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Expansion of 3*(2 - x)/((1 - x)*(1 + x)^2).
1

%I #9 Jun 07 2016 13:53:22

%S 6,-9,15,-18,24,-27,33,-36,42,-45,51,-54,60,-63,69,-72,78,-81,87,-90,

%T 96,-99,105,-108,114,-117,123,-126,132,-135,141,-144,150,-153,159,

%U -162,168,-171,177,-180,186,-189,195,-198,204,-207,213,-216,222,-225,231,-234,240

%N Expansion of 3*(2 - x)/((1 - x)*(1 + x)^2).

%C Alternating sum of A017233.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (-1,1,1).

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

%F a(n) = -a(n-1) + a(n-2) + a(n-3).

%F a(n) = Sum_{k=0..n} (-1)^k*3*(3*k + 2).

%F a(n) = 3*((-1)^n*6*n + (-1)^n*7 + 1)/4.

%F Sum_{n>=0} 1/a(n) = log(3)/6 - Pi/(18*sqrt(3)) = 0.082335416765006179088425414... . - _Vaclav Kotesovec_, Feb 25 2016

%F a(n) = 3*(-1)^n*A007494(n+1). - _R. J. Mathar_, Jun 07 2016

%e a(0) = 1 + 2 + 3 = 6;

%e a(1) = 1 + 2 + 3 - 4 - 5 - 6 = -9;

%e a(2) = 1 + 2 + 3 - 4 - 5 - 6 + 7 + 8 + 9 = 15;

%e a(3) = 1 + 2 + 3 - 4 - 5 - 6 + 7 + 8 + 9 - 10 - 11 - 12 = -18;

%e a(4) = 1 + 2 + 3 - 4 - 5 - 6 + 7 + 8 + 9 - 10 - 11 - 12 + 13 + 14 + 15 = 24, etc.

%t LinearRecurrence[{-1, 1, 1}, {6, -9, 15}, 53]

%t Table[3 ((6 (-1)^n n + 7 (-1)^n + 1)/4), {n, 0, 52}]

%Y Cf. A000027, A017233.

%K sign,easy

%O 0,1

%A _Ilya Gutkovskiy_, Feb 25 2016