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%I #14 Jul 02 2018 14:23:43
%S 3,120,2040,21420,183330,1320480,8691480,52727400,303958710,
%T 1674472800,8928735816,46280581620,234611247780,1166708558400,
%U 5710351190400,27565250985360,131495088522060,620771489730000,2903870526350640,13473567673441260,62061657617625204,283995655732351200
%N a(n) is the number of all four-color bracelets (necklaces with turning over allowed) with n beads and the four colors are from a repertoire of n distinct colors, for n >= 4.
%C This is the fourth column (m=4) of triangle A214306.
%C Each 4 part partition of n, with the parts written in nonincreasing order, defines a color signature. For a given color signature, say [p[1], p[2], p[3], p[4]], with p[1] >= p[2] >= p[3] >= p[4] >= 1, there are A213941(n,k)= A035206(n,k)*A213939(n,k) bracelets if this signature corresponds (with the order of the parts reversed) to the k-th partition of n in Abramowitz-Stegun (A-St) order. See A213941 for more details. Here all p(n,4)= A008284(n,4) partitions of n with 4 parts are considered. The color repertoire for a bracelet with n beads is [c[1], ..., c[n]].
%C Compare this with A032275 where also bracelets with less than four colors are included, and the color repertoire is only [c[1], c[2], c[3], c[4]] for all n.
%H Andrew Howroyd, <a href="/A214312/b214312.txt">Table of n, a(n) for n = 4..100</a>
%F a(n) = A214306(n,4), n >= 4.
%F a(n) = sum(A213941(n,k),k = A214314(n,4) .. (A214314(n,4) - 1 + A008284(n,4))), n >= 4.
%F a(n) = binomial(n,4) * A056344(n). - _Andrew Howroyd_, Mar 25 2017
%e a(5) = A213941(5,6) = 120 from the bracelet (with colors j for c[j], j=1, 2, ..., 5) 11234, 11243, 11324, 12134, 13124 and 14123, all six taken cyclically, each representing a class of order A035206(5,6) = 20 (if all 5 colors are used). For example, cyclic(11342) becomes equivalent to cyclic(11243) by turning over or reflection. The multiplicity 20 depends only on the color signature.
%t t[n_, k_] := (For[t1 = 0; d = 1, d <= n, d++, If[Mod[n, d] == 0, t1 = t1 + EulerPhi[d]*k^(n/d)]]; If[EvenQ[n], (t1 + (n/2)*(1 + k)*k^(n/2))/(2*n), (t1 + n*k^((n + 1)/2))/(2*n)]);
%t a56344[n_, k_] := Sum[(-1)^i*Binomial[k, i]*t[n, k - i], {i, 0, k - 1}];
%t a[n_] := Binomial[n, 4]*a56344[n, 4];
%t Table[a[n], {n, 4, 25}] (* _Jean-François Alcover_, Jul 02 2018, after _Andrew Howroyd_ *)
%Y Cf. A213941, A214306, A214309 (m=4, representative bracelets), A214313 (m=5).
%K nonn
%O 4,1
%A _Wolfdieter Lang_, Jul 31 2012
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