A309784 T(n,k) is the number of non-equivalent distinguishing coloring partitions of the cycle on n vertices with exactly k parts. Regular triangle read by rows, n >= 1, 1 <= k <= n.
0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 4, 2, 1, 0, 1, 8, 10, 3, 1, 0, 1, 25, 32, 16, 3, 1, 0, 4, 62, 129, 84, 27, 4, 1, 0, 7, 176, 468, 433, 171, 37, 4, 1, 0, 18, 470, 1806, 2260, 1248, 338, 54, 5, 1, 0, 31, 1311, 6780, 11515, 8388, 3056, 590, 70, 5, 1, 0, 70, 3620, 25917, 58312, 56065, 26695, 6907, 1014, 96, 6, 1
Offset: 1
Examples
The triangle begins: 0; 0, 0; 0, 0, 1; 0, 0, 1, 1; 0, 0, 4, 2, 1; 0, 1, 8, 10, 3, 1; 0, 1, 25, 32, 16, 3, 1; 0, 4, 62, 129, 84, 27, 4, 1; 0, 7, 176, 468, 433, 171, 37, 4, 1; 0, 18, 470, 1806, 2260, 1248, 338, 54, 5, 1; ... For n=6, we can partition the vertices of C_6 into exactly 3 parts in 8 ways such that all these partitions induce distinguishing colorings for C_6 and that all the 8 partitions are non-equivalent. The partitions are as follows: { { 1 }, { 2 }, { 3, 4, 5, 6 } } { { 1 }, { 2, 3 }, { 4, 5, 6 } } { { 1 }, { 2, 3, 4, 6 }, { 5 } } { { 1 }, { 2, 3, 5 }, { 4, 6 } } { { 1 }, { 2, 3, 6 }, { 4, 5 } } { { 1 }, { 2, 4, 5 }, { 3, 6 } } { { 1, 2 }, { 3, 4 }, { 5, 6 } } { { 1, 2 }, { 3, 5 }, { 4, 6 } } For n=6, the above 8 partitions can be written as the following 3 colored bracelet structures: ABCCCC, ABBCCC, ABBBCB, ABBCBC, ABBCCB, ABCBBC, AABBCC, AABCBC. - _Andrew Howroyd_, Sep 22 2019
Links
- Andrew Howroyd, Table of n, a(n) for n = 1..1275
- B. Ahmadi, F. Alinaghipour and M. H. Shekarriz, Number of Distinguishing Colorings and Partitions, arXiv:1910.12102 [math.CO], 2019.
- Mohammad Hadi Shekarriz, GAP Program
Crossrefs
Programs
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PARI
\\ Ach is A304972 and R is A152175 as square matrices. Ach(n)={my(M=matrix(n, n, i, k, i>=k)); for(i=3, n, for(k=2, n, M[i, k]=k*M[i-2, k] + M[i-2, k-1] + if(k>2, M[i-2, k-2]))); M} R(n)={Mat(Col([Vecrev(p/y, n) | p<-Vec(intformal(sum(m=1, n, eulerphi(m) * subst(serlaplace(-1 + exp(sumdiv(m, d, y^d*(exp(d*x + O(x*x^(n\m)))-1)/d))), x, x^m))/x))]))} T(n)={my(A=Ach(n), M=R(n), S=matrix(n, n, n, k, stirling(n, k, 2))); Mat(vectorv(n, n, sumdiv(n, d, moebius(d)*(M[n/d,] + A[n/d,])/2 - moebius(d)*(S[(n/d+1)\2, ] + S[n/d\2+1, ] + if((n-d)%2, A[(n/d+1)\2, ] + A[n/d\2+1, ]))/if(d%2, 2, 1) )))} { my(A=T(12)); for(n=1, #A, print(A[n, 1..n])) } \\ Andrew Howroyd, Oct 02 2019
Formula
Extensions
T(10,6) corrected by Mohammad Hadi Shekarriz, Sep 28 2019
a(56)-a(78) from Andrew Howroyd, Sep 28 2019
Comments