A159758 Positive numbers y such that y^2 is of the form x^2+(x+79)^2 with integer x.

65, 79, 101, 289, 395, 541, 1669, 2291, 3145, 9725, 13351, 18329, 56681, 77815, 106829, 330361, 453539, 622645, 1925485, 2643419, 3629041, 11222549, 15406975, 21151601, 65409809, 89798431, 123280565, 381236305, 523383611, 718531789

Offset: 1

Author

Klaus Brockhaus, Apr 30 2009

Keywords

• nonn
• easy

Comments

(-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) + 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

Examples

			(-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.
		

Links

• Harvey P. Dale, 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. 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).

Programs

Magma

I:=[65,79,101,289,395,541]; [n le 6 select I[n] else 6*Self(n-3) - Self(n-6): n in [1..30]]; // G. C. Greubel, May 22 2018

Mathematica

RecurrenceTable[{a[1]==65,a[2]==79,a[3]==101,a[4]==289,a[5]==395, a[6]== 541, a[n]==6a[n-3]-a[n-6]},a[n],{n,30}] (* or *) LinearRecurrence[ {0,0,6,0,0,-1},{65,79,101,289,395,541},30] (* Harvey P. Dale, Oct 03 2011 *)

PARI

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

Formula

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.