A193068 Generating primitive Pythagorean triangles by using (n, n+1) gives perimeters for each n. This sequence list 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

1,1

Comments

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

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).

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

Example

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
(4*5^3 + 15*5^2 + 17*5)/3 = 320

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 *)

Prog

(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

Crossrefs

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

Sequence in context: A005901 A090554 A009948 * A007586 A228391 A334277

Adjacent sequences: A193065 A193066 A193067 * A193069 A193070 A193071

Keyword

nonn,easy

Author

J. M. Bergot, Jul 15 2011

Status

approved