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A285522
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Array read by antidiagonals: T(m,n) = number of circulant digraphs up to Cayley isomorphism on n vertices with edges colored according to step value using a maximum of m-1 colors.
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6
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1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 6, 6, 4, 1, 1, 6, 18, 10, 5, 1, 1, 20, 24, 40, 15, 6, 1, 1, 14, 135, 70, 75, 21, 7, 1, 1, 48, 130, 544, 165, 126, 28, 8, 1, 1, 52, 648, 700, 1625, 336, 196, 36, 9, 1, 1, 140, 1137, 4480, 2635, 3996, 616, 288, 45, 10, 1
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OFFSET
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1,5
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COMMENTS
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For the base case of m=2 the sequence counts circulant digraphs up to Cayley isomorphism. Two circulant graphs are Cayley isomorphic if there is a d, which is necessarily prime to n, that transforms through multiplication modulo n the step values of one graph into those of the other. For squarefree n this is the only way that two circulant graphs can be isomorphic. (See Liskovets reference for a proof.)
Alternatively, the number of mappings with domain {1..n-1} and codomain {1..m} up to equivalence. Mappings A and B are equivalent if there is a d, prime to n, such that A(i) = B(i*d mod n) for i in {1..n-1}. This sequence differs from A132191 only in that sequence also includes 0 in the domain which introduces an extra factor of m into the results since zero multiplied by anything is zero.
All column sequences are polynomials of order n-1 and these are the cycle index polynomials.
This sequence is also related to A075195(n, m) which counts necklaces and A285548(m, n) which is the sequence described in the Titsworth reference. In particular, A075195 is the analogous array with equivalence determined through the additive group instead of by multiplication whereas A285548 allows for both addition and multiplication.
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LINKS
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FORMULA
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EXAMPLE
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Table starts:
\n 1 2 3 4 5 6 7 8 9 10
m\ ---------------------------------------------------
1 | 1 1 1 1 1 1 1 1 1 1 ...
2 | 1 2 3 6 6 20 14 48 52 140 ...
3 | 1 3 6 18 24 135 130 648 1137 4995 ...
4 | 1 4 10 40 70 544 700 4480 11056 65824 ...
5 | 1 5 15 75 165 1625 2635 20625 65425 489125 ...
6 | 1 6 21 126 336 3996 7826 72576 280596 2521476 ...
...
Case n=10:
Only 1, 3, 7, 9 are prime to 10.
Multiplication modulo 10 is described by the following multiplication table.
1, 2, 3, 4, 5, 6, 7, 8, 9 => (1)(2)(3)(4)(5)(6)(7)(8)(9) => m^9
3, 6, 9, 2, 5, 8, 1, 4, 7 => (1397)(2684)(5) => m^3
7, 4, 1, 8, 5, 2, 9, 6, 3 => (1793)(2486)(5) => m^3
9, 8, 7, 6, 5, 4, 3, 2, 1 => (19)(28)(37)(46)(5) => m^5
Each row of the multiplication table can be viewed as a permutation and together these form a commutative group on 4 elements. In this case the group is isomorphic to the cyclic group C_4. Each permutation can be represented in cycle notation. (shown above to the right of the corresponding multiplication table row). In order to count the equivalence classes using Polya's enumeration theorem only the number of cycles in each permutation is needed.
This gives the cycle index polynomial (1/4)*(m^9 + m^5 + 2*m^3). Putting m = 1..4 gives 1, 140, 4995, 65824.
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MATHEMATICA
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A132191[m_, n_] := (1/EulerPhi[n])*Sum[If[GCD[k, n] == 1, m^DivisorSum[n, EulerPhi[#] / MultiplicativeOrder[k, #] &], 0], {k, 1, n}];
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PROG
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(PARI)
a(n, x)=sum(k=1, n, if(gcd(k, n)==1, x^(sumdiv(n, d, eulerphi(d)/znorder(Mod(k, d)))-1), 0))/eulerphi(n);
for(m=1, 6, for(n=1, 10, print1( a(n, m), ", ") ); print(); );
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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