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A008811
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Expansion of x*(1+x^4)/((1-x)^2*(1-x^4)).
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10
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0, 1, 2, 3, 4, 7, 10, 13, 16, 21, 26, 31, 36, 43, 50, 57, 64, 73, 82, 91, 100, 111, 122, 133, 144, 157, 170, 183, 196, 211, 226, 241, 256, 273, 290, 307, 324, 343, 362, 381, 400, 421, 442, 463, 484, 507, 530, 553, 576
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OFFSET
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0,3
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COMMENTS
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Number of 0..n-1 arrays of 5 elements with zero 2nd differences. - R. H. Hardin, Nov 15 2011
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REFERENCES
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Pach and Agarwal, Combinatorial Geometry, p. 220, Problem 13.10.
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LINKS
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G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,-1,0,1,-2,1).
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FORMULA
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G.f.: x*(1+x^4)/((1-x)^2*(1-x^4)).
a(n) = 2*a(n-1) -a(n-2) +a(n-4) -2*a(n-5) +a(n-6). - R. H. Hardin, Nov 15 2011
a(n) = (-2*(1+(-1)^n)*(-1)^floor(n/2) + 2*n^2 + 5 - (-1)^n)/8. - Tani Akinari, Jul 24 2013
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MAPLE
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f := n->n^2/4+3*n/2+g(n);
g := n->if n mod 2 = 0 then 3 elif n mod 4 = 1 then 9/4 else 13/4; fi;
seq(f(n), n=-3..50);
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MATHEMATICA
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CoefficientList[Series[x*(1+x^4)/((1-x)^2*(1-x^4)), {x, 0, 60}], x] (* G. C. Greubel, Sep 12 2019 *)
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PROG
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(PARI) concat([0], Vec(x*(1+x^4)/((1-x)^2*(1-x^4))+O(x^60))) \\ Charles R Greathouse IV, Sep 26 2012, modified by G. C. Greubel, Sep 12 2019
(MAGMA) R<x>:=PowerSeriesRing(Integers(), 60); [0] cat Coefficients(R!( x*(1+x^4)/((1-x)^2*(1-x^4)) )); // G. C. Greubel, Sep 12 2019
(Sage)
def A008811_list(prec):
P.<x> = PowerSeriesRing(ZZ, prec)
return P(x*(1+x^4)/((1-x)^2*(1-x^4))).list()
A008811_list(60) # G. C. Greubel, Sep 12 2019
(GAP) a:=[0, 1, 2, 3, 4, 7];; for n in [7..60] do a[n]:=2*a[n-1]-a[n-2] +a[n-4]-2*a[n-5]+a[n-6]; od; a; # G. C. Greubel, Sep 12 2019
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CROSSREFS
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Cf. A129756 (first differences).
Cf. Expansions of the form (1+x^m)/((1-x)^2*(1-x^m)): A000290 (m=1), A000982 (m=2), A008810 (m=3), this sequence (m=4), A008812 (m=5), A008813 (m=6), A008814 (m=7), A008815 (m=8), A008816 (m=9), A008817 (m=10).
Sequence in context: A073627 A062042 A107817 * A144678 A309678 A279225
Adjacent sequences: A008808 A008809 A008810 * A008812 A008813 A008814
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KEYWORD
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nonn,easy
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AUTHOR
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N. J. A. Sloane
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STATUS
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approved
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