A038163 G.f.: 1/((1-x)*(1-x^2))^3.
1, 3, 9, 19, 39, 69, 119, 189, 294, 434, 630, 882, 1218, 1638, 2178, 2838, 3663, 4653, 5863, 7293, 9009, 11011, 13377, 16107, 19292, 22932, 27132, 31892, 37332, 43452, 50388, 58140, 66861, 76551, 87381, 99351, 112651, 127281
Offset: 0
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (3,0,-8,6,6,-8,0,3,-1).
Crossrefs
Programs
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Haskell
import Data.List (inits, intersperse) a038163 n = a038163_list !! n a038163_list = map (sum . zipWith (*) (intersperse 0 $ tail a000217_list) . reverse) $ tail $ inits $ tail a000217_list where -- Reinhard Zumkeller, Feb 27 2015
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Maple
G := 1/((1-x)^3*(1-x^2)^3): Gser := series(G, x = 0, 42): seq(coeff(Gser, x, n), n = 0 .. 37); # Emeric Deutsch, Jun 26 2009 # alternative A038163 := proc(n) (4*n^5+90*n^4+760*n^3+2970*n^2+5266*n+3285+(-1)^n*(30*n^2+270*n+555))/3840 ; end proc: seq(A038163(n),n=0..30) ; # R. J. Mathar, Feb 22 2021
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Mathematica
CoefficientList[Series[1/((1-x)*(1-x^2))^3, {x, 0, 40}], x] (* Jean-François Alcover, Mar 11 2014 *) LinearRecurrence[{3,0,-8,6,6,-8,0,3,-1},{1,3,9,19,39,69,119,189,294},50] (* Harvey P. Dale, Nov 24 2022 *)
Formula
a(2*k) = (4*k + 5)*binomial(k + 4, 4)/5 = A034263(k); a(2*k + 1) = binomial(k + 4, 4)*(15 + 4*k)/5 = A059599(k), k >= 0.
a(n) = (1/3840)*(4*n^5 + 90*n^4 + 760*n^3 + 2970*n^2 + 5266*n + 3285 + (-1)^n*(30*n^2 + 270*n + 555)). Recurrence: a(n) = 3*a(n-1) - 8*a(n-3) + 6*a(n-4) + 6*a(n-5) - 8*a(n-6) + 3*a(n-8) - a(n-9). - Vladeta Jovovic, Apr 24 2002
a(n+1) - a(n) = A096338(n+2). - R. J. Mathar, Nov 04 2008
Comments