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%I #27 Sep 08 2022 08:45:03
%S 1,0,1,1,1,0,1,-1,-1,-3,-4,-7,-8,-13,-15,-20,-24,-31,-35,-44,-50,-60,
%T -68,-80,-89,-104,-115,-131,-145,-164,-179,-201,-219,-243,-264,-291,
%U -314,-345,-371,-404,-434,-471,-503,-544,-580,-624,-664,-712,-755,-808,-855,-911,-963,-1024,-1079,-1145
%N Expansion of (1-x-x^N)/((1-x)(1-x^2)(1-x^3)...(1-x^N)) for N = 4.
%C Difference of the number of partitions of n+3 into 3 parts and the number of partitions of n+3 into 4 parts. - _Wesley Ivan Hurt_, Apr 16 2019
%H Colin Barker, <a href="/A060023/b060023.txt">Table of n, a(n) for n = 0..1000</a>
%H P. A. MacMahon, <a href="https://doi.org/10.1112/plms/s1-17.1.139">Perpetual reciprocants</a>, Proc. London Math. Soc., 17 (1886), 139-151; Coll. Papers II, pp. 584-596.
%H <a href="/index/Par#part">Index entries for sequences related to partitions</a>
%H <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (1, 1, 0, 0, -2, 0, 0, 1, 1, -1).
%F a(n) = A069905(n+3) - A026810(n+3). - _Wesley Ivan Hurt_, Apr 16 2019
%F From _Colin Barker_, Apr 17 2019: (Start)
%F G.f.: (1 - x - x^4) / ((1 - x)^4*(1 + x)^2*(1 + x^2)*(1 + x + x^2)).
%F a(n) = a(n-1) + a(n-2) - 2*a(n-5) + a(n-8) + a(n-9) - a(n-10) for n>9.
%F (End)
%t CoefficientList[Series[(1-x-x^4)/Times@@(1-x^Range[4]),{x,0,60}],x] (* or *) LinearRecurrence[{1,1,0,0,-2,0,0,1,1,-1},{1,0,1,1,1,0,1,-1,-1,-3},70] (* _Harvey P. Dale_, Jan 14 2015 *)
%o (Magma) I:=[1,0,1,1,1,0,1,-1,-1,-3]; [n le 10 select I[n] else Self(n-1)+Self(n-2)-2*Self(n-5)+Self(n-8)+Self(n-9)-Self(n-10): n in [1..60]]; // _Vincenzo Librandi_, Jun 23 2015
%o (PARI) Vec((1 - x - x^4) / ((1 - x)^4*(1 + x)^2*(1 + x^2)*(1 + x + x^2)) + O(x^40)) \\ _Colin Barker_, Apr 17 2019
%Y Cf. A026810, A069905.
%Y Cf. For other values of N: A060022 (N=3), this sequence (N=4), A060024 (N=5), A060025 (N=6), A060026 (N=7), A060027 (N=8), A060028 (N=9), A060029 (N=10).
%K sign,easy
%O 0,10
%A _N. J. A. Sloane_, Mar 17 2001
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