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A078007
Expansion of (1-x)/(1-x-2*x^2-x^3).
4
1, 0, 2, 3, 7, 15, 32, 69, 148, 318, 683, 1467, 3151, 6768, 14537, 31224, 67066, 144051, 309407, 664575, 1427440, 3065997, 6585452, 14144886, 30381787, 65257011, 140165471, 301061280, 646649233, 1388937264, 2983297010, 6407820771, 13763352055, 29562290607
OFFSET
0,3
COMMENTS
Let X = the 3x3 matrix [0,1,0; 0,0,1; 1,2,1]. a(n) = center term of X^n; but A002478(n) = term (3,3) of X^n. - Gary W. Adamson, May 30 2008
First bisection of A058278. - Oboifeng Dira, Aug 04 2016
FORMULA
a(n) = a(n-1) + 2*a(n-2) + a(n-3). - Ilya Gutkovskiy, Aug 06 2016
MATHEMATICA
LinearRecurrence[{1, 2, 1}, {1, 0, 2}, 40] (* or *) CoefficientList[Series[(1 -x)/(1-x-2*x^2-x^3), {x, 0, 40}], x] (* G. C. Greubel, Jun 28 2019 *)
PROG
(PARI) Vec((1-x)/(1-x-2*x^2-x^3)+O(x^40)) \\ Charles R Greathouse IV, Sep 26 2012
(Magma) R<x>:=PowerSeriesRing(Integers(), 40); Coefficients(R!( (1-x)/(1-x-2*x^2-x^3) )); // G. C. Greubel, Jun 28 2019
(Sage) ((1-x)/(1-x-2*x^2-x^3)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Jun 28 2019
(GAP) a:=[1, 0, 2];; for n in [4..40] do a[n]:=a[n-1]+2*a[n-2]+a[n-3]; od; a; # G. C. Greubel, Jun 28 2019
CROSSREFS
First differences of A002478.
Cf. A058278.
Sequence in context: A076993 A076698 A323598 * A368410 A358734 A198683
KEYWORD
nonn,easy
AUTHOR
N. J. A. Sloane, Nov 17 2002
STATUS
approved