A160451 a(n) = (4/3)*u*(u^3+6*u^2+8*u-3) where u=floor((3*n+5)/2).

1008, 2080, 6440, 10208, 22360, 31416, 57408, 75208, 122816, 153680, 232408, 281520, 402600, 476008, 652400, 757016, 1003408, 1147008, 1479816, 1671040, 2108408, 2356760, 2918560, 3234408, 3942240, 4336816, 5214008, 5699408, 6771016, 7360200, 8653008, 9359800

Offset: 1

Author

John W. Layman, May 14 2009

Keywords

• nonn
• easy

Comments

It appears that the 4-tuple (3, (u^2-1)/3, (floor((3*n+11)/2)^2-1)/3, a(n)) has the Diophantus' property that the product of any two distinct terms plus one is a square.

Examples

			For n=1 we get the 4-tuple (3,5,16,1008), and 3*5+1=16=4^2, 3*16+1=49=7^2, 3*1008+1=3025=55^2, 5*16+1=81=9^2, 5*1008+1=5041=71^2, 16*1008+1=16129=127^2.
		

Links

• Lenny Jones, A polynomial Approach to a Diophantine Problem, Math. Mag. 72 (1999) 52-55.
• Eric Weisstein's World of Mathematics, Diophantus Property.
• Index entries for linear recurrences with constant coefficients, signature (1,4,-4,-6,6,4,-4,-1,1).

Crossrefs

Cf. A086302, A160372.

Programs

Mathematica

Table[u=Floor[(3n+5)/2];4/3 u(u^3+6u^2+8u-3),{n,30}] (* or *) LinearRecurrence[{1,4,-4,-6,6,4,-4,-1,1},{1008,2080,6440,10208,22360,31416,57408,75208,122816},30] (* Harvey P. Dale, Nov 19 2013 *)

Formula

From R. J. Mathar, May 15 2009: (Start)

a(n) = a(n-1)+4*a(n-2)-4*a(n-3)-6*a(n-4)+6*a(n-5)+4*a(n-6)-4*a(n-7)-a(n-8)+a(n-9).

G.f.: -8*x*(126+134*x+41*x^2-65*x^3+95*x^4+52*x^5-61*x^6-13*x^7+15*x^8)/((1+x)^4*(x-1)^5). (End)