OFFSET
1,1
COMMENTS
(-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) +2*p with a(1)=0, a(2)= 2*m+2, a(3)= 3*m^2 -10*m +8, a(4)= 3*p, a(5)= 3*m^2 +10*m +8, a(6)= 20*m^2 -58*m +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 +2*m +2, b(3)= 5*m^2 -14*m +10, b(4)= 5*p, b(5)= 5*m^2 +14*m +10, b(6)= 29*m^2 -82*m +58. - Mohamed Bouhamida, Sep 09 2009
LINKS
G. C. Greubel, Table of n, a(n) for n = 1..3895
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)=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.
EXAMPLE
MATHEMATICA
LinearRecurrence[{0, 0, 6, 0, 0, -1}, {401, 439, 485, 1921, 2195, 2509}, 50] (* G. C. Greubel, May 17 2018 *)
PROG
(PARI) {forstep(n=-40, 10000000, [1, 3], if(issquare(2*n^2+878*n+192721, &k), print1(k, ", ")))}
(Magma) I:=[401, 439, 485, 1921, 2195, 2509]; [n le 6 select I[n] else 6*Self(n-3) - Self(n-6): n in [1..30]]; // G. C. Greubel, May 17 2018
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Klaus Brockhaus, Apr 30 2009
STATUS
approved