Takeuchi Number

Author

Eric W. Weisstein
August 5, 2008

This notebook downloaded from http://mathworld.wolfram.com/notebooks/IntegerSequences/TakeuchiNumber.nb.

For more information, see Eric's MathWorld entry http://mathworld.wolfram.com/TakeuchiNumber.html.

©2008 Wolfram Research, Inc. except for portions noted otherwise

Procedural

Clear[TakeuchiNumber];
TakeuchiNumber[0] = 0;
TakeuchiNumber[1] = 1;
TakeuchiNumber[n_Integer] := TakeuchiNumber[n] = 
   Module[{k, res = 0, tmp = 1},
(
Do[res += tmp; 
tmp = Quotient[tmp*2*(2 k + 1), k + 2];, {k, n}];
res) +Total[tmp = 2 Binomial[-2 + 2 n, -1 + n] - Binomial[-1 + 2 n, -1 + n];
      Table[res = tmp*TakeuchiNumber[n - 1 - k];
       tmp = Quotient[tmp*k (-1 + k - n), (k - n) (-1 + k + n)];
       res, {k, n - 1, 0, -1}]]]

Timing[res = Table[TakeuchiNumber[n]/BellB[n]/Exp[1/2 ProductLog[n] 2],{n, 10000}] // N[#, 20] &;]

Recurrence

Wrong in Finch

(Subscript[t, n])==(∑_(k=0)^(n-2))(2 (Binomial[k+n-1, k])-(Binomial[k+n, k])) (Subscript[t, -k+n-1])+(∑_k^n)(1/(k+1))(Binomial[2 k, k])

t[0]=0;
t[1]=1;
t[2]=4;
t[3]=14;
t[n_]:=t[n]=Sum[(2Binomial[n+k-1,k]-Binomial[n+k,k])t[n-k-1],{k,0,n-2}]+Sum[CatalanNumber[k],{k,n}]

t/@Range[0,20]

Formula

A(z-z^2)/z-A(z)=1/(1-z)+z/(1-z+z^2)

A[z_]:=Sum[t[n]z^n,{n,0,30}]

Series[((A[z-z^2])/z)-A[z],{z,0,20}]

Series[(1/(1-z))+(z/(1-z+(z^2))),{z,0,20}]

Sum

FullSimplify[CatalanNumber[k]//FunctionExpand,k∈Integers&&k>0]

Accumulate[Table[CatalanNumber[k],{k,10}]]

Sum[CatalanNumber[k],{k,n}]

FunctionExpand[%]

FullSimplify[Sum[Binomial[2k,k](1/(k+1)),{k,n}],n∈Integers&&n>0]

Table[%,{n,10}]//Expand

Generating Function

Series[(1 - Sqrt[1 - 4 x])/(2 x (1 - x)), {x, 0, 10}]

SeriesCoefficient[(1 - Sqrt[1 - 4 x])/(2 x (1 - x)),{x,0,n}]//FunctionExpand//FullSimplify