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Highest power of 12 dividing n!.
4

%I #20 Mar 27 2018 10:58:13

%S 0,0,0,0,1,1,2,2,2,3,4,4,5,5,5,5,6,6,8,8,8,9,9,9,10,10,10,11,12,12,13,

%T 13,14,15,15,15,17,17,17,17,18,18,19,19,19,20,21,21,22,22,22,23,23,23,

%U 25,25,26,26,27,27,28,28,28,28,30,30,31,31,31,32,32,32,34,34,34,35,35

%N Highest power of 12 dividing n!.

%C Most sequences of the form "highest power of k dividing n!" essentially depend on one of the primes or prime powers dividing k. But in this case, the sequences with k=3 (A054861) and k=4 (A090616) are both close to n/2 and vary in which one is lower for different values of n.

%C a(2^n) = A090616(2^n) and a(3^n-1) = A090616(3^n-1) while a(2^n-1) = A054861(2^n-1) and a(3^n) = A054861(3^n). - _Robert Israel_, Mar 25 2018

%H Robert Israel, <a href="/A090619/b090619.txt">Table of n, a(n) for n = 0..10000</a>

%F a(n) =A090622(n, 12) =min(A054861(n), A090616(n)). Close to n/2, indeed for n>3: n/2-log3(n+1) <= a(n) < n/2.

%e a(6)=2 since 6!=720=12^2*5.

%p f2:= n -> n - convert(convert(n,base,2),`+`):

%p f3:= n -> (n - convert(convert(n,base,3),`+`))/2:

%p f:= n -> min(f3(n), floor(f2(n)/2)):

%p f(0):= 0:

%p map(f, [$0..100]); # _Robert Israel_, Mar 23 2018

%t Table[IntegerExponent[n!, 12], {n, 0, 100}] (* _Jean-François Alcover_, Mar 26 2018 *)

%o (PARI) a(n) = valuation(n!, 12); \\ _Michel Marcus_, Mar 24 2018

%Y Cf. A011371, A054861, A090616, A027868, A054896, A090617, A090618.

%K nonn

%O 0,7

%A _Henry Bottomley_, Dec 06 2003