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%I #24 Mar 14 2020 07:05:23
%S 0,0,1,2,3,10,20,91,207,792,2589,17749,52997
%N Number of representations of the n-th factorial group as a (nondecreasing) product of (nontrivial) cyclic groups.
%C The algorithm given which generates and counts a(n) goes as follows:
%C 1. Consider the product [1,2,3,...,n] as Z_1 x Z_2 x ... x Z_n and "refine" wherever possible to create a multiset. For example, Z_6 ~= Z_2 x Z_3, so [1,2,3,4,5,6] refines as the multiset [2,2,3,3,4,5] or {2:2,3:2,4:1,5:1}.
%C 2. Place the above multiset at the top of a (ranked) poset, row 0.
%C 3. Set i=0.
%C 4. While there exists an element on the i-th row which, generate the (i+1)-th row of elements: those which are coarsed from elements on the i-th row.
%C 5. Find the number of representations generated.
%F With Z_k denoting the cyclic group on k letters, let G_0:=Z_1 and for all positive integers i, set G_i:=G_(i-1) x Z_i. Then a(n) is the number of (isomorphic) representations of G_n as a (nondecreasing) product of (nontrivial) cyclic groups.
%e Note: in the following ~= denotes isomorphism.
%e For example, G_0=Z_1 which cannot be represented as a product of nontrivial cyclic groups. Hence, a(0)=0. Likewise, G_1=G_0 x Z_1~=Z_1, so a(1)=0.
%e However G_2~=Z_2 is the only such representation of G_2.
%e For G_5=Z_1 x Z_2 x Z_3 x Z_4 x Z_5, we have exactly the following representations, sorted by the number of terms:
%e *Z_2 x Z_3 x Z_4 x Z_5,
%e *Z_4 x Z_5 x Z_6, Z_3 x Z_4 x Z_10, Z_2 x Z_5 x Z_12, Z_2 x Z_4 x Z_15, Z_2 x Z_3 x Z_20, and
%e *Z_6 x Z_20, Z_4 x Z_30, Z_10 x Z_12, Z_2 x Z_60.
%e Hence, a(5)=10.
%o (Sage)
%o #NOTE: by uncommenting the second return argument, the reader is given the array of representations.
%o def d_split(prod):
%o p_counts={}
%o for term in prod:
%o for p, m in term.factor():
%o pm = p^m
%o if pm in p_counts:
%o p_counts[pm]+=1
%o else:
%o p_counts[pm]=1
%o return p_counts
%o def factorial_group_reps(m):
%o if m<2:
%o return 0
%o i=0
%o widest_rep=d_split([Integer(n) for n in range(1,m+1)])
%o w_max=sum([widest_rep[p] for p in widest_rep])
%o rep_poset=[[widest_rep]]
%o r_count=1
%o while w_max-i>m//2:
%o row_new=[]
%o for rep in rep_poset[i]:
%o for [a,b] in Combinations(rep,2):
%o if gcd([a,b])==1:
%o rep_new=rep.copy()
%o if rep_new[a]==1:
%o rep_new.pop(a)
%o else:
%o rep_new[a]-=1
%o if rep_new[b]==1:
%o rep_new.pop(b)
%o else:
%o rep_new[b]-=1
%o if a*b in rep_new:
%o rep_new[a*b]+=1
%o else:
%o rep_new[a*b]=1
%o if not rep_new in row_new:
%o r_count+=1
%o row_new.append(rep_new)
%o rep_poset.append(row_new)
%o i+=1
%o return r_count#,rep_poset
%o for i in range(11):
%o # for i>10, a(i) is a very tedious computation for this algorithm
%o print(i,factorial_group_reps(i))
%K nonn,more
%O 0,4
%A _Alexander Riasanovsky_, Jun 04 2013
%E a(11) and a(12) added by _Alexander Riasanovsky_, Jun 06 2013
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