A334240
a(n) = exp(-n) * Sum_{k>=0} (k + 1)^n * n^k / k!.
Original entry on oeis.org
1, 2, 11, 103, 1357, 23031, 478207, 11741094, 332734521, 10689163687, 383851610331, 15236978883127, 662491755803269, 31311446539427926, 1598351161031967063, 87638233726766111731, 5136809177699534717169, 320521818480481139673919, 21212211430440994022892019
Offset: 0
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Table[n! SeriesCoefficient[Exp[x + n (Exp[x] - 1)], {x, 0, n}], {n, 0, 18}]
Table[Sum[Binomial[n, k] BellB[k, n], {k, 0, n}], {n, 0, 18}]
A338282
a(n) = (1/e^n) * Sum_{j>=3} j^n * n^j / (j-3)!.
Original entry on oeis.org
0, 4, 216, 7371, 239424, 8127875, 296315496, 11685617608, 498593804800, 22959117809685, 1137033860419000, 60338078785131785, 3418430599382500800, 206053517402599981504, 13172124530670958537160, 890361160360138336174875, 63463906792476058870550528, 4758276450884470061869230823
Offset: 0
a(3) = 7371 = (1/e^3) * Sum_{j>=3} j^3 * 3^j / factorial(j-3).
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seq(add(n^(k+3)*A143495(n+3, k+3), k = 0..n), n = 0..17); # Peter Luschny, Oct 21 2020
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a[n_] := Exp[-n] * Sum[j^n * n^j/(j - 3)!, {j, 3, Infinity}]; Array[a, 17, 0] (* Amiram Eldar, Oct 20 2020 *)
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a(n)={sum(k=0, n+3, n^k*(stirling(n+3,k,2) - 3*stirling(n+2,k,2) + 2*stirling(n+1,k,2)))} \\ Andrew Howroyd, Oct 20 2020
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# Increase precision for larger n!
R = RealField(100)
t = 3
sol = [0]*18
for n in range(0, 18):
suma = R(0)
for j in range(t, 1000):
suma += (j^n * n^j) / factorial(j - t)
suma *= exp(-n)
sol[n] = round(suma)
print(sol) # Peter Luschny, Oct 20 2020
A332188
a(n) = (1/e^n) * Sum_{j>=2} j^n * n^j / (j-2)!.
Original entry on oeis.org
0, 3, 72, 1557, 36928, 986550, 29641608, 994006209, 36887753216, 1502798312547, 66730937637400, 3209318261685690, 166242143849148864, 9229638177763268395, 546842961612529341032, 34443269219453881669425
Offset: 0
a(3) = 1557 = (1/e^3) * Sum_{j>=2} j^3 * 3^j / factorial(j-2).
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a[n_] := Sum[n^k*(StirlingS2[n + 2, k] - StirlingS2[n + 1, k]), {k, 2, n + 2}]; Array[a, 16, 0] (* Amiram Eldar, Oct 30 2020 *)
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a(n) = sum(k=0, n+2, n^k*(stirling(n+2,k,2) - stirling(n+1,k,2))); \\ Michel Marcus, Oct 30 2020
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# Increase precision for larger n!
R = RealField(100)
t = 2
sol = [0]*18
for n in range(0, 18):
suma = R(0)
for j in range(t, 1000):
suma += (j^n * n^j) / factorial(j - t)
suma *= exp(-n)
sol[n] = round(suma)
print(sol) # Thanks to Peter Luschny for his example in A338282.
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