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Partial sums of A193911.
1

%I #32 Oct 22 2024 05:40:00

%S 1,4,11,25,50,93,162,272,439,694,1069,1627,2432,3611,5292,7730,11181,

%T 16156,23167,33237,47390,67673,96134,136868,193971,275634,390049,

%U 553599,782668,1110023,1568432,2223430,3140553,4450872,6285459,8906457,12576010,17818405

%N Partial sums of A193911.

%H G. C. Greubel, <a href="/A193912/b193912.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (3,0,-8,7,3,-6,2).

%F a(n) = Sum_{i=1..n} 1/8*(2^(i/2+2)*((10-7*sqrt(2))*(-1)^(i) + 10 + 7*sqrt(2))-(-1)^(i)-2*i*(i+12)-79).

%F G.f.: x*(1+x-x^2)/((1-x)^4*(1+x)*(1-2*x^2)). - _Alexander R. Povolotsky_, Aug 12 2011

%F a(n) = (1/32)*( (-1/2)^n + 32*(41*sqrt(2)-58)*(sqrt(2)-2)^n - 32*(58+41*sqrt(2))*(-2-sqrt(2))^n ).

%e We have A193911(1)=1, A193911(2)=3, and A193911(3)=7. Thus a(1)=1, a(2)=4, and a(3)=11.

%t LinearRecurrence[{3,0,-8,7,3,-6,2},{1,4,11,25,50,93,162},40] (* _Harvey P. Dale_, Sep 09 2015 *)

%t CoefficientList[Series[(1 + x - x^2)/((1 - x)^4*(1 + x)*(1 - 2*x^2)), {x, 0, 50}], x] (* _G. C. Greubel_, Feb 25 2017 *)

%o (PARI) my(x='x+O('x^50)); Vec((1+x-x^2)/((1-x)^4*(1+x)*(1-2*x^2))) \\ _G. C. Greubel_, Feb 25 2017

%K nonn,easy

%O 1,2

%A _Jeffrey R. Goodwin_, Aug 08 2011