|
|
A156308
|
|
Inverse of triangle S(n,m) defined by sequence A156290, n >= 1, 1 <= m <= n.
|
|
11
|
|
|
1, 4, 1, 9, 6, 1, 16, 20, 8, 1, 25, 50, 35, 10, 1, 36, 105, 112, 54, 12, 1, 49, 196, 294, 210, 77, 14, 1, 64, 336, 672, 660, 352, 104, 16, 1, 81, 540, 1386, 1782, 1287, 546, 135, 18, 1, 100, 825, 2640, 4290, 4004, 2275, 800, 170, 20, 1
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
This triangle S(n,m) appears as U_m(n) in the Knuth reference on p. 285. It is related to the Riordan triangle T_m(n) = A111125(n,m) by S(n,m) = A111125(n,m) - A111125(n-1,m), n >= m >= 1 (identity on p. 286).
Also, S(n,m)-S(n-1,m) = A111125(n-1,m-1), n >= 2, m >= 1 (identity on p. 286).
(End)
|
|
LINKS
|
|
|
FORMULA
|
S(n, m) = (n/m) * binomial(n + m - 1, 2*m - 1).
The n-th row o.g.f. is polynomial R(n,x) = 2/x*( T(n,(x + 2)/2) - 1 ), where T(n,x) is Chebyshev polynomial of the first kind. They form a divisibility sequence: if n divides m then R(n,x) divides R(m,x) in the ring Z[x].
R(2*n,x) = (x + 4)*U(n-1,(x + 2)/2)^2;
R(2*n + 1,x) = ( U(n,(x + 2)/2) + U(n-1,(x + 2)/2) )^2.
O.g.f.: Sum_{n >= 0} R(n,x)*z^n = z*(1 + z)/( (1 - z)*(1 - (x + 2)*z + z^2) ). (End)
The polynomial R(n,x) defined above satisfies (x + 1/x - 2) * R(n, x + 1/x - 2) = x^n + 1/x^n - 2. - Alexander Burstein, May 23 2021
|
|
EXAMPLE
|
Triangle starts:
n=1: 1;
n=2: 4, 1;
n=3: 9, 6, 1;
n=4: 16, 20, 8, 1;
...
|
|
MATHEMATICA
|
S[m_] := Flatten[Table[k/j Binomial[k + j - 1, 2 j - 1], {k, 1, m}, {j, 1, k}]]
|
|
PROG
|
(Sage) flatten([[(n/k)*binomial(n+k-1, 2*k-1) for k in (1..n)] for n in (1..12)]) # G. C. Greubel, Feb 01 2022
(Magma) [(n/k)*Binomial(n+k-1, 2*k-1): k in [1..n], n in [1..12]]; // G. C. Greubel, Feb 01 2022
|
|
CROSSREFS
|
Same as triangle A208513 with the first column truncated.
|
|
KEYWORD
|
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|