A154708 Numbers a such that b and c exist with b <= a < c and a*(a+1) + b^2 = c^2.

3, 4, 7, 8, 11, 12, 15, 16, 19, 20, 23, 24, 27, 28, 31, 32, 35, 36, 39, 40, 43, 44, 47, 48, 51, 52, 55, 56, 59, 60, 63, 64, 67, 68, 71, 72, 75, 76, 79, 80, 83, 84, 87, 88, 91, 92, 95, 96, 99, 100, 103, 104, 107, 108, 111, 112, 115, 116, 119, 120, 123, 124, 127, 128, 131, 132, 135, 136, 139, 140

Offset: 1

Author

Pierre CAMI, Jan 14 2009, Jan 15 2009

Keywords

• nonn
• easy

Comments

Numbers a such that a*(a+1) = c2 - b2 with b <= a < c let a(1)=3 then a(2*n) = a(2*n-1) + 1 and a(2*n+1) = a(2*n) + 3. [Pierre CAMI, Jan 15 2009]

Examples

			4*5 + 4*4 = 6*6; 7*8 + 5*5 = 9*9; 8*9 + 7*7 = 11*11;
3*4 = 4*4 - 2*2; a(1)=3; 4*5 = 6*6 - 4*4; a(2) = 4; 7*6 = 9*9 - 5*5; a(3)=7. [_Pierre CAMI_, Jan 15 2009]
		

Links

• Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Crossrefs

Essentially the same as A014601. [R. J. Mathar, Nov 01 2009]

Programs

Mathematica

LinearRecurrence[{1,1,-1},{3,4,7},70] (* Harvey P. Dale, Jan 07 2016 *)

Formula

a(1)=4; thereafter a(2*n) = a(2*n-1) + 3, a(2*n+1) = a(2*n) + 1.

a(n) = 4*n - a(n-1) - 1 (with a(1)=3). [Vincenzo Librandi, Nov 26 2010]

From Colin Barker, Mar 06 2013: (Start)

a(n) = (1 - (-1)^n + 4*n)/2.

a(n) = a(n-1) + a(n-2) - a(n-3).

G.f.: x*(x+3) / ((x-1)^2*(x+1)). (End)

Extensions

More terms from Vincenzo Librandi, Nov 26 2010