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A234959
Highest power of 6 dividing n.
10
1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 36, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 36, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 6
OFFSET
1,6
COMMENTS
The generalized binomial coefficients produced by this sequence provide an analog to Kummer's Theorem using arithmetic in base 6.
LINKS
Tyler Ball, Tom Edgar, and Daniel Juda, Dominance Orders, Generalized Binomial Coefficients, and Kummer's Theorem, Mathematics Magazine, Vol. 87, No. 2, April 2014, pp. 135-143.
FORMULA
a(n) = 6^(valuation(n,6)).
a(n) = 6^A122841(n). - Joerg Arndt, Jan 02 2014
G.f.: x/(1 - x) + 5 * Sum_{k>=1} 6^(k-1)*x^(6^k)/(1 - x^(6^k)). - Ilya Gutkovskiy, Jul 10 2019
EXAMPLE
Since 12 = 6 * 2, a(12) = 6. Likewise, since 6 does not divide 13, a(13) = 1.
MATHEMATICA
6^Table[IntegerExponent[n, 6], {n, 84}] (* Alonso del Arte, Jan 01 2014 *)
PROG
(Sage)
n=200 #change n for more terms
[6^(valuation(i, 6)) for i in [1..n]]
(Haskell)
a234959 = f 1 where
f y x = if m == 0 then f (y * 6) x' else y where (x', m) = divMod x 6
-- Reinhard Zumkeller, Feb 09 2015
(PARI) a(n)=6^valuation(n, 6) \\ Charles R Greathouse IV, Aug 05 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Tom Edgar, Jan 01 2014
STATUS
approved