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A090619
Highest power of 12 dividing n!.
4
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
0,7
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
LINKS
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.
EXAMPLE
a(6)=2 since 6!=720=12^2*5.
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 *)
PROG
(PARI) a(n) = valuation(n!, 12); \\ Michel Marcus, Mar 24 2018
KEYWORD
nonn
AUTHOR
Henry Bottomley, Dec 06 2003
STATUS
approved