A193068 Generating primitive Pythagorean triangles by using (n, n+1) gives perimeters for each n. This sequence lists the sum of these perimeters for each n triangles.

12, 42, 98, 188, 320, 502, 742, 1048, 1428, 1890, 2442, 3092, 3848, 4718, 5710, 6832, 8092, 9498, 11058, 12780, 14672, 16742, 18998, 21448, 24100, 26962, 30042, 33348, 36888, 40670, 44702, 48992, 53548, 58378, 63490, 68892, 74592, 80598, 86918, 93560

Offset: 1

Author

J. M. Bergot, Jul 15 2011

Keywords

• nonn
• easy

Comments

Partial sums of A002939 starting at A002939(2). - R. J. Mathar, Aug 23 2011

Examples

			The perimeters of the first five triangles produced by pairs (1,2), (2,3), (3,4), (4,5), (5,6) are in order 12, 30, 56, 90, 132 with sum 320.
From the formula, a(5) = 5*(4*5^2 + 15*5 + 17)/3 = 320.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Crossrefs

Cf. A083374 (sum of areas for the first n triangles), A002412.

Programs

Magma

I:=[12, 42, 98, 188]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..40]]; // Vincenzo Librandi, Jul 04 2012

Mathematica

CoefficientList[Series[(2*(6-3*x+x^2))/((x-1)^4),{x,0,50}],x] (* Vincenzo Librandi, Jul 04 2012 *)
LinearRecurrence[{4,-6,4,-1},{12,42,98,188},40] (* Harvey P. Dale, Oct 29 2022 *)

Formula

a(n) = n*(4*n^2 + 15*n + 17)/3.

G.f.: ( 2*x*(6-3*x+x^2) ) / ( (x-1)^4 ). - R. J. Mathar, Aug 23 2011

a(n) = 4*a(n-1) -6*a(n-2) +4*a(n-3) -a(n-4). - Vincenzo Librandi, Jul 04 2012

a(n) = 2*(A002412(n+1) - 1). - Hugo Pfoertner, Oct 22 2024