OFFSET
1,1
COMMENTS
(-13,a(1)) and (A129836(n), a(n+1)) are solutions (x, y) to the Diophantine equation x^2 + (x+97)^2 = y^2.
Lim_{n -> infinity} a(n)/a(n-3) = 3+2*sqrt(2).
Lim_{n -> infinity} a(n)/a(n-1) = (99+14*sqrt(2))/97 for n mod 3 = {0, 2}.
Lim_{n -> infinity} a(n)/a(n-1) = (19491+12070*sqrt(2))/97^2 for n mod 3 = 1.
For the generic case x^2+(x+p)^2=y^2 with p=2*m^2-1 a prime number in A066436, m>=2, 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+1, a(3)=6m^2-10m+4, a(4)=3p, a(5)=6m^2+10m+4, a(6)=40m^2-58m+21.Y values are given by the sequence defined by: b(n)=6*b(n-3)-b(n-6) with b(1)=p, b(2)=2m^2+2m+1, b(3)=10m^2-14m+5, b(4)=5p, b(5)=10m^2+14m+5, b(6)=58m^2-82m+29. - Mohamed Bouhamida, Sep 09 2009
LINKS
G. C. Greubel, Table of n, a(n) for n = 1..1000
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)=85, a(2)=97, a(3)=113, a(4)=397, a(5)=485, a(6)=593.
G.f.: (1-x)*(85 + 182*x + 295*x^2 + 182*x^3 + 85*x^4)/(1-6*x^3+x^6).
a(3*k-1) = 97*A001653(k) for k >= 1.
EXAMPLE
MATHEMATICA
LinearRecurrence[{0, 0, 6, 0, 0, -1}, {85, 97, 113, 397, 485, 593}, 30] (* Harvey P. Dale, Apr 04 2013 *)
PROG
(PARI) {forstep(n=-20, 800000000, [3, 1], if(issquare(2*n^2+194*n+9409, &k), print1(k, ", ")))};
(Magma) I:=[85, 97, 113, 397, 485, 593]; [n le 6 select I[n] else 6*Self(n-3) - Self(n-6): n in [1..50]]; // G. C. Greubel, Mar 31 2018
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Klaus Brockhaus, Mar 12 2009
STATUS
approved