A214313 a(n) is the number of all five-color bracelets (necklaces with turning over allowed) with n beads and the four colors are from a repertoire of n distinct colors, for n >= 5.

12, 900, 25200, 442680, 5846400, 64420272, 622175400, 5466166200, 44611306740, 343916472900, 2531921456064, 17956666859040, 123458676825120, 827056125453600, 5419508203393200, 34847210197637424, 220424306985639540, 1374479672119161300, 8463477229726134000, 51536194734146965920, 310706598354410079360

Offset: 5

Author

Wolfdieter Lang, Aug 08 2012

Keywords

• nonn

Comments

This is the fifth column (m=5) of triangle A214306.

Each 5 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[5]], with p[1] >= p[2] >= .. >= p[5] >= 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,5)= A008284(n,5) partitions of n with 5 parts are considered. The color repertoire for a bracelet with n beads is [c[1], ..., c[n]].

It appears that this sequence is divisible by 12, producing 1, 75, 2100, 36890, 487200, 5368356, 51847950, 455513850, ...

Compare this with A056345 where only 5 colors are used for all n >= 5.

Examples

			a(6) = A213941(6,10) = 900 from the bracelet with color signature [2,1,1,1,1] and color repertoire [c[j], j=1, 2, ..., 6]. There are A213939(6,10) = 30 bracelets with representative color multinomials c[1]^2 c[2] c[3] c[4] c[5]. If the colors c[j] are taken as j, e.g., 112345, 112354, 112435, 112453, 112534, 112543, 113245, 113254, 113425, (113452 is equivalent to 112543 by turning over), 113524, (113542 ==112453), 114235, ..., 121345, ... (all taken cyclically). Each of these 30 bracelets represents a class of A035206(6,10) = 30 bracelets when all six colors are used. Thus a(6) = 30*30 = 900 = 12*75.
		

Links

• Andrew Howroyd, Table of n, a(n) for n = 5..100

Crossrefs

Cf. A213941, A214306, A214311 (m=5, representative bracelets), A214312 (m=4).

Formula

a(n) = A214306(n,5), n >= 5.

a(n) = sum(A213941(n,k),k = A214314(n,5) .. (A214314(n,5) - 1 + A008284(n,5))), n >= 5.

a(n) = binomial(n,5) * A056345(n). - Andrew Howroyd, Mar 25 2017