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A352219 a(n) is the least k such that A051037(n) | 60^k.

%I A352219 #16 May 25 2022 08:42:13
%S A352219 0,1,1,1,1,1,2,2,1,1,1,2,2,1,2,2,3,1,3,2,2,2,2,2,3,1,3,2,2,2,4,2,3,2,
%T A352219 3,2,3,4,3,2,2,3,4,2,3,2,3,2,2,5,3,4,3,3,2,3,4,2,3,4,2,4,3,2,3,5,3,5,
%U A352219 3,3,2,4,4,4,3,2,6,3,4,3,4,3,2,3,5,3,5
%N A352219 a(n) is the least k such that A051037(n) | 60^k.
%C A352219 Also, number of digits in the sexagesimal expansion of terminating unit fractions 1/A051037.
%D A352219 G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Chapter IX: The Representation of Numbers by Decimals, Theorem 136. 8th ed., Oxford Univ. Press, 2008, 144-145.
%H A352219 Michael De Vlieger, <a href="/A352219/b352219.txt">Table of n, a(n) for n = 1..10540</a> (A051037(10540) = 60^10)
%H A352219 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/Sexagesimal.html">Sexagesimal</a>
%H A352219 Wikipedia, <a href="http://en.wikipedia.org/wiki/Regular_number">Regular number</a>.
%F A352219 a(n) ≍ n^(1/3), with lim sup a(n)/n^(1/3) being (6*log(2)*log(3)*log(5))^(1/3)/log(3)  = 1.770... where A051037(n) is a power of 3 and the lim inf being (6*log(2)*log(3)*log(5))^(1/3)/log(60) = 0.4749... where A051037(n) is a power of 60. - _Charles R Greathouse IV_, Mar 08 2022
%e A352219 a(1) = 0 since A051037(1) = 1 | 60^0.
%e A352219 a(2) = 1 since A051037(2) = 2 | 60^1; 1/2 in base 60 is represented by digit 30 after the radix point ":", i.e., :30.
%e A352219 a(7) = 2 since A051037(7) = 8 | 60^2; 1/8 in base 60 is :7,30, etc.
%t A352219 With[{nn = 1024}, Sort[Flatten[Table[{2^a * 3^b * 5^c, Max[Ceiling[a/2], b, c]}, {a, 0, Log2[nn]}, {b, 0, Log[3, nn/(2^a)]}, {c, 0, Log[5, nn/(2^a*3^b)]}], 2]][[All, -1]] ]
%Y A352219 Cf. A051037, A086415, A117920.
%K A352219 nonn,easy,base
%O A352219 1,7
%A A352219 _Michael De Vlieger_, Mar 08 2022