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 A208825 T(n,k) is the number of n-bead necklaces labeled with numbers -k..k allowing reversal, with sum zero. 13
 1, 1, 2, 1, 3, 2, 1, 4, 5, 5, 1, 5, 8, 16, 7, 1, 6, 13, 38, 45, 18, 1, 7, 18, 75, 155, 167, 32, 1, 8, 25, 131, 415, 828, 609, 84, 1, 9, 32, 210, 905, 2821, 4390, 2471, 185, 1, 10, 41, 316, 1755, 7582, 19657, 25202, 10143, 486, 1, 11, 50, 453, 3085, 17339, 65134, 144871 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 1,3 COMMENTS Table starts ..1....1.....1......1......1.......1.......1........1........1........1 ..2....3.....4......5......6.......7.......8........9.......10.......11 ..2....5.....8.....13.....18......25......32.......41.......50.......61 ..5...16....38.....75....131.....210.....316......453......625......836 ..7...45...155....415....905....1755....3085.....5077.....7891....11761 .18..167...828...2821...7582...17339...35288....65769...114442...188463 .32..609..4390..19657..65134..177097..417204...883409..1720628..3135633 .84.2471.25202.144871.587682.1888153.5134796.12322101.26828152.54037203 LINKS R. H. Hardin, Table of n, a(n) for n = 1..165 FORMULA Empirical for row n: n=2: a(k) = k + 1. n=3: a(k) = 2*a(k-1) - 2*a(k-3) + a(k-4). n=4: a(k) = (2/3)*k^3 + (3/2)*k^2 + (11/6)*k + 1. n=5: a(k) = 3*a(k-1) - a(k-2) - 5*a(k-3) + 5*a(k-4) + a(k-5) - 3*a(k-6) + a(k-7). n=6: a(k) = (22/15)*k^5 + (11/3)*k^4 + (14/3)*k^3 + (13/3)*k^2 + (43/15)*k + 1. n=7: a(k) = 4*a(k-1) - 3*a(k-2) - 8*a(k-3) + 14*a(k-4) - 14*a(k-6) + 8*a(k-7) + 3*a(k-8) - 4*a(k-9) + a(k-10). EXAMPLE All solutions for n=3, k=3: .-2....0...-1...-1...-3...-2...-3...-2 .-1....0...-1....0....1....1....0....0 ..3....0....2....1....2....1....3....2 CROSSREFS Row 3 is A000982(n+1). Row 4 is A174723(n+1). Sequence in context: A167237 A277813 A200154 * A344391 A089353 A136451 Adjacent sequences: A208822 A208823 A208824 * A208826 A208827 A208828 KEYWORD nonn,tabl AUTHOR R. H. Hardin, Mar 01 2012 STATUS approved

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Last modified June 3 09:57 EDT 2023. Contains 363107 sequences. (Running on oeis4.)