%I #14 Dec 03 2017 17:53:38
%S 1,1,3,18,170,2220,37149,758814,18301950,508907970,16023271660,
%T 563336380167,21870097514265,929083211885920,42864992943282825,
%U 2134180938818369850,114045356939250133338,6510275427522198876426
%N Column 2 of triangular matrix T=A101479, in which row n equals row (n-1) of T^(n-1) followed by '1'.
%H Vaclav Kotesovec, <a href="/A101483/b101483.txt">Table of n, a(n) for n = 0..370</a>
%F a(n) ~ (2 + LambertW(-2*exp(-2))) * exp(n+2) * n^(n - 3/2) / (sqrt(Pi) * 2^(n + 1/2)). - _Vaclav Kotesovec_, Dec 03 2017
%e This sequence can also be generated in the following manner.
%e Start a table with the all 1's sequence in row 0; from then on, row n+1 can be formed from row n by dropping the initial n+1 terms of row n and taking partial sums of the remaining terms to obtain row n+1.
%e The table below illustrates this method:
%e [1], 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...;
%e [1, 2], 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, ...;
%e [3, 7, 12], 18, 25, 33, 42, 52, 63, 75, 88, 102, 117, 133, ...;
%e [18, 43, 76, 118], 170, 233, 308, 396, 498, 615, 748, 898, ...;
%e [170, 403, 711, 1107, 1605], 2220, 2968, 3866, 4932, 6185, ...;
%e [2220, 5188, 9054, 13986, 20171, 27816], 37149, 48420, ...;
%e [37149, 85569, 147471, 225363, 322075, 440785, 585046], 758814, ...; ...
%e In the above table, drop the initial n+1 terms in row n (enclosed in square brackets) and then take partial sums to obtain row n+1 for n>=0;
%e this sequence then forms the first column of the resultant table.
%e Note: column k of the above table equals column 2 of matrix power T^(k+1) where T=A101479, for k>=0.
%o (PARI) {a(n)=local(A=Mat(1),B);for(m=1,n+1,B=matrix(m,m);for(i=1,m, for(j=1,i, if(j==i,B[i,j]=1,B[i,j]=(A^i)[i-1,j]);));A=B);return(A[n+1,1])}
%o for(n=0,25,print1(a(n),", "))
%o (PARI) {a(n)=local(A=[1]);for(i=1,n,A=Vec(Ser(A)/(1-x)^(#A));A=concat(A,A[#A]));A[#A]}
%o for(n=0,25,print1(a(n),", "))
%Y Cf. A101479, A101481, A101482.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Jan 21 2005
|