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

0,2

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!

Formula

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

Mathematica

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

Prog

(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

Crossrefs

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

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

Sequence in context: A011383 A329513 A329213 * A224523 A159785 A057338

Adjacent sequences: A007841 A007842 A007843 * A007845 A007846 A007847

Keyword

nonn

Author

Bruce Dearden and Jerry Metzger, R. Muller

Status

approved