A159844 Positive numbers y such that y^2 is of the form x^2+(x+359)^2 with integer x.

325, 359, 401, 1549, 1795, 2081, 8969, 10411, 12085, 52265, 60671, 70429, 304621, 353615, 410489, 1775461, 2061019, 2392505, 10348145, 12012499, 13944541, 60313409, 70013975, 81274741, 351532309, 408071351, 473703905, 2048880445

Offset: 1

Author

Klaus Brockhaus, Apr 30 2009

Keywords

• nonn
• easy

Comments

(-36, a(1)) and (A130610(n), a(n+1)) are solutions (x, y) to the Diophantine equation x^2+(x+359)^2 = y^2.

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

lim_{n -> infinity} a(n)/a(n-1) = (363+38*sqrt(2))/359 for n mod 3 = {0, 2}.

lim_{n -> infinity} a(n)/a(n-1) = (293619+186550*sqrt(2))/359^2 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. [Mohamed Bouhamida, Sep 09 2009]

Examples

			(-36, a(1)) = (-36, 325) is a solution: (-36)^2+(-36+359)^2 = 1296+104329 = 105625 = 325^2.
(A130610(1), a(2)) = (0, 359) is a solution: 0^2+(0+359)^2 = 128881 = 359^2.
(A130610(3), a(4)) = (901, 1549) is a solution: 901^2+(901+359)^2 = 811801+1587600 = 2399401 = 1549^2.
		

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).

Crossrefs

Cf. A130610, A001653, A156035 (decimal expansion of 3+2*sqrt(2)), A159845 (decimal expansion of (363+38*sqrt(2))/359), A159846 (decimal expansion of (293619+186550*sqrt(2))/359^2).

Programs

Magma

I:=[325, 359, 401, 1549, 1795, 2081]; [n le 6 select I[n] else 6*Self(n-3) - Self(n-6): n in [1..30]]; // G. C. Greubel, May 19 2018

Mathematica

t={325,359,401,1549,1795,2081}; Do[AppendTo[t, 6*t[[-3]]-t[[-6]]], {25}]; t
CoefficientList[Series[(325+359 x+401 x^2-401 x^3-359 x^4-325 x^5)/(1-6 x^3+x^6),{x,0,30}],x]  (* Harvey P. Dale, Feb 16 2011 *)
LinearRecurrence[{0,0,6,0,0,-1}, {325, 359, 401, 1549, 1795, 2081}, 50] (* G. C. Greubel, May 19 2018 *)

PARI

{forstep(n=-36, 10000000, [1, 3], if(issquare(2*n^2+718*n+128881, &k), print1(k, ",")))}

PARI

V=[]; v=[[-323,-325], [-323,325], [0,-359], [-359,359], [-399,-401], [399,401]]; for(n=1,100,u=[]; for(i=1,#v,if(v[i][2]>0, u=concat(u,v[i][2])); t=3*v[i][1]+2*v[i][2]+359; v[i][2]=4*v[i][1]+3*v[i][2]+718; v[i][1]=t); V=concat(V,u)); vecsort(V,,8) \\ Charles R Greathouse IV, Feb 14 2011

Formula

a(n) = 6*a(n-3)-a(n-6) for n > 6; a(1)=325, a(2)=359, a(3)=401, a(4)=1549, a(5)=1795, a(6)=2081.

G.f.: (1-x)*(325+684*x+1085*x^2+684*x^3+325*x^4) / (1-6*x^3+x^6).

a(3*k-1) = 359*A001653(k) for k >= 1.