cp's OEIS Frontend

A234959 Highest power of 6 dividing n.

%I A234959 #35 Jul 10 2019 17:58:08
%S A234959 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 A234959 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 A234959 1,1,1,36,1,1,1,1,1,6,1,1,1,1,1,6
%N A234959 Highest power of 6 dividing n.
%C A234959 The generalized binomial coefficients produced by this sequence provide an analog to Kummer's Theorem using arithmetic in base 6.
%H A234959 Reinhard Zumkeller, <a href="/A234959/b234959.txt">Table of n, a(n) for n = 1..10000</a>
%H A234959 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 A234959 a(n) = 6^(valuation(n,6)).
%F A234959 a(n) = 6^A122841(n). - _Joerg Arndt_, Jan 02 2014
%F A234959 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 A234959 Since 12 = 6 * 2, a(12) = 6. Likewise, since 6 does not divide 13, a(13) = 1.
%t A234959 6^Table[IntegerExponent[n, 6], {n, 84}] (* _Alonso del Arte_, Jan 01 2014 *)
%o A234959 (Sage)
%o A234959 n=200 #change n for more terms
%o A234959 [6^(valuation(i,6)) for i in [1..n]]
%o A234959 (Haskell)
%o A234959 a234959 = f 1 where
%o A234959    f y x = if m == 0 then f (y * 6) x' else y  where (x', m) = divMod x 6
%o A234959 -- _Reinhard Zumkeller_, Feb 09 2015
%o A234959 (PARI) a(n)=6^valuation(n,6) \\ _Charles R Greathouse IV_, Aug 05 2015
%Y A234959 Cf. A006519, A038500, A122841, A234957, A243758.
%K A234959 nonn,easy
%O A234959 1,6
%A A234959 _Tom Edgar_, Jan 01 2014