A220452 Number of unordered full binary trees with labels from a set of n labels.
1, 3, 9, 37, 225, 1881, 19873, 251889, 3712257, 62286625, 1171487361, 24402416193, 557542291969, 13861636770177, 372514645389825, 10759590258589441, 332386419622387713, 10935312198369141249, 381705328034883127297, 14089260601787531469825, 548302210950105933701121
Offset: 1
Examples
For n=3, each of the three pairs of ancestors produces one descendant, and each of these descendants produces one more descendant with the respective remaining ancestor; three ancestors, three first-order descendants and three second-order descendants makes a population of a(3)=9.
Links
- Felix A. Pahl, Table of n, a(n) for n = 1..400
- Mathematics Stack Exchange, The n immortals problem
- V. Kotesovec, Interesting asymptotic formulas for binomial sums, Jun 09 2013
Programs
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Java
import java.math.BigInteger; public class A220452 { public static void main (String [] args) { int max = Integer.parseInt (args [0]); BigInteger [] doubleFactorials = new BigInteger [max + 1]; BigInteger [] [] binomialCoefficients = new BigInteger [max + 1] [max + 1]; doubleFactorials [0] = BigInteger.ONE; for (int n = 1;n <= max;n++) { binomialCoefficients [n] [0] = BigInteger.ONE; BigInteger sum = BigInteger.ZERO; for (int k = 1;k <= n;k++) { binomialCoefficients [n] [k] = k == n ? BigInteger.ONE : binomialCoefficients [n - 1] [k - 1].add (binomialCoefficients [n - 1] [k]); sum = sum.add (binomialCoefficients [n] [k].multiply (doubleFactorials [k - 1])); } System.out.println (n + " " + sum); doubleFactorials [n] = doubleFactorials [n - 1].multiply (BigInteger.valueOf (2 * n - 1)); } } }
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Mathematica
Table[Sum[Binomial[n,k]*(2*k-3)!!, {k, 1, n}], {n, 1, 20}] (* Vaclav Kotesovec, Dec 17 2012 *)
Formula
a(n) = Sum_{k=1..n} binomial(n,k)*(2k-3)!!.
a(n) ~ (2n-3)!!*sqrt(e) ~ (2n)!/(n!*2^n*(2n-1))*sqrt(e) ~ n^(n-1)*2^(n-1/2)*exp(1/2-n). - Vaclav Kotesovec, Dec 17 2012
Comments