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Maximal value of m such that 19^m <= n! : a(n) = floor( log(n!) / log(19) ).
10

%I #4 Sep 04 2015 10:25:06

%S 0,0,0,1,1,2,2,3,4,5,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,23,

%T 24,25,26,27,28,30,31,32,33,34,36,37,38,39,41,42,43,45,46,47,49,50,51,

%U 53,54,55,57,58,59,61,62,64

%N Maximal value of m such that 19^m <= n! : a(n) = floor( log(n!) / log(19) ).

%o (PARI) a(n)=lngamma(n+1)\log(19) \\ _Charles R Greathouse IV_, Sep 04 2015

%o (PARI) a(n)=logint(n!,19) \\ _Charles R Greathouse IV_, Sep 04 2015

%Y Cf. A067850, A127031, A127032, A127033, A127034, A127035, A127036, A127037, A127039.

%K nonn

%O 1,6

%A _Artur Jasinski_, Jan 03 2007