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
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
KEYWORD
nonn
AUTHOR
Bruce Dearden and Jerry Metzger, R. Muller
STATUS
approved