Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).
%I #53 Aug 23 2024 03:25:20
%S 1,49,9800,4412401,3530881200,4414129955298,7945866428953600,
%T 19467894010226044005,62298157203907977632000,
%U 252309651689367225339613486,1261554846529199611110022246400,7632433016288078444696820350362442,54953647052313016042619300361129676800
%N 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).
%C 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
%H S. R. Finch, <a href="/A062980/a062980.pdf">Shapes of binary trees</a>, June 24, 2004. [Cached copy, with permission of the author]
%H S. R. Finch, <a href="https://arxiv.org/abs/2408.12440">An exceptional convolutional recurrence</a>, arXiv:2408.12440 [math.CO], 22 Aug 2024.
%H S. Janson, <a href="https://dx.doi.org/10.1002/rsa.10074">The Wiener index of simply generated random trees</a>, Random Structures Algorithms 22 (2003) 337-358.
%H S. Janson and P. Chassaing, <a href="http://arXiv.org/abs/math/0309284">The center of mass of the ISE and the Wiener index of trees</a>, arXiv:math/0309284 [math.PR], 2003.
%H Jian Zhou, <a href="http://arxiv.org/abs/1503.08546">On a Mean Field Theory of Topological 2D Gravity</a>, arXiv:1503.08546 [math.AG], 30 Mar 2015.
%F 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).
%F 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
%e a(2) = 2*(10-4)*(10-6)*a(1)+a(1) = 49 since a(1)=1.
%t 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 *)
%Y Cf. A062980.
%K nonn
%O 1,2
%A _Steven Finch_, May 25 2004
%E Name corrected by _Steven Finch_, Aug 12 2024