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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.
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%I #27 Oct 22 2024 03:15:00

%S 12,42,98,188,320,502,742,1048,1428,1890,2442,3092,3848,4718,5710,

%T 6832,8092,9498,11058,12780,14672,16742,18998,21448,24100,26962,30042,

%U 33348,36888,40670,44702,48992,53548,58378,63490,68892,74592,80598,86918,93560

%N 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.

%C Partial sums of A002939 starting at A002939(2). - _R. J. Mathar_, Aug 23 2011

%H Vincenzo Librandi, <a href="/A193068/b193068.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).

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

%F G.f.: ( 2*x*(6-3*x+x^2) ) / ( (x-1)^4 ). - _R. J. Mathar_, Aug 23 2011

%F a(n) = 4*a(n-1) -6*a(n-2) +4*a(n-3) -a(n-4). - _Vincenzo Librandi_, Jul 04 2012

%F a(n) = 2*(A002412(n+1) - 1). - _Hugo Pfoertner_, Oct 22 2024

%e 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.

%e From the formula, a(5) = 5*(4*5^2 + 15*5 + 17)/3 = 320.

%t CoefficientList[Series[(2*(6-3*x+x^2))/((x-1)^4),{x,0,50}],x] (* _Vincenzo Librandi_, Jul 04 2012 *)

%t LinearRecurrence[{4,-6,4,-1},{12,42,98,188},40] (* _Harvey P. Dale_, Oct 29 2022 *)

%o (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

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

%K nonn,easy

%O 1,1

%A _J. M. Bergot_, Jul 15 2011