STATUS
reviewed
approved
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Revision History for A090582
(Bold, blue-underlined text is an addition; faded, red-underlined text is aShowing entries 1-10 | older changes
T(n, k) = Sum_{j=0..n-k} (-1)^j*binomial(n - k + 1, j)*(n - k + 1 - j)^n. Triangle read by rows, T(n, k) for 1 <= k <= n.
(history; published version)
(history; published version)
STATUS
proposed
reviewed
STATUS
editing
proposed
COMMENTS
Triangle T(n,k), 1 <= k <= n, read by rows given by [1, 1, 2, 2, 3, 3, 4, 4, 5, 5, ...] DELTA [[0, 1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Nov 10 2006
FORMULA
Generated T(n, k) = (n - k + 1)!*Stirling2(n, n - k + 1), generated by Stirling numbers of the second kind multiplied by a factorial. - Victor Adamchik, Oct 05 2005
Triangle T(n,k), 1 <= k <= n, read by rows given by [1, 1, 2, 2, 3, 3, 4, 4, 5, 5, ...] DELTA [[0, 1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Nov 10 2006
G(x,t) = 1/ (1 + (1-exp(x*t))/t) = 1 + 1*x + (2 + t)*x^2/2! + (6 + 6*t + t^2)*x^3/3! + ... gives row polynomials of A090582, the f-polynomials for the permutohedra (see A019538).
G(x,t-1) = 1 + 1*x + (1 + t)*x^2/2! + (1 + 4*t + t^2)*x^3/3! + ... gives row polynomials of A090582, for A008292, the fh-polynomials for the of permutohedra (see A019538).
G[(x,t-+1) = x,-1 /(t+ 1*x )] = 1 + (1 + t)*x^2/2! + (1 + 43*t + 2*t^2)*x^32/32! + ... gives row polynomials of A028246. (End)
gives row polynomials for A008292, the h-polynomials of permutohedra.
G[(t+1)x,-1/(t+1)] = 1 + (1 + t)*x + (1 + 3*t + 2*t^2)*x^2/2! + ...
gives row polynomials of A028246. (End)
With e.g.f. A(x,t) = G(x,t) - 1, the compositional inverse in x is
B(x,t) = log((t+1)-t/(1+x))/t. Let h(x,t) = 1/(dB/dx) = (1+x)*(1+(1+t)x), then the row polynomial P(n,t) is given by (1/n!)*(h(x,t)*d/dx)^n x, evaluated at x=0, A=exp(x*h(y,t)*d/dy) y, eval. at y=0, and dA/dx = h(A(x,t),t). (End)
k <= 0 or k > n yields Q(n, k) = 0; Q(1,1) = 1; Q(n, k) = k * (Q(n-1, k) + Q(n-1, k-1)). - David A. Corneth, Feb 17 2014
MAPLE
# Or:
T := (n, k) -> (n - k + 1)!*Stirling2(n, n - k + 1):
for n from 1 to 9 do seq( T(n, k), k = 1..n) od; # Peter Luschny, May 21 2021
NAME
x
T(n, k) = Sum_{j=0..n-k} (-1)^j*binomial(n - k + 1, j)*(n - k + 1 - j)^n. Triangle read by rows, T(n, k) for 1 <= k <= n.
MAPLE
T := (n, k) -> add((-1)^j*binomial(n - k + 1, j)*(n - k + 1 - j)^n, j = 0..n-k):
for n from 1 to 9 do seq(T(n, k), k = 1..n) od; # Peter Luschny, May 21 2021
NAME
Numerator Q(m,n) of probability P(m,n) = Q(m,n)/n^m of seeing each card at least once if m >= n cards are drawn with replacement from a deck of n cards, written in a two-dimensional array read by antidiagonals with Q(m,m) as first row and Q(m,1) as first column.
x
COMMENTS
Let Q(m, n) = Sum_(k=0..n-1) (-1)^k * binomial(n, k) * (n-k)^m. Then Q(m,n) is the numerator of the probability P(m,n) = Q(m,n)/n^m of seeing each card at least once if m >= n cards are drawn with replacement from a deck of n cards, written in a two-dimensional array read by antidiagonals with Q(m,m) as first row and Q(m,1) as first column.
FORMULA
Q(m, n) = Sum_(k=0..n-1) (-1)^k * binomial(n, k) * (n-k)^m.
EXAMPLE
For m = 4, n = 2, we draw 4 times from a deck of two cards. Call the cards "a" and "b" - of the 16 possible combinations of draws (each of which is equally likely to occur), only two do not contain both a and b: a, a, a, a and b, b, b, b. So the probability of seeing both a and b is 14/16. Therefore Q(m, n) = 14.
Table starts:
[1] 1;
[2] 2, 1;
[3] 6, 6, 1;
[4] 24, 36, 14, 1;
[5] 120, 240, 150, 30, 1;
[6] 720, 1800, 1560, 540, 62, 1;
[7] 5040, 15120, 16800, 8400, 1806, 126, 1;
[8] 40320, 141120, 191520, 126000, 40824, 5796, 254, 1;
[9] 362880, 1451520, 2328480, 1905120, 834120, 186480, 18150, 510, 1.
STATUS
approved
editing
STATUS
reviewed
approved
STATUS
proposed
reviewed
STATUS
editing
proposed