A133385 Number of permutations of n elements divided by the number of (binary) heaps on n+1 elements.

1, 1, 1, 2, 3, 6, 9, 24, 45, 108, 189, 504, 945, 2268, 3969, 12096, 25515, 68040, 130977, 381024, 773955, 2000376, 3750705, 11430720, 24111675, 64297800, 123773265, 360067680, 731387475, 1890355320, 3544416225, 11522165760, 25823603925, 72913705200, 148156598205

Offset: 0

Author

Alois P. Heinz, Nov 22 2007

Keywords

• nonn

Comments

In a min-heap on (n+1) distinct elements only n elements can change places, since the first element is determined to be the minimum. a(n) gives the number of all possibilities divided by the number of legal possibilities to do this.

Is this the sequence mentioned on page 360 of Motzkin (1948)? - N. J. A. Sloane, Jul 04 2015

Examples

			a(4) = 3 because 3 = 24/8 and there are 4! = 24 permutations on 4 elements and 8 min-heaps on 5 elements, namely (0,1,2,3,4), (0,1,2,4,3), (0,1,3,2,4), (0,1,3,4,2), (0,1,4,2,3), (0,1,4,3,2), (0,2,1,3,4), and (0,2,1,4,3). In every (min-) heap, the element at position i has to be larger than the element at position floor(i/2) for all i=2..n. The minimum is always found at position 1.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..2450
• T. Motzkin, The hypersurface cross ratio, Bull. Amer. Math. Soc., 51 (1945), 976-984.
• T. S. Motzkin, Relations between hypersurface cross ratios and a combinatorial formula for partitions of a polygon, for permanent preponderance and for non-associative products, Bull. Amer. Math. Soc., 54 (1948), 352-360.
• Eric Weisstein's World of Mathematics, Heap
• Wikipedia, Binary heap

Crossrefs

Cf. A000142, A056971, A132862.

Column k=2 of A273730.

Programs

Maple

h:= proc(n) option remember; `if`(n=0, 1, (b-> (f->
      h(f)*n*h(n-1-f))(min(b-1, n-b/2)))(2^ilog2(n)))
    end:
a:= n-> h(n+1)/(n+1):
seq(a(n), n=0..50);

Mathematica

aa[n_] := aa[n] = Module[{b, nl}, If[n<2, 1, b = 2^Floor[Log[2, n]]; nl = Min[b-1, n-b/2]; n*aa[nl]*aa[n-1-nl]]]; a[n_] := aa[n+1]/(n+1); Table[a[i], {i, 0, 50}] (* Jean-François Alcover, Mar 05 2014, after Alois P. Heinz *)

Formula

a(n) = A132862(n+1)/(n+1) = A000142(n)/A056971(n+1).