A230282 Largest k such that (k*n)! >= (k!)^(n+1).

1, 1, 6, 64, 679, 8468, 126784, 2238565, 45605124, 1053117974, 27182818156, 775557529509, 24236473829015, 823299898542083, 30205566231626957, 1190319005015526817, 50143449209799256306, 2248672171655330927835

Offset

0,3

Links

Table of n, a(n) for n=0..17.

Formula

For n > 1, a(n) = floor(e*(n^n) - ((n^2-1)*log(n) + n*(1+log(2*Pi)))/2) [conjectural, but verified for all n in 2..5000]. - Jon E. Schoenfield, Oct 22 2013

Example

Biggest k such that (3*k)! >= k!^4 is k = 64, so a(3) = 64.
a(10) = 27182818156 because k = 27182818156 satisfies the inequality (k*10)! >= (k!)^11, but k = 27182818157 does not. To verify this, note that taking the logarithm of each side of the inequality gives log((k*10)!) >= 11*log(k!), and use the series expression log(m!) = log(2*Pi*m)/2 + m*log(m) - m + (1/12)/m - (1/360)/m^3 + (1/1260)/m^5 - ... (where the numerators and denominators of the fractions 1/12, -1/360, 1/1260, etc., are from A046968 and A046969, respectively), to get, at k = 27182818156, log(271828181560!) = 6884982704601.26... for the left hand side of the inequality, and the slightly smaller result 11*log(27182818156!) = 6884982704600.83... for the right hand side; then repeat the calculations using k = 27182818157, and observe that this makes the right hand side slightly larger than the left hand side. - Jon E. Schoenfield, Oct 23 2013

Mathematica

Table[k = 0; While[(k n)! >= (k!)^(n + 1), k++]; k - 1, {n, 0, 4}] (* T. D. Noe, Oct 18 2013 *)

Prog

(Python)
import math
for n in range(8):
  for k in range(10000000):
    if math.factorial(n*k) < math.factorial(k)**(n+1):
      print k-1,
      break

Crossrefs

Cf. A000142, A065027, A136432, A230319.

Sequence in context: A083225 A320528 A237357 * A186668 A025609 A309186

Adjacent sequences: A230279 A230280 A230281 * A230283 A230284 A230285

Keyword

nonn

Author

Alex Ratushnyak, Oct 14 2013

Extensions

a(7)-a(17) from Jon E. Schoenfield, Oct 22 2013

Status

approved