+10m+8, a(6)=20m^2-58m+42.Y values are given by the sequence defined by: b(n)=6*b(n-3)-b(n-6) with b(1)=p, b(2)=m^2+2m+2, b(3)=5m^2-14m+10, b(4)=5p, b(5)=5m^2+14m+10, b(6)=29m^2-82m+58. [From Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Sep 09 2009]

a(n) = 6*a(n-3)-a(n-6) for n > 6; a(1)=401, a(2)=439, a(3)=485, a(4)=1921, a(5)=2195, a(6)=2509.

G.f.: (1-x)*(401+840*x+1325*x^2+840*x^3+401*x^4) / (1-6*x^3+x^6).

a(3*k-1) = 439*A001653(k) for k >= 1.

(-40, a(1)) = (-40, 401) is a solution: (-40)^2+(-40+439)^2 = 1600+159201 = 160801 = 401^2.

(A130645(1), a(2)) = (0, 439) is a solution: 0^2+(0+439)^2 = 192721 = 439^2.

(A130645(3), a(4)) = (1121, 1921) is a solution: 1121^2+(1121+439)^2 = 1256641+2433600 = 3690241 = 1921^2.

(PARI) {forstep(n=-40, 10000000, [1, 3], if(issquare(2*n^2+878*n+192721, &k), print1(k, ", ")))}

Cf. A130645, A001653, A156035 (decimal expansion of 3+2*sqrt(2)), A159891 (decimal expansion of (443+42*sqrt(2))/439), A159892 (decimal expansion of (450483+287918*sqrt(2))/439^2).

Sequence in context: A151652 A013767 A013895 this_sequence A029705 A096991 A141026

Adjacent sequences: A159887 A159888 A159889 this_sequence A159891 A159892 A159893

nonn

Klaus Brockhaus (klaus-brockhaus(AT)t-online.de), Apr 30 2009