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
Keywords
Examples
a(2) = 2*(10-4)*(10-6)*a(1)+a(1) = 49 since a(1)=1.
Links
- 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.
Crossrefs
Cf. A062980.
Programs
-
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 *)
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
Extensions
Name corrected by Steven Finch, Aug 12 2024
Comments