A122841 Greatest k such that 6^k divides n.

0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0

Offset

1,36

Comments

See A054895 for the partial sums. - Hieronymus Fischer, Jun 08 2012

Links

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Formula

From Hieronymus Fischer, Jun 03 2012: (Start)

With m = floor(log_6(n)), frac(x) = x-floor(x):

a(n) = Sum_{j=1..m} (1 - ceiling(frac(n/6^j))).

a(n) = m + Sum_{j=1..m} (floor(-frac(n/6^j))).

a(n) = A054895(n) - A054895(n-1).

G.f.: Sum_{j>0} x^6^j/(1-x^6^j). (End)

a(A047253(n)) = 0; a(A008588(n)) > 0; a(A044102(n)) > 1. - Reinhard Zumkeller, Nov 10 2013

6^a(n) = A234959(n), n >= 1. - Wolfdieter Lang, Jun 30 2014

Asymptotic mean: lim_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1/5. - Amiram Eldar, Jan 17 2022

a(n) = min(A007814(n), A007949(n)). - Jianing Song, Jul 23 2022

Mathematica

Table[IntegerExponent[n, 6], {n, 1, 100}] (* Amiram Eldar, Sep 14 2020 *)

Prog

(Haskell)
a122841 = f 0 where
   f y x = if r > 0 then y else f (y + 1) x'
           where (x', r) = divMod x 6
-- Reinhard Zumkeller, Nov 10 2013
(PARI) a(n) = valuation(n, 6); \\ Michel Marcus, Jan 17 2022

Crossrefs

Cf. A122840, A007814, A007949, A054895, A234959.

Sequence in context: A070205 A138363 A080225 * A326075 A060862 A325801

Adjacent sequences: A122838 A122839 A122840 * A122842 A122843 A122844

Keyword

nonn

Author

Reinhard Zumkeller, Sep 13 2006

Status

approved