A094416 Array read by antidiagonals: generalized ordered Bell numbers Bo(r,n).

1, 2, 3, 3, 10, 13, 4, 21, 74, 75, 5, 36, 219, 730, 541, 6, 55, 484, 3045, 9002, 4683, 7, 78, 905, 8676, 52923, 133210, 47293, 8, 105, 1518, 19855, 194404, 1103781, 2299754, 545835, 9, 136, 2359, 39390, 544505, 5227236, 26857659, 45375130, 7087261

Offset

1,2

Comments

Also, r times the number of (r+1)-level labeled linear rooted trees with n leaves.

"AIJ" (ordered, indistinct, labeled) transform of {r,r,r,...}.

Stirling transform of r^n*n!, i.e. of e.g.f. 1/(1-r*x).

Also, Bo(r,s) is ((x*d/dx)^n)(1/(1+r-r*x)) evaluated at x=1.

r-th ordered Bell polynomial (A019538) evaluated at n.

Bo(r,n) is the n-th moment of a geometric distribution with probability parameter = 1/(r+1). Here, geometric distribution is the number of failures prior to the first success. - Geoffrey Critzer, Jan 01 2019

Row r (starting at r=0), Bo(r+1, n), is the Akiyama-Tanigawa algorithm applied to the powers of r+1. See Python program below. - Shel Kaphan, May 03 2024

Links

G. C. Greubel, Antidiagonals n = 1..50, flattened

Paul Barry, Three Études on a sequence transformation pipeline, arXiv:1803.06408 [math.CO], 2018.

P. Blasiak, K. A. Penson and A. I. Solomon, Dobinski-type relations and the log-normal distribution, arXiv:quant-ph/0303030, 2003.

C. G. Bower, Transforms

Formula

E.g.f.: 1/(1 + r*(1 - exp(x))).

Bo(r, n) = Sum_{k=0..n} k!*r^k*Stirling2(n, k) = 1/(r+1) * Sum_{k>=1} k^n * (r/(r+1))^k, for r>0, n>0.

Recurrence: Bo(r, n) = r * Sum_{k=1..n} C(n, k)*Bo(r, n-k), with Bo(r, 0) = 1.

Bo(r,0) = 1, Bo(r,n) = r*Bo(r,n-1) - (r+1)*Sum_{j=1..n-1} (-1)^j * binomial(n-1,j) * Bo(r,n-j). - Seiichi Manyama, Nov 17 2023

Example

Array begins as:
  1,  3,   13,    75,     541,     4683,      47293, ...
  2, 10,   74,   730,    9002,   133210,    2299754, ...
  3, 21,  219,  3045,   52923,  1103781,   26857659, ...
  4, 36,  484,  8676,  194404,  5227236,  163978084, ...
  5, 55,  905, 19855,  544505, 17919055,  687978905, ...
  6, 78, 1518, 39390, 1277646, 49729758, 2258233998, ...

Mathematica

Bo[_, 0]=1; Bo[r_, n_]:= Bo[r, n]= r*Sum[Binomial[n, k] Bo[r, n-k], {k, n}];
Table[Bo[r-n+1, n], {r, 10}, {n, r}] // Flatten (* Jean-François Alcover, Nov 03 2018 *)

Prog

(Magma)
A094416:= func< n, k | (&+[Factorial(j)*n^j*StirlingSecond(k, j): j in [0..k]]) >;
[A094416(n-k+1, k): k in [1..n], n in [1..12]]; // G. C. Greubel, Jan 12 2024
(SageMath)
def A094416(n, k): return sum(factorial(j)*n^j*stirling_number2(k, j) for j in range(k+1)) # array
flatten([[A094416(n-k+1, k) for k in range(1, n+1)] for n in range(1, 13)]) # G. C. Greubel, Jan 12 2024
(Python)
# The Akiyama-Tanigawa algorithm applied to the powers of r + 1
# generates the rows. Adds one row (r=0) and one column (n=0).
# Adapted from Peter Luschny on A371568.
def f(n, r): return (r + 1)**n
def ATtransform(r, len, f):
  A = [0] * len
  R = [0] * len
  for n in range(len):
      R[n] = f(n, r)
      for j in range(n, 0, -1):
          R[j - 1] = j * (R[j] - R[j - 1])
      A[n] = R[0]
  return A
for r in range(8): print([r], ATtransform(r, 8, f)) # Shel Kaphan, May 03 2024

Crossrefs

Rows 1-10 are A000670, A004123, A032033, A094417, A094418, A094419, A238464, A238465, A238466, A238467.

Columns include A014105, A094421.

Main diagonal is A094420.

Antidiagonal sums are A094422.

Cf. A019538, A131689, A344499.

Sequence in context: A337432 A123027 A100652 * A218868 A329874 A152300

Adjacent sequences: A094413 A094414 A094415 * A094417 A094418 A094419

Keyword

nonn,tabl

Author

Ralf Stephan, May 02 2004

Extensions

Offset corrected by Geoffrey Critzer, Jan 01 2019

Status

approved