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

%I #41 Sep 12 2023 16:53:37

%S 1,9,35,105,279,693,1651,3825,8687,19437,42987,94185,204775,442341,

%T 950243,2031585,4325343,9175005,19398619,40894425,85983191,180355029,

%U 377487315,788529105,1644167119,3422552013,7113539531,14763950025,30601641927,63350767557,130996502467,270582939585

%N a(n) = (2*n - 1) * (2^n - 1).

%C Row sums of triangle A118413.

%C For fixed n, define a triangle T(r,c) counting down the first n odd numbers on the left side, T(r,1) = 2*(n-r)+1, and counting up odd numbers on the right side, T(r,r) = 2*(n+r)-3, r>1. The interior elements are set by T(r,c)=T(r-1,c-1) + T(r-1,c). The sum of all members in this triangle is a(n). - _J. M. Bergot_, Oct 12 2012

%C Row sums of triangle A277046. - _Miquel Cerda_, Sep 28 2016

%H Altug Alkan, <a href="/A118414/b118414.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 (6,-13,12,-4)

%F a(n) = A005408(n-1)*(A000079(n) - 1). Corrected by _Omar E. Pol_, Sep 26 2016

%F G.f. -x*(-1-3*x+6*x^2) / ( (2*x-1)^2*(x-1)^2 ). - _R. J. Mathar_, Oct 15 2012

%F a(n) = A005408(n-1)*A000225(n). - _Miquel Cerda_, Sep 26 2016

%e The triangle T(r,c) for n=4 has row(1)=7; row(2) = 5, 9; row(3) = 3, 14, 11; row(4) = 1, 17, 25, 13, and a sum of 7+5+9+...+13 = 105 = a(4). - _J. M. Bergot_, Oct 12 2012

%t Table[(2 n - 1) (2^n - 1), {n, 32}] (* or *)

%t Rest@ CoefficientList[Series[-x (-1 - 3 x + 6 x^2)/((2 x - 1)^2*(x - 1)^2), {x, 0, 32}], x] (* _Michael De Vlieger_, Sep 26 2016 *)

%t LinearRecurrence[{6,-13,12,-4},{1,9,35,105},40] (* _Harvey P. Dale_, Sep 12 2023 *)

%o (Magma)[(2*n-1)*(2^n-1): n in [1..40]]; // _Vincenzo Librandi_, Dec 26 2010

%o (PARI) a(n)=(2*n-1)*(2^n-1) \\ _Charles R Greathouse IV_, Oct 12 2012

%Y Cf. A014477, A000225, A005408, A118413, A277046.

%K nonn,easy

%O 1,2

%A _Reinhard Zumkeller_, Apr 27 2006