A226356 Number of representations of the n-th factorial group as a (nondecreasing) product of (nontrivial) cyclic groups.

0, 0, 1, 2, 3, 10, 20, 91, 207, 792, 2589, 17749, 52997

Offset: 0

Author

Alexander Riasanovsky, Jun 04 2013

Keywords

• nonn
• more

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.

Examples

			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.
		

Programs

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))

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.

Extensions

a(11) and a(12) added by Alexander Riasanovsky, Jun 06 2013