A320955 Square array read by ascending antidiagonals: A(n, k) (n >= 0, k >= 0) = Sum_{j=0..n-1} (!j/j!)*((n - j)^k/(n - j)!) if k > 0 and 1 if k = 0. Here !n denotes the subfactorial of n.
1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 2, 1, 0, 1, 1, 2, 4, 1, 0, 1, 1, 2, 5, 8, 1, 0, 1, 1, 2, 5, 14, 16, 1, 0, 1, 1, 2, 5, 15, 41, 32, 1, 0, 1, 1, 2, 5, 15, 51, 122, 64, 1, 0, 1, 1, 2, 5, 15, 52, 187, 365, 128, 1, 0, 1, 1, 2, 5, 15, 52, 202, 715, 1094, 256, 1, 0
Offset: 0
Examples
Array starts: n\k 0 1 2 3 4 5 6 7 8 9 ... ---------------------------------------------------- [0] 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ... A000007 [1] 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ... A000012 [2] 1, 1, 2, 4, 8, 16, 32, 64, 128, 256, ... A011782 [3] 1, 1, 2, 5, 14, 41, 122, 365, 1094, 3281, ... A124302 [4] 1, 1, 2, 5, 15, 51, 187, 715, 2795, 11051, ... A124303 [5] 1, 1, 2, 5, 15, 52, 202, 855, 3845, 18002, ... A056272 [6] 1, 1, 2, 5, 15, 52, 203, 876, 4111, 20648, ... A056273, ?A284727 [7] 1, 1, 2, 5, 15, 52, 203, 877, 4139, 21110, ... [8] 1, 1, 2, 5, 15, 52, 203, 877, 4140, 21146, ... [9] 1, 1, 2, 5, 15, 52, 203, 877, 4140, 21147, ... ---------------------------------------------------- Seen as a triangle given by the descending antidiagonals: [0] 1 [1] 0, 1 [2] 0, 1, 1 [3] 0, 1, 1, 1 [4] 0, 1, 2, 1, 1 [5] 0, 1, 4, 2, 1, 1 [6] 0, 1, 8, 5, 2, 1, 1 [7] 0, 1, 16, 14, 5, 2, 1, 1
Links
- Joerg Arndt and N. J. A. Sloane, Counting Words that are in "Standard Order"
Crossrefs
Programs
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Maple
A := (n, k) -> if k = 0 then 1 else add(A008290(n, n-j)*(n-j)^k, j=0..n-1)/n! fi: seq(lprint(seq(A(n, k), k=0..9)), n=0..9); # Prints the array row-wise. seq(seq(A(n-k, k), k=0..n), n=0..11); # Gives the array as listed.
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Mathematica
T[n_, 0] := 1; T[n_, k_] := Sum[(Subfactorial[j]/Factorial[j])((n - j)^k/(n - j)!), {j, 0, n - 1}]; Table[T[n - k, k], {n, 0, 11}, {k, 0, n}] // Flatten Table[Sum[StirlingS2[k, j], {j, 0, n-k}], {n, 0, 11}, {k, 0, n}] // Flatten (* Robert A. Russell, Nov 10 2018 *)
Formula
A(n, k) = (1/n!)*Sum_{j=0..n-1} A008290(n, n-j)*(n-j)^k if k > 0.
If one drops the special case A(n, 0) = 1 from the definition then column 0 becomes Sum_{k=0..n} (-1)^k/k! = A103816(n)/A053556(n).
Row n is given for k >= 1 by a_n(k), where
a_0(k) = 0^k/0!.
a_1(k) = 1^k/1!.
a_2(k) = (2^k)/2!.
a_3(k) = (3^k + 3)/3!.
a_4(k) = (6*2^k + 4^k + 8)/4!.
a_5(k) = (20*2^k + 10*3^k + 5^k + 45)/5!.
a_6(k) = (135*2^k + 40*3^k + 15*4^k + 6^k + 264)/6!.
a_7(k) = (924*2^k + 315*3^k + 70*4^k + 21*5^k + 7^k + 1855)/7!.
a_8(k) = (7420*2^k + 2464*3^k + 630*4^k + 112*5^k + 28*6^k + 8^k + 14832)/8!.
Note that the coefficients of the generating functions a_n are the recontres numbers A000240, A000387, A000449, ...
Rewriting the formulas with exponential generating functions for the rows we have egf(n) = Sum_{k=0..n} !k*binomial(n,k)*exp(x*(n-k)) and A(n, k) = (k!/n!)*[x^k] egf(n). In this formulation no special rule for the case k = 0 is needed.
The rows converge to the Bell numbers. Convergence here means that for every fixed k the terms in column k differ from A000110(k) only for finitely many indices.
A(n, n) are the Bell numbers A000110(n) for n >= 0.
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