A090619 Highest power of 12 dividing n!.

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, 13, 14, 15, 15, 15, 17, 17, 17, 17, 18, 18, 19, 19, 19, 20, 21, 21, 22, 22, 22, 23, 23, 23, 25, 25, 26, 26, 27, 27, 28, 28, 28, 28, 30, 30, 31, 31, 31, 32, 32, 32, 34, 34, 34, 35, 35

Offset: 0

Author

Henry Bottomley, Dec 06 2003

Keywords

• nonn

Comments

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.

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

Examples

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

Links

• Robert Israel, Table of n, a(n) for n = 0..10000

Crossrefs

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

Programs

Maple

f2:= n -> n - convert(convert(n,base,2),`+`):
f3:= n -> (n - convert(convert(n,base,3),`+`))/2:
f:= n -> min(f3(n), floor(f2(n)/2)):
f(0):= 0:
map(f, [$0..100]); # Robert Israel, Mar 23 2018

Mathematica

Table[IntegerExponent[n!, 12], {n, 0, 100}] (* Jean-François Alcover, Mar 26 2018 *)

PARI

a(n) = valuation(n!, 12); \\ Michel Marcus, Mar 24 2018

Formula

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.