A230282 - OEIS

%I #25 Jul 06 2019 15:29:22

%S 1,1,6,64,679,8468,126784,2238565,45605124,1053117974,27182818156,

%T 775557529509,24236473829015,823299898542083,30205566231626957,

%U 1190319005015526817,50143449209799256306,2248672171655330927835

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

%F 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

%e Biggest k such that (3*k)! >= k!^4 is k = 64, so a(3) = 64.

%e 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

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

%o (Python)

%o import math

%o for n in range(8):

%o   for k in range(10000000):

%o     if math.factorial(n*k) < math.factorial(k)**(n+1):

%o       print k-1,

%o       break

%Y Cf. A000142, A065027, A136432, A230319.

%K nonn

%O 0,3

%A _Alex Ratushnyak_, Oct 14 2013

%E a(7)-a(17) from _Jon E. Schoenfield_, Oct 22 2013