OFFSET
0,4
COMMENTS
The column for k=0 is A000110 (Bell or exponential numbers). The column for k=1 is A000110 starting at offset 1. The column for k=2 is A005493 (Sum_{k=0..n} k*Stirling2(n,k).). The column for k=3 is A005494 (E.g.f.: exp(3*z+exp(z)-1).). The column for k=4 is A045379 (E.g.f.: exp(4*z+exp(z)-1).). The row for n=0 is 1's sequence, the row for n=1 is the natural numbers. The row for n=2 is A002522 (n^2 + 1.). The row for n=3 is A005491 (n^3 + 3n + 1.). The row for n=4 is A005492.
Number of ways of placing n labeled balls into n+k boxes, where k of the boxes are labeled and the rest are indistinguishable. - Bradley Austin (artax(AT)cruzio.com), Apr 24 2006
The column for k = -1 (not shown) is A000296 (Number of partitions of an n-set into blocks of size >1. Also number of cyclically spaced (or feasible) partitions.). - Gerald McGarvey, Oct 08 2006
Equals antidiagonals of an array in which (n+1)-th column is the binomial transform of n-th column, with leftmost column = the Bell sequence, A000110. - Gary W. Adamson, Apr 16 2009
Number of partitions of [n+k] where at least k blocks contain their own index element. A(2,2) = 10: 134|2, 13|24, 13|2|4, 14|23, 1|234, 1|23|4, 14|2|3, 1|24|3, 1|2|34, 1|2|3|4. - Alois P. Heinz, Jan 07 2022
REFERENCES
F. Ruskey, Combinatorial Generation, preprint, 2001.
LINKS
Alois P. Heinz, Antidiagonals n = 0..140, flattened
I. Mezo, The r-Bell numbers, J. Int. Seq. 14 (2011) # 11.1.1, Figure 1.
J. Riordan, Letter, Oct 31 1977, The array is on the second page.
F. Ruskey, Combinatorial Generation, 2003.
F. Ruskey, Lexicographic Algorithms [Broken link]
FORMULA
For n> 1, A(n, k) = k^n + sum_{i=0..n-2} A086659(n, i)*k^i. (A086659 is set partitions of n containing k-1 blocks of length 1, with e.g.f: exp(x*y)*(exp(exp(x)-1-x)-1).)
A(n, k) = k * A(n-1, k) + A(n-1, k+1), A(0, k) = 1. - Bradley Austin (artax(AT)cruzio.com), Apr 24 2006
A(n,k) = Sum_{i=0..n} C(n,i) * k^i * Bell(n-i). - Alois P. Heinz, Jul 18 2012
Sum_{k=0..n-1} A(n-k,k) = A005490(n). - Alois P. Heinz, Jan 05 2022
From G. C. Greubel, Dec 02 2022: (Start)
T(n, n) = A000012(n).
T(n, n-1) = A000027(n).
T(n, n-2) = A002522(n-1).
T(n, n-3) = A005491(n-2).
T(n, n-4) = A005492(n+1).
T(2*n, n) = A134980(n).
T(2*n, n+1) = A124824(n), n >= 1.
Sum_{k=0..n} T(n, k) = A347420(n). (End)
EXAMPLE
Array A(n,k) begins:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ... A000012;
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, ... A000027;
2, 5, 10, 17, 26, 37, 50, 65, 82, 101, ... A002522;
5, 15, 37, 77, 141, 235, 365, 537, 757, 1031, ... A005491;
15, 52, 151, 372, 799, 1540, 2727, 4516, 7087, 10644, ... A005492;
52, 203, 674, 1915, 4736, 10427, 20878, 38699, 67340, 111211, ... ;
Antidiagonal triangle, T(n, k), begins as:
1;
1, 1;
2, 2, 1;
5, 5, 3, 1;
15, 15, 10, 4, 1;
52, 52, 37, 17, 5, 1;
203, 203, 151, 77, 26, 6, 1;
877, 877, 674, 372, 141, 37, 7, 1;
4140, 4140, 3263, 1915, 799, 235, 50, 8, 1;
MAPLE
with(combinat):
A:= (n, k)-> add(binomial(n, i) * k^i * bell(n-i), i=0..n):
seq(seq(A(d-k, k), k=0..d), d=0..12); # Alois P. Heinz, Jul 18 2012
MATHEMATICA
Unprotect[Power]; 0^0 = 1; A[n_, k_] := Sum[Binomial[n, i] * k^i * BellB[n - i], {i, 0, n}]; Table[Table[A[d - k, k], {k, 0, d}], {d, 0, 12}] // Flatten (* Jean-François Alcover, Nov 05 2015, after Alois P. Heinz *)
PROG
(PARI) f(n, k)=round (suminf(i=0, (i+k)^n/i!)/exp(1));
g(n, k)=for(k=0, k, print1(f(n, k), ", ")) \\ prints k+1 terms of n-th row
(Magma)
A108087:= func< n, k | (&+[Binomial(n-k, j)*k^j*Bell(n-k-j): j in [0..n-k]]) >;
[A108087(n, k): k in [0..n], n in [0..13]]; // G. C. Greubel, Dec 02 2022
(SageMath)
def A108087(n, k): return sum( k^j*bell_number(n-k-j)*binomial(n-k, j) for j in range(n-k+1))
flatten([[A108087(n, k) for k in range(n+1)] for n in range(14)]) # G. C. Greubel, Dec 02 2022
KEYWORD
nonn,tabl
AUTHOR
Gerald McGarvey, Jun 05 2005
STATUS
approved