A208544 - OEIS

%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