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
Examples
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, ...
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
Crossrefs
Programs
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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
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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 *)
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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
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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
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
Extensions
Offset corrected by Geoffrey Critzer, Jan 01 2019
Comments