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Highest power of 6 dividing n.
10

%I #35 Jul 10 2019 17:58:08

%S 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,

%T 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,

%U 1,1,1,36,1,1,1,1,1,6,1,1,1,1,1,6

%N Highest power of 6 dividing n.

%C The generalized binomial coefficients produced by this sequence provide an analog to Kummer's Theorem using arithmetic in base 6.

%H Reinhard Zumkeller, <a href="/A234959/b234959.txt">Table of n, a(n) for n = 1..10000</a>

%H Tyler Ball, Tom Edgar, and Daniel Juda, <a href="http://dx.doi.org/10.4169/math.mag.87.2.135">Dominance Orders, Generalized Binomial Coefficients, and Kummer's Theorem</a>, Mathematics Magazine, Vol. 87, No. 2, April 2014, pp. 135-143.

%F a(n) = 6^(valuation(n,6)).

%F a(n) = 6^A122841(n). - _Joerg Arndt_, Jan 02 2014

%F G.f.: x/(1 - x) + 5 * Sum_{k>=1} 6^(k-1)*x^(6^k)/(1 - x^(6^k)). - _Ilya Gutkovskiy_, Jul 10 2019

%e Since 12 = 6 * 2, a(12) = 6. Likewise, since 6 does not divide 13, a(13) = 1.

%t 6^Table[IntegerExponent[n, 6], {n, 84}] (* _Alonso del Arte_, Jan 01 2014 *)

%o (Sage)

%o n=200 #change n for more terms

%o [6^(valuation(i,6)) for i in [1..n]]

%o (Haskell)

%o a234959 = f 1 where

%o f y x = if m == 0 then f (y * 6) x' else y where (x', m) = divMod x 6

%o -- _Reinhard Zumkeller_, Feb 09 2015

%o (PARI) a(n)=6^valuation(n,6) \\ _Charles R Greathouse IV_, Aug 05 2015

%Y Cf. A006519, A038500, A122841, A234957, A243758.

%K nonn,easy

%O 1,6

%A _Tom Edgar_, Jan 01 2014