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%I #24 Oct 30 2017 10:28:27
%S 1,2,0,3,1,0,4,3,0,0,5,6,1,1,0,6,10,4,6,0,0,7,15,10,21,3,1,0,8,21,20,
%T 55,24,13,0,0,9,28,35,120,102,92,9,1,0,10,36,56,231,312,430,156,30,0,
%U 0,11,45,84,406,777,1505,1170,498,29,1,0,12,55,120,666,1680,4291,5580,4435
%N T(n,k) = Number of n-bead necklaces of k colors allowing reversal, with no adjacent beads having the same color.
%C Table starts
%C .1.2..3...4....5.....6......7......8.......9......10......11.......12.......13
%C .0.1..3...6...10....15.....21.....28......36......45......55.......66.......78
%C .0.0..1...4...10....20.....35.....56......84.....120.....165......220......286
%C .0.1..6..21...55...120....231....406.....666....1035....1540.....2211.....3081
%C .0.0..3..24..102...312....777...1680....3276....5904....9999....16104....24882
%C .0.1.13..92..430..1505...4291..10528...23052...46185...86185...151756...254618
%C .0.0..9.156.1170..5580..19995..58824..149796..341640..714285..1391940..2559414
%C .0.1.30.498.4435.25395.107331.365260.1058058.2707245.6278140.13442286.26942565
%H Andrew Howroyd, <a href="/A208544/b208544.txt">Table of n, a(n) for n = 1..1275</a> (first 264 terms from R. H. Hardin)
%F T(2n+1,k) = A208535(2n+1,k)/2 for n > 0, T(2n,k) = (A208535(2n,k) + (k*(k-1)^n)/2)/2. - _Andrew Howroyd_, Mar 12 2017
%F Empirical for row n:
%F n=1: a(k) = k
%F n=2: a(k) = (1/2)*k^2 - (1/2)*k
%F n=3: a(k) = (1/6)*k^3 - (1/2)*k^2 + (1/3)*k
%F n=4: a(k) = (1/8)*k^4 - (1/4)*k^3 + (3/8)*k^2 - (1/4)*k
%F n=5: a(k) = (1/10)*k^5 - (1/2)*k^4 + k^3 - k^2 + (2/5)*k
%F n=6: a(k) = (1/12)*k^6 - (1/2)*k^5 + (3/2)*k^4 - (7/3)*k^3 + (23/12)*k^2 - (2/3)*k
%F n=7: a(k) = (1/14)*k^7 - (1/2)*k^6 + (3/2)*k^5 - (5/2)*k^4 + (5/2)*k^3 - (3/2)*k^2 + (3/7)*k
%e All solutions for n=7, k=3:
%e ..1....1....1....1....1....1....1....1....1
%e ..2....2....2....2....2....2....2....2....2
%e ..3....3....1....1....3....1....3....1....3
%e ..1....1....2....2....1....2....2....3....2
%e ..2....3....3....3....3....1....3....1....3
%e ..3....1....1....2....2....2....2....2....1
%e ..2....3....3....3....3....3....3....3....3
%t T[n_, k_] := If[n == 1, k, (DivisorSum[n, EulerPhi[n/#]*(k-1)^#&]/n + If[ OddQ[n], 1-k, k*(k-1)^(n/2)/2])/2]; Table[T[n-k+1, k], {n, 1, 12}, {k, n, 1, -1}] // Flatten (* _Jean-François Alcover_, Oct 30 2017, after _Andrew Howroyd_ *)
%o (PARI)
%o T(n, k) = if(n==1, k, (sumdiv(n, d, eulerphi(n/d)*(k-1)^d)/n + if(n%2, 1-k, k*(k-1)^(n/2)/2))/2);
%o for(n=1, 10, for(k=1, 10, print1(T(n, k), ", ")); print) \\ _Andrew Howroyd_, Oct 14 2017
%Y Cf. A081720, A208535, A106512.
%Y Main diagonal is A208538.
%Y Columns 3..7 are A208539, A208540, A208541, A208542, A208543.
%Y Row 2 is A000217(n-1).
%Y Row 3 is A000292(n-2).
%Y Row 4 is A002817(n-1).
%Y Row 5 is A164938(n-1).
%Y Row 6 is A027670(n-1).
%K nonn,tabl
%O 1,2
%A _R. H. Hardin_, Feb 27 2012
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