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T(n,k) is the number of n-bead necklaces labeled with numbers -k..k allowing reversal, with sum zero.
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%I #10 Mar 13 2017 04:24:45

%S 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,

%T 1,8,25,131,415,828,609,84,1,9,32,210,905,2821,4390,2471,185,1,10,41,

%U 316,1755,7582,19657,25202,10143,486,1,11,50,453,3085,17339,65134,144871

%N T(n,k) is the number of n-bead necklaces labeled with numbers -k..k allowing reversal, with sum zero.

%C Table starts

%C ..1....1.....1......1......1.......1.......1........1........1........1

%C ..2....3.....4......5......6.......7.......8........9.......10.......11

%C ..2....5.....8.....13.....18......25......32.......41.......50.......61

%C ..5...16....38.....75....131.....210.....316......453......625......836

%C ..7...45...155....415....905....1755....3085.....5077.....7891....11761

%C .18..167...828...2821...7582...17339...35288....65769...114442...188463

%C .32..609..4390..19657..65134..177097..417204...883409..1720628..3135633

%C .84.2471.25202.144871.587682.1888153.5134796.12322101.26828152.54037203

%H R. H. Hardin, <a href="/A208825/b208825.txt">Table of n, a(n) for n = 1..165</a>

%F Empirical for row n:

%F n=2: a(k) = k + 1.

%F n=3: a(k) = 2*a(k-1) - 2*a(k-3) + a(k-4).

%F n=4: a(k) = (2/3)*k^3 + (3/2)*k^2 + (11/6)*k + 1.

%F 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).

%F n=6: a(k) = (22/15)*k^5 + (11/3)*k^4 + (14/3)*k^3 + (13/3)*k^2 + (43/15)*k + 1.

%F 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).

%e All solutions for n=3, k=3:

%e .-2....0...-1...-1...-3...-2...-3...-2

%e .-1....0...-1....0....1....1....0....0

%e ..3....0....2....1....2....1....3....2

%Y Row 3 is A000982(n+1).

%Y Row 4 is A174723(n+1).

%K nonn,tabl

%O 1,3

%A _R. H. Hardin_, Mar 01 2012