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a(n) = n*(2^n + 1).
1

%I #25 May 06 2021 15:28:33

%S 0,3,10,27,68,165,390,903,2056,4617,10250,22539,49164,106509,229390,

%T 491535,1048592,2228241,4718610,9961491,20971540,44040213,92274710,

%U 192938007,402653208,838860825,1744830490,3623878683,7516192796

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

%C Odd terms are multiples of 3. - _Dario Piazzalunga_, Jan 10 2013

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (6, -13, 12, -4).

%F Recurrence: a(n) = 6*a(n-1) - 13*a(n-2) + 12*a(n-3) - 4*a(n-4).

%F G.f.: x*(3 - 8*x + 6*x^2)/(1-x)^2/(1 - 2*x)^2.

%F a(n) = n+A036289(n). - _R. J. Mathar_, Jun 17 2020

%t Table[n*2^n+n,{n,0,3*4!}] (* _Vladimir Joseph Stephan Orlovsky_, Apr 25 2010 *)

%t LinearRecurrence[{6,-13,12,-4},{0,3,10,27},30] (* _Harvey P. Dale_, May 06 2021 *)

%Y Cf. A066524.

%K nonn,easy

%O 0,2

%A _Vladeta Jovovic_, Apr 12 2002