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A208544
T(n,k) = Number of n-bead necklaces of k colors allowing reversal, with no adjacent beads having the same color.
10
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, 55, 24, 13, 0, 0, 9, 28, 35, 120, 102, 92, 9, 1, 0, 10, 36, 56, 231, 312, 430, 156, 30, 0, 0, 11, 45, 84, 406, 777, 1505, 1170, 498, 29, 1, 0, 12, 55, 120, 666, 1680, 4291, 5580, 4435
OFFSET
1,2
COMMENTS
Table starts
.1.2..3...4....5.....6......7......8.......9......10......11.......12.......13
.0.1..3...6...10....15.....21.....28......36......45......55.......66.......78
.0.0..1...4...10....20.....35.....56......84.....120.....165......220......286
.0.1..6..21...55...120....231....406.....666....1035....1540.....2211.....3081
.0.0..3..24..102...312....777...1680....3276....5904....9999....16104....24882
.0.1.13..92..430..1505...4291..10528...23052...46185...86185...151756...254618
.0.0..9.156.1170..5580..19995..58824..149796..341640..714285..1391940..2559414
.0.1.30.498.4435.25395.107331.365260.1058058.2707245.6278140.13442286.26942565
LINKS
Andrew Howroyd, Table of n, a(n) for n = 1..1275 (first 264 terms from R. H. Hardin)
FORMULA
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
Empirical for row n:
n=1: a(k) = k
n=2: a(k) = (1/2)*k^2 - (1/2)*k
n=3: a(k) = (1/6)*k^3 - (1/2)*k^2 + (1/3)*k
n=4: a(k) = (1/8)*k^4 - (1/4)*k^3 + (3/8)*k^2 - (1/4)*k
n=5: a(k) = (1/10)*k^5 - (1/2)*k^4 + k^3 - k^2 + (2/5)*k
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
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
EXAMPLE
All solutions for n=7, k=3:
..1....1....1....1....1....1....1....1....1
..2....2....2....2....2....2....2....2....2
..3....3....1....1....3....1....3....1....3
..1....1....2....2....1....2....2....3....2
..2....3....3....3....3....1....3....1....3
..3....1....1....2....2....2....2....2....1
..2....3....3....3....3....3....3....3....3
MATHEMATICA
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 *)
PROG
(PARI)
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);
for(n=1, 10, for(k=1, 10, print1(T(n, k), ", ")); print) \\ Andrew Howroyd, Oct 14 2017
CROSSREFS
Main diagonal is A208538.
Columns 3..7 are A208539, A208540, A208541, A208542, A208543.
Row 2 is A000217(n-1).
Row 3 is A000292(n-2).
Row 4 is A002817(n-1).
Row 5 is A164938(n-1).
Row 6 is A027670(n-1).
Sequence in context: A144257 A257232 A321980 * A208535 A284856 A276550
KEYWORD
nonn,tabl
AUTHOR
R. H. Hardin, Feb 27 2012
STATUS
approved