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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.
1
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
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
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, a(5) = 5*(4*5^2 + 15*5 + 17)/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), A002412.
Sequence in context: A005901 A090554 A009948 * A007586 A228391 A334277
KEYWORD
nonn,easy
AUTHOR
J. M. Bergot, Jul 15 2011
STATUS
approved