A108087 Array, read by antidiagonals, where A(n,k) = exp(-1)*Sum_{i>=0} (i+k)^n/i!.
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, 21147, 21147, 17007, 10481, 4736, 1540, 365, 65, 9, 1, 115975, 115975, 94828, 60814, 29371, 10427, 2727, 537, 82, 10, 1
Offset: 0
Examples
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;
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]
Crossrefs
Programs
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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
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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
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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 *)
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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
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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
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)
Comments