A193912 Partial sums of A193911.
1, 4, 11, 25, 50, 93, 162, 272, 439, 694, 1069, 1627, 2432, 3611, 5292, 7730, 11181, 16156, 23167, 33237, 47390, 67673, 96134, 136868, 193971, 275634, 390049, 553599, 782668, 1110023, 1568432, 2223430, 3140553, 4450872, 6285459, 8906457, 12576010, 17818405
Offset: 1
Examples
We have A193911(1)=1, A193911(2)=3, and A193911(3)=7. Thus a(1)=1, a(2)=4, and a(3)=11.
Links
- G. C. Greubel, Table of n, a(n) for n = 1..1000
- Index entries for linear recurrences with constant coefficients, signature (3,0,-8,7,3,-6,2).
Programs
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Mathematica
LinearRecurrence[{3,0,-8,7,3,-6,2},{1,4,11,25,50,93,162},40] (* Harvey P. Dale, Sep 09 2015 *) 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 *) -
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
Formula
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).
G.f.: x*(1+x-x^2)/((1-x)^4*(1+x)*(1-2*x^2)). - Alexander R. Povolotsky, Aug 12 2011
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 ).