A007844 Least positive integer k for which 3^n divides k!.

1, 3, 6, 9, 9, 12, 15, 18, 18, 21, 24, 27, 27, 27, 30, 33, 36, 36, 39, 42, 45, 45, 48, 51, 54, 54, 54, 57, 60, 63, 63, 66, 69, 72, 72, 75, 78, 81, 81, 81, 81, 84, 87, 90, 90, 93, 96, 99, 99, 102, 105, 108, 108, 108, 111, 114, 117, 117, 120, 123, 126, 126, 129, 132, 135, 135, 135

Offset: 0

Author

Bruce Dearden and Jerry Metzger, R. Muller

Keywords

• nonn

Comments

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]

Also least positive integer k for which 6^n divides k!. - Michel Marcus, Aug 20 2013

References

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

Links

• T. D. Noe, Table of n, a(n) for n = 0..1000
• F. Smarandache, Only Problems, Not Solutions!

Crossrefs

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

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

Programs

Mathematica

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

PARI

a(n) = {k = 1; while (valuation(k!, 3) < n, k++); k;} \\ Michel Marcus, Aug 19 2013

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

Formula

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