A093578 Take n sheets of paper, arrange them into piles, write on each sheet the cardinality (number of sheets) of its pile. Do this again, so each sheet is labeled by an ordered pair of positive integers. How many ways can this be done so that every sheet has a unique label? (Only distinct sets of labels count, not every permutation of the labels or sheets.).

1, 1, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 2, 1, 1, 2, 1, 1, 1, 0, 0, 2, 2, 0, 2, 2, 0, 1, 1, 0, 0, 1, 2, 1, 1, 2, 2, 1, 1, 2, 1, 0, 1, 1, 1, 2, 1, 2, 2, 0, 2, 2, 0, 0, 1, 1, 0, 2, 2, 1, 2, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1

Offset: 0

Author

Howard A. Landman, Apr 01 2004

Keywords

• easy
• nonn

Comments

If n is a triangular number (A000217), then there is a trivial solution using piles of 1,2,3,...,k, where n = k(k+1)/2. All solutions are based on sums of triangular numbers, but not all such sums are legal. No indices of the triangular numbers can have a ratio smaller than 2; if they do then labels from the two triangles are not disjoint. a(28) = 2 because we can either use the trivial T(7) = 28 solution or the T(6) + T(3) + T(1) = 21 + 6 + 1 = 28 solution. A093579 gives the integers for which there is a solution, so that a(A093579(n)) > 0 for all n.

Examples

			a(1) = 1 because the only possible label is (1,1); a(2) = 0 because there is no way to prevent both pieces of paper from getting labeled identically.
		

Crossrefs

Cf. A000217, A093579.