1,1

(-16, a(1)) and (A118676(n), a(n+1)) are solutions (x, y) to the Diophantine equation x^2+(x+79)^2 = y^2.

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

lim_{n -> infinity} a(n)/a(n-1) = (83+18*sqrt(2))/79 for n mod 3 = {0, 2}.

lim_{n -> infinity} a(n)/a(n-1) = (10659+6110*sqrt(2))/79^2 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)=65, a(2)=79, a(3)=101, a(4)=289, a(5)=395, a(6)=541.

G.f.: (1-x)*(65+144*x+245*x^2+144*x^3+65*x^4) / (1-6*x^3+x^6).

a(3*k-1) = 79*A001653(k) for k >= 1.

(-16, a(1)) = (-16, 65) is a solution: (-16)^2+(-16+79)^2 = 256+3969 = 4225 = 65^2.

(A118676(1), a(2)) = (0, 79) is a solution: 0^2+(0+79)^2 = 6241 = 79^2.

(A118676(3), a(4)) = (161, 289) is a solution: 161^2+(161+79)^2 = 25921+57600 = 83521 = 289^2.

(PARI) {forstep(n=-16, 10000000, [1, 3], if(issquare(2*n^2+158*n+6241, &k), print1(k, ", ")))}

Cf. A118676, A001653, A156035 (decimal expansion of 3+2*sqrt(2)), A159759 (decimal expansion of (83+18*sqrt(2))/79), A159760 (decimal expansion of (10659+6110*sqrt(2))/79^2).

Adjacent sequences: A159755 A159756 A159757 this_sequence A159759 A159760 A159761

Sequence in context: A095523 A060877 A113688 this_sequence A056693 A164282 A025312

nonn

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