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Factor n; replace each prime(i) with i, take the conjugate partition, replace parts i with prime(i) and multiply out.

From Antti Karttunen, May 12-19 2014: (Start)

For all n >= 1, A001222(a(n)) = A061395(n), and vice versa, A061395(a(n)) = A001222(n).

Because the partition conjugation doesn't change the partition's total sum, this permutation preserves A056239, i.e., A056239(a(n)) = A056239(n) for all n.

(Similarly, for all n, A001221(a(n)) = A001221(n), because the number of steps in the Ferrers/Young-diagram stays invariant under the conjugation. - Note added Apr 29 2022).

Because this permutation commutes with A241909, in other words, as a(A241909(n)) = A241909(a(n)) for all n, from which follows, because both permutations are self-inverse, that a(n) = A241909(a(A241909(n))), it means that this is also induced when partitions are conjugated in the partition enumeration system A241918. (Not only in A112798.)

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From Antti Karttunen, Jul 31 2014: (Start)

Rows in arrays A243060 and A243070 converge towards this sequence, and also, assuming no surprises at the rate of that convergence, this sequence occurs also as the central diagonal of both.

Each even number is mapped to a unique term of A102750 and vice versa.

Conversely, each odd number (larger than 1) is mapped to a unique term of A070003, and vice versa. The permutation pair A243287-A243288 has the same property. This is also used to induce the permutations A244981-A244984.

Taking the odd bisection and dividing out the largest prime factor results in the permutation A243505.

Shares with A245613 the property that each term of A028260 is mapped to a unique term of A244990 and each term of A026424 is mapped to a unique term of A244991.

Conversely, with A245614 (the inverse of above), shares the property that each term of A244990 is mapped to a unique term of A028260 and each term of A244991 is mapped to a unique term of A026424.

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The Maple program follows the steps described in the first comment. The subprogram C yields the conjugate partition of a given partition. - Emeric Deutsch, May 09 2015

The Heinz number of the partition that is conjugate to the partition with Heinz number n. The Heinz number of a partition p = [p_1, p_2, ..., p_r] is defined as Product(p_j-th prime, j=1...r). Example: a(3) = 4. Indeed, the partition with Heinz number 3 is [2]; its conjugate is [1,1] having Heinz number 4. - Emeric Deutsch, May 19 2015

Like A129594 this sequence utilizes the fact that compositions (i.e., ordered partitions) can be bijectively mapped to (unordered) partitions by taking the partial sums of the list of composants after one has been subtracted from each except the first one. Compositions in turn are mapped to nonnegative integers via the runlength encoding, where the lengths of maximum runs of 0's or 1's in binary representation of n give the composants. See the OEIS Wiki page and the example below.

Each n occurs A000041(n) times in total and occurs for the first time at A227368(n) and for the last time at position A000225(n). See further comments and conjectures at A227368 and A227370.

This sequence is based on the fact that compositions (i.e. ordered partitions) can be mapped 1-to-1 to partitions by taking the partial sums of the list where one is subtracted from each composant except the first. (See table A227189 where the parts for each partition are listed).

The inverse process, from partitions to compositions, occurs by inserting the first (i.e. smallest) element of a partition sorted into ascending order to the front of the list obtained by adding one to the first differences of the elements.

Compositions map bijectively to nonnegative integers by assigning each run of k consecutive 1's (or 0's) in binary expansion of n with summand k in the composition.

The graph of this sequence is quite interesting.

a(2n) = 1 or 2 mod 4 and a(2n+1) = 0 or 3 mod 4 for all n > 1

For this sequence the partitions are encoded in the binary expansion of n in the same way as in A129594.

In "Bulgarian solitaire" a deck of cards or another finite set of objects is divided into one or more piles, and the "Bulgarian operation" is performed by taking one card from each pile, and making a new pile of them. The question originally posed was: on what condition the resulting partitions will eventually reach a fixed point, that is, a collection of piles that will be unchanged by the operation. See Martin Gardner reference and the Wikipedia-page.

A037481 gives the fixed points of this sequence, which are numbers that encode triangular partitions: 1 + 2 + 3 + ... + n.

A227752(n) tells how many times n occurs in this sequence, and A227753 gives the terms that do not occur here.

Of further interest: among each A000041(n) numbers j_i: j1, j2, ..., jk for which A227183(j_i)=n, how many cycles occur and what is the size of the largest one? (Both are 1 when n is in A000217, as then the fixed points are the only cycles.) Cf. A185700, A188160.

Also, A123975 answers how many Garden of Eden partitions there are for the deck of size n in Bulgarian Solitaire, corresponding to values that do not occur as the terms of this sequence.