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 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. 3
 12, 900, 25200, 442680, 5846400, 64420272, 622175400, 5466166200, 44611306740, 343916472900, 2531921456064, 17956666859040, 123458676825120, 827056125453600, 5419508203393200, 34847210197637424, 220424306985639540, 1374479672119161300, 8463477229726134000, 51536194734146965920, 310706598354410079360 (list; graph; refs; listen; history; text; internal format)
 OFFSET 5,1 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, p, ..., p], with p >= p >= .. >= p >= 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, ..., 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. LINKS Andrew Howroyd, Table of n, a(n) for n = 5..100 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 EXAMPLE 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^2 c c c c. 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. CROSSREFS Cf. A213941, A214306, A214311 (m=5, representative bracelets), A214312 (m=4). Sequence in context: A140412 A276013 A116225 * A306642 A283570 A159870 Adjacent sequences: A214310 A214311 A214312 * A214314 A214315 A214316 KEYWORD nonn AUTHOR Wolfdieter Lang, Aug 08 2012 STATUS approved

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Last modified November 29 08:15 EST 2023. Contains 367429 sequences. (Running on oeis4.)