A094199 a(0) = -1/2; for n > 0, a(n) = 2*(5*n-4)*(5*n-6)*a(n-1) + Sum_{k=1..n-1} a(k)*a(n-k).

1, 49, 9800, 4412401, 3530881200, 4414129955298, 7945866428953600, 19467894010226044005, 62298157203907977632000, 252309651689367225339613486, 1261554846529199611110022246400, 7632433016288078444696820350362442, 54953647052313016042619300361129676800

Offset

1,2

Comments

The unknown constant in the article "Shapes of binary trees" by S. Finch (page 3, unsolved problem) is C = 0.0196207628432398766811334785902747944894235476341... = sqrt(15)/(20*Pi^2). - Vaclav Kotesovec, Jan 19 2015

Links

Table of n, a(n) for n=1..13.

S. R. Finch, Shapes of binary trees, June 24, 2004. [Cached copy, with permission of the author]

S. R. Finch, An exceptional convolutional recurrence, arXiv:2408.12440 [math.CO], 22 Aug 2024.

S. Janson, The Wiener index of simply generated random trees, Random Structures Algorithms 22 (2003) 337-358.

S. Janson and P. Chassaing, The center of mass of the ISE and the Wiener index of trees, arXiv:math/0309284 [math.PR], 2003.

Jian Zhou, On a Mean Field Theory of Topological 2D Gravity, arXiv:1503.08546 [math.AG], 30 Mar 2015.

Formula

With a(0) = -1/2 one has for n > 0 the recurrence a(n) = 2*(5*n-4)*(5*n-6)*a(n-1)+sum(a(k)*a(n-k), k=1..n-1).

a(n) ~ sqrt(3) * 2^(n-1) * 5^(2*n-1/2) * n^(2*n-1) / (Pi * exp(2*n)). The unknown constant in theorem 4.2. in the article by S. Janson and P. Chassaing is beta = 5*sqrt(15)/(2*Pi^2) = 0.981038142161993834... . - Vaclav Kotesovec, Jan 19 2015

Example

a(2) = 2*(10-4)*(10-6)*a(1)+a(1) = 49 since a(1)=1.

Mathematica

a[1] = 1; a[n_] := a[n] = 2*(5*n - 4)*(5*n - 6)*a[n - 1] + Sum[a[k]*a[n - k], {k, 1, n - 1}]; Table[a[n], {n, 1, 10}] (* Jean-François Alcover, Jun 20 2013 *)

Crossrefs

Cf. A062980.

Sequence in context: A351598 A014801 A187406 * A194023 A195273 A222459

Adjacent sequences: A094196 A094197 A094198 * A094200 A094201 A094202

Keyword

nonn

Author

Steven Finch, May 25 2004

Extensions

Name corrected by Steven Finch, Aug 12 2024

Status

approved