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A132030 a(n) = Product_{k=0..floor(log_6(n))} floor(n/6^k), n>=1.

%I A132030 #13 Dec 21 2015 13:12:31
%S A132030 1,2,3,4,5,6,7,8,9,10,11,24,26,28,30,32,34,54,57,60,63,66,69,96,100,
%T A132030 104,108,112,116,150,155,160,165,170,175,216,222,228,234,240,246,294,
%U A132030 301,308,315,322,329,384,392,400,408,416,424,486,495,504,513,522,531,600
%N A132030 a(n) = Product_{k=0..floor(log_6(n))} floor(n/6^k), n>=1.
%C A132030 If n is written in base 6 as n = d(m)d(m-1)d(m-2)...d(2)d(1)d(0) (where d(k) is the digit at position k) then a(n) is also the product d(m)d(m-1)d(m-2)...d(2)d(1)d(0)*d(m)d(m-1)d(m-2)...d(2)d(1)*d(m)d(m-1)d(m-2)...d(2)*...*d(m)d(m-1)d(m-2)*d(m)d(m-1)*d(m).
%H A132030 G. C. Greubel, <a href="/A132030/b132030.txt">Table of n, a(n) for n = 1..10000</a>
%F A132030 Recurrence: a(n)=n*a(floor(n/6)); a(n*6^m)=n^m*6^(m(m+1)/2)*a(n).
%F A132030 a(k*6^m) = k^(m+1)*6^(m(m+1)/2), for 0<k<6.
%F A132030 Asymptotic behavior: a(n) = O(n^((1+log_6(n))/2)); this follows from the inequalities below.
%F A132030 a(n) <= b(n), where b(n) = n^(1+floor(log_6(n)))/6^((1+floor(log_6(n)))*floor(log_6(n))/2); equality holds for n=k*6^m, 0<k<6, m>=0. b(n) can also be written n^(1+floor(log_6(n)))/6^A000217(floor(log_6(n))).
%F A132030 Also: a(n) <= 2^((1-log_6(2))/2)*n^((1+log_6(n))/2) = 1.236766885...*6^A000217(log_6(n)), equality holds for n=2*6^m and for n=3*6^m, m>=0 (consider 2^((1-log_6(2))/2)=3^((1-log_6(3))/2) since 6=2*3).
%F A132030 a(n) > c*b(n), where c = 0.45071262522603913... (see constant A132022).
%F A132030 Also: a(n) > c*(sqrt(2)/2^log_6(sqrt(2)))*n^((1+log_6(n))/2) = 0.557426449...*6^A000217(log_6(n)).
%F A132030 lim inf a(n)/b(n) = 0.45071262522603913..., for n-->oo.
%F A132030 lim sup a(n)/b(n) = 1, for n-->oo.
%F A132030 lim inf a(n)/n^((1+log_6(n))/2) = 0.45071262522603913...*sqrt(2)/2^log_6(sqrt(2)), for n-->oo.
%F A132030 lim sup a(n)/n^((1+log_6(n))/2) = sqrt(3)/3^log_6(sqrt(3))=1.236766885..., for n-->oo.
%F A132030 lim inf a(n)/a(n+1) = 0.45071262522603913... for n-->oo (see constant A132022).
%F A132030 G.f. g(x) satisfies g(x) = (x+2x^2+3x^3+4x^4+5x^5)*(1 + g(x^6)) + 6*(x^6+x^7+x^8+x^9+x^10+x^11)*g'(x^6). - _Robert Israel_, Dec 20 2015
%e A132030 a(52) = floor(52/6^0)*floor(52/6^1)*floor(52/6^2) = 52*8*1 = 416;
%e A132030 a(58) = 522 since 58 = 134_6 and so a(58) = 134_6 * 13_6 * 1_6 = 58*9*1 = 522.
%p A132030 f:= proc(n) option remember; n*procname(floor(n/6)) end proc:
%p A132030 f(0):= 1:
%p A132030 seq(f(i),i=1..100); # _Robert Israel_, Dec 20 2015
%t A132030 Table[Product[Floor[n/6^k], {k, 0, Floor[Log[6, n]]}], {n, 1, 100}] (* _G. C. Greubel_, Dec 20 2015 *)
%Y A132030 Cf. A048651, A132022, A132034, A000217.
%Y A132030 For formulas regarding a general parameter p (i.e., terms floor(n/p^k)) see A132264.
%Y A132030 For the product of terms floor(n/p^k) for p=2 to p=12 see A098844(p=2), A132027(p=3)-A132033(p=9), A067080(p=10), A132263(p=11), A132264(p=12).
%Y A132030 For the products of terms 1+floor(n/p^k) see A132269-A132272, A132327, A132328.
%K A132030 nonn
%O A132030 1,2
%A A132030 _Hieronymus Fischer_, Aug 20 2007