login
A101483
Column 2 of triangular matrix T=A101479, in which row n equals row (n-1) of T^(n-1) followed by '1'.
13
1, 1, 3, 18, 170, 2220, 37149, 758814, 18301950, 508907970, 16023271660, 563336380167, 21870097514265, 929083211885920, 42864992943282825, 2134180938818369850, 114045356939250133338, 6510275427522198876426
OFFSET
0,3
LINKS
FORMULA
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
EXAMPLE
This sequence can also be generated in the following manner.
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.
The table below illustrates this method:
[1], 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...;
[1, 2], 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, ...;
[3, 7, 12], 18, 25, 33, 42, 52, 63, 75, 88, 102, 117, 133, ...;
[18, 43, 76, 118], 170, 233, 308, 396, 498, 615, 748, 898, ...;
[170, 403, 711, 1107, 1605], 2220, 2968, 3866, 4932, 6185, ...;
[2220, 5188, 9054, 13986, 20171, 27816], 37149, 48420, ...;
[37149, 85569, 147471, 225363, 322075, 440785, 585046], 758814, ...; ...
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;
this sequence then forms the first column of the resultant table.
Note: column k of the above table equals column 2 of matrix power T^(k+1) where T=A101479, for k>=0.
PROG
(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])}
for(n=0, 25, print1(a(n), ", "))
(PARI) {a(n)=local(A=[1]); for(i=1, n, A=Vec(Ser(A)/(1-x)^(#A)); A=concat(A, A[#A])); A[#A]}
for(n=0, 25, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Jan 21 2005
STATUS
approved