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Least positive integer k for which 3^n divides k!.
6

%I #27 Dec 30 2019 12:22:02

%S 1,3,6,9,9,12,15,18,18,21,24,27,27,27,30,33,36,36,39,42,45,45,48,51,

%T 54,54,54,57,60,63,63,66,69,72,72,75,78,81,81,81,81,84,87,90,90,93,96,

%U 99,99,102,105,108,108,108,111,114,117,117,120,123,126,126,129,132,135,135,135

%N Least positive integer k for which 3^n divides k!.

%C It appears than for n>0, a(n) is divisible by 3, and that the resulting sequence a(n)/3 is A120503 (checked up to n=1000). - _Michel Marcus_, Aug 19 2013. [This is true: see A007843 for the idea of the proof. - _M. F. Hasler_, Dec 27 2019]

%C Also least positive integer k for which 6^n divides k!. - _Michel Marcus_, Aug 20 2013

%D H. Ibstedt, Smarandache Primitive Numbers, Smarandache Notions Journal, Vol. 8, No. 1-2-3, 1997, 216-229.

%H T. D. Noe, <a href="/A007844/b007844.txt">Table of n, a(n) for n = 0..1000</a>

%H F. Smarandache, <a href="http://www.gallup.unm.edu/~smarandache/OPNS.pdf">Only Problems, Not Solutions!</a>

%F a(n) = 3*A120503(n) for n > 0, cf. A007843. - _M. F. Hasler_, Dec 27 2019

%t Array[Block[{k = 1}, While[Mod[k!, 3^#] != 0, k++]; k] &, 67, 0] (* _Michael De Vlieger_, Dec 29 2019 *)

%o (PARI) a(n) = {k = 1; while (valuation(k!, 3) < n, k++); k;} \\ _Michel Marcus_, Aug 19 2013

%o (PARI) apply( A007844(n)={my(s=sumdigits(n*=2,3)\2); n-=n%3; while(s>0, s-=valuation(n+=3,3)); n+!n}, [0..99]) \\ _M. F. Hasler_, Dec 27 2019

%Y Cf. A007843 (analog for 2), A007845 (analog for 5).

%Y Cf. A120503 (Meta-Fibonacci, k = 3).

%K nonn

%O 0,2

%A Bruce Dearden and _Jerry Metzger_, R. Muller