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 A226356 Number of representations of the n-th factorial group as a (nondecreasing) product of (nontrivial) cyclic groups. 0
 0, 0, 1, 2, 3, 10, 20, 91, 207, 792, 2589, 17749, 52997 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,4 COMMENTS The algorithm given which generates and counts a(n) goes as follows: 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}. 2. Place the above multiset at the top of a (ranked) poset, row 0. 3. Set i=0. 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. 5. Find the number of representations generated. LINKS FORMULA 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. EXAMPLE Note: in the following ~= denotes isomorphism. 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. However G_2~=Z_2 is the only such representation of G_2. 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: *Z_2 x Z_3 x Z_4 x Z_5, *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 *Z_6 x Z_20, Z_4 x Z_30, Z_10 x Z_12, Z_2 x Z_60. Hence, a(5)=10. PROG (Sage) #NOTE: by uncommenting the second return argument, the reader is given the array of representations. def d_split(prod):     p_counts={}     for term in prod:         for p, m in term.factor():             pm = p^m             if pm in p_counts:                 p_counts[pm]+=1             else:                 p_counts[pm]=1     return p_counts def factorial_group_reps(m):     if m<2:         return 0     i=0     widest_rep=d_split([Integer(n) for n in range(1, m+1)])     w_max=sum([widest_rep[p] for p in widest_rep])     rep_poset=[[widest_rep]]     r_count=1     while w_max-i>m//2:         row_new=[]         for rep in rep_poset[i]:             for [a, b] in Combinations(rep, 2):                 if gcd([a, b])==1:                     rep_new=rep.copy()                     if rep_new[a]==1:                         rep_new.pop(a)                     else:                         rep_new[a]-=1                     if rep_new[b]==1:                         rep_new.pop(b)                     else:                         rep_new[b]-=1                     if a*b in rep_new:                         rep_new[a*b]+=1                     else:                         rep_new[a*b]=1                     if not rep_new in row_new:                         r_count+=1                         row_new.append(rep_new)         rep_poset.append(row_new)         i+=1     return r_count#, rep_poset for i in range(11):     # for i>10, a(i) is a very tedious computation for this algorithm     print(i, factorial_group_reps(i)) CROSSREFS Sequence in context: A089791 A297872 A298135 * A141050 A252865 A252868 Adjacent sequences:  A226353 A226354 A226355 * A226357 A226358 A226359 KEYWORD nonn,more AUTHOR Alexander Riasanovsky, Jun 04 2013 EXTENSIONS a(11) and a(12) added by Alexander Riasanovsky, Jun 06 2013 STATUS approved

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