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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

Table of n, a(n) for n=0..12.

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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Last modified August 2 21:30 EDT 2021. Contains 346429 sequences. (Running on oeis4.)