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%I #22 Apr 22 2019 18:09:35
%S 0,0,0,0,0,1,0,0,0,1,1,0,0,2,1,1,0,3,2,1,1,5,3,2,1,7,5,3,2,10,7,5,3,
%T 13,10,7,5,18,13,10,7,23,18,13,10,30,23,18,13,37,30,23,18,47,37,30,23,
%U 57,47,37,30,70,57,47,37,84,70,57,47,101,84,70,57,119
%N Expansion of x^5/((1-x^5)*(1-x^4)*(1-x^8)*(1-x^12)*(1-x^16)).
%H Seiichi Manyama, <a href="/A288166/b288166.txt">Table of n, a(n) for n = 0..10000</a>
%H Daniel Panario, Murat Sahin and Qiang Wang, <a href="http://www.combinatorics.org/ojs/index.php/eljc/article/view/v19i4p55/pdf">Generalized Alcuin’s Sequence</a>, The Electronic Journal of Combinatorics, Volume 19, Issue 4 (2012).
%H <a href="/index/Rec#order_45">Index entries for linear recurrences with constant coefficients</a>, signature (0, 0, 0, 1, 1, 0, 0, 1, -1, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, -2, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, -1, 0, 0, -1, -1, 0, 0, 0, 1).
%F a(n) = p_5(n/4) if n == 0 mod 4,
%F a(n) = p_5((n+15)/4) if n == 1 mod 4,
%F a(n) = p_5((n+10)/4) if n == 2 mod 4,
%F a(n) = p_5((n+5)/4) if n == 3 mod 4,
%F where p_5(n) is the number of partitions of n into exactly 5 parts.
%e a(56) = p_5(56/4) = p_5(14) = A001401(9) = 23,
%e a(57) = p_5((57+15)/4) = p_5(18) = A001401(13) = 57,
%e a(58) = p_5((58+10)/4) = p_5(17) = A001401(12) = 47,
%e a(59) = p_5((59+5)/4) = p_5(16) = A001401(11) = 37,
%e a(60) = p_5(60/4) = p_5(15) = A001401(10) = 30,
%e a(61) = p_5((61+15)/4) = p_5(19) = A001401(14) = 70,
%e a(62) = p_5((62+10)/4) = p_5(18) = A001401(13) = 57,
%e a(63) = p_5((63+5)/4) = p_5(17) = A001401(12) = 47.
%t CoefficientList[Series[x^5/((1-x^4)(1-x^5)(1-x^8)(1-x^12)(1-x^16)),{x,0,120}],x] (* or *) LinearRecurrence[ {0,0,0,1,1,0,0,1,-1,0,0,0,-1,0,0,0,0,0,0,-2,0,0,0,0,2,0,0,0,0,0,0,1,0,0,0,1,-1,0,0,-1,-1,0,0,0,1},{0,0,0,0,0,1,0,0,0,1,1,0,0,2,1,1,0,3,2,1,1,5,3,2,1,7,5,3,2,10,7,5,3,13,10,7,5,18,13,10,7,23,18,13,10},120] (* _Harvey P. Dale_, Apr 22 2019 *)
%Y Cf. A001401.
%Y Cf. A005044 (k=3), A288165 (k=4), this sequence (k=5).
%K nonn,easy
%O 0,14
%A _Seiichi Manyama_, Jun 06 2017
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