cp's OEIS Frontend

This also counts a restricted set of plane partitions of n. Each element of the set which contains the A000041(n) partitions of n can be converted into plane partitions by insertion of line feeds at some or all places of the "pluses." Since the length of rows in plane partitions must be nonincreasing, there are only A000041(L(P)) ways to comply with this rule, where L(P) is the number of terms in that particular partition. Example for n=4: consider all five partitions 4 = 3+1 = 2+2 = 2+1+1 = 1+1+1+1 of four. The associated a(4)=13 plane partitions are 4, 31, 3|1, 22, 2|2, 211, 21|1, 2|1|1, 1111, 111|1, 11|11, 11|1|1 and 1|1|1|1, where the bar represents start of the next row, where a(4) = A000041(L(4)) + A000041(L(3+1)) + A000041(L(2+2)) + A000041(L(2+1+1))+ A000041(L(1+1+1+1)) = A000041(1) + A000041(2) + A000041(2) + A000041(3) + A000041(4). By construction from sorted partitions, all the plane partitions are strictly decreasing along each row and each column. - R. J. Mathar, Aug 12 2008

Also the number of pairs of integer partitions, the first with sum n and the second with sum equal to the length of the first. - Gus Wiseman, Jul 19 2018

Consider a set N={1,2,3,...n}. We can apply the operation S~(N) on N which gives us the set partitions S~(N)=SP(N) of N. Let denote SP_i(N) such a set partition, then SP(N)={SP_1(N), SP_2(N)...,SP_B(n)} (There are B(n) set partitions of N with B(n) as the Bell number). Observe that in each SP(N) we have SP(1)={{1,2,3,...,n}} and SP(B(n))={{1},{2},{3},...,{n}} and their magnitudes are |SP(1)|=1 and |SP(B(n)|=n.

Now we perform an iteration on the set partitions SP_i(N). We set partition each SP_i(N), thus we perform S~(SP_i(N), but we exclude SP(1)={{1,2,3,...,n}} and SP(B(n))={{1},{2},{3},...,{n}} from this repetition. Otherwise an infinite recursion arises. Thus if 1 < |SP_i(N)|=m < n, then we apply S~ on SP_i(N) again and get S~(SP_i(N))= SP(SP_i(N))={SP_1(SP_i(N)),...,SP_B(m)(SP_i(N))}. We repeat this partition operation S~ on every set partition we encounter. Let denote U_k a subset of SP_i(X) were X is a set. X may be any of the subsequent set partitions. Since |U_k| < X (under the condition above on m) the repeated application of S~ will end in set partitions SP(X) with |SP(X)| = 1.

Let us consider the example N={1,2,3}. The S~(N) gives us {{1,2,3}}, {{1,2},{3}}, {{1,3},{2}}, {{2,3},{13}} and {{1},{2},{1}}. We exclude {{1,2,3}} and {{1},{2},{1}} from further partitioning. From {{1,2},{3}} we get {{{1,2},{3}}} and {{{1,2}},{{3}}}. Consider the last two partitions. They correspond to N'={1',2'} and are thus {{1',2'}} and {{1'},{2'}}. Since |{{1',2'}}|=1 and |{{1'},{2'}}|=2 these last two set partitions cannot be partitioned any further according to our condition above. In total we get {{1,2,3}}, {{1,2},{3}}, {{1,3},{2}}, {{2,3},{1}}, {{{1,2},{3}}}, {{{1,2}},{{3}}}, {{{1,3},{2}}}, {{{1,3}},{{2}}}, {{{2,3},{1}}}, {{{2,3}},{{1}}}, {{1},{2},{3}} and we have a(3)=11.

A possible application are the number of coalitions among the set N={1,2,...,n} of n persons. These persons will split into parties = subsets U_k of N. Then coalitions will form among these parties, thus we encounter sets of subsets. It is even possible that coalitions form coalitions in turn. We thus define a coalition structure as a set of repeated set partitions. For example if n=6 we could have {{1,2},{3}},{{4,5,6}}, the parties {1,2} and {3} form the coalition {{1,2},{3}}. Since {{456}}={4,5,6} one might not want to consider a single set as a coalition, but formally it is possible to do so. However, if in the example all three parties are patriotic, they may stand together in questions of national interest and the coalition structure would be {{{1,2},{3}},{{4,5,6}}}.

However, in my opinion, the usual definition of a coalition as a partition of a set falls too short.

See also A005121 = Ultradissimilarity relations on an n-set. The paper "On the Asymptotic Analysis of a Class of Linear Recurrences" (by Thomas Prellberg) outlines how to find an asymptotic formula for A005121. Perhaps this method is applicable to the present sequence as well, but one needs to have the generating function as starting point.

An integer n can be partitioned into P(n) partitions P([n],i) where i=1,...,P(n) counts the partitions. The partition P([n],i) consists of T(n,i) integer parts t(i,j) with j=1,...,T(n,i). Now we perform on each t(i,j) an integer partition again and arrive at new partitions. Their parts can be partitioned again and so forth. We count such repeated partitions of n. One convention is necessary to avoid an infinite loop: The trivial partition P([n],1)=[n] will not be partitioned again but just counted once (and therefore we also have a(1)=1).

Start from the A000041(n) integer partitions P(n,i,s) of the integer n at stage s=1.

The index i=1,...,A000041(n) denotes the different partitions.

We call the index s the partition stage and increase it by one as we sub-partition the partitions of a previous stage.

Each P(n,i,s) is a set P(n,i,s)={t(n,1,j,s)),...,t(P,i,j,s),...,t(P,i,J,s)} of parts t(P,i,j,s) of S.

The index j is attached to the parts of a partition P(n,i,s). 1<=j<=n since there are at most n parts.

Now apply the set partition process on every P(n,i,s=1).

That is, each t(n,i,j,s=1) is subjected to a further partitioning.

We get partitions P(t'(n,i,j,1),i',j',2)={t'(t(n,i,j,1),i',1,2),...,t'(t(n,i,j,1), i',j',2),...,t'(t(n,i,j,1),i',J',2)} of the second partition stage.

We repeat this partitioning process on each part t'(i,j',2) until we arrive at parts equal to 1 which cannot be partitioned any further.

We may speak of the full decomposition F of n into parts.

The sequence counts the total number of partitions of all stages of the full decomposition of n.

Note that n is its own partition, e.g. P(n=3,i=1,s=1)={3} is an integer partition of n=3.

We do not apply the repeated partitioning on the partition P(n,i,s)={n} (otherwise an infinite loop would arise).

For n=1 and n=2 there is no second partition stage: s stays at s=1.

The corresponding labeled counterpart is sequence A143140.

a(n) = 1 + Sum_{i=2..n-1} 1*a(i) = 2^n; a(n) = 1 + Sum_{i=2..n-1} 2*a(i) = 3^n; etc. It seems that a(n+1)/(n*a(n)) -> 1 for n -> oo. [Comment corrected by Emeric Deutsch, Aug 10 2007]

Let M(n) denote the n X n matrix with ones along the subdiagonal, ones everywhere above the main diagonal, the integers 4, 5, etc., along the main diagonal, and zeros everywhere else. Then a(n) equals the permanent of M(n-2) for n >= 3. - John M. Campbell, Apr 20 2021