A132027 a(n) = Product_{k=0..floor(log_3(n))} floor(n/3^k), n>=1.

1, 2, 3, 4, 5, 12, 14, 16, 27, 30, 33, 48, 52, 56, 75, 80, 85, 216, 228, 240, 294, 308, 322, 384, 400, 416, 729, 756, 783, 900, 930, 960, 1089, 1122, 1155, 1728, 1776, 1824, 2028, 2080, 2132, 2352, 2408, 2464, 3375, 3450, 3525, 3840, 3920, 4000, 4335

Offset: 1

Author

Hieronymus Fischer, Aug 13 2007, Aug 20 2007

Keywords

• nonn

Comments

If n is written in base 3 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).

Examples

			a(11)=floor(11/3^0)*floor(11/3^1)*floor(11/3^2)=11*3*1=33;
a(13)=52 since 13=111(base-3) and so a(13)=111*11*1(base-3)=13*4*1=52.
		

Links

• Ivan Neretin, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A048651, A098844, A067080, A132019, A100220, A000217.

For formulas regarding a general parameter p (i.e. terms floor(n/p^k)) see A132264.

For the product of terms floor(n/p^k) for p=2 to p=12 see A098844(p=2), A132028(p=4)-A132033(p=9), A067080(p=10), A132263(p=11), A132264(p=12).

For the products of terms 1+floor(n/p^k) see A132269-A132272, A132327, A132328.

Programs

Mathematica

Table[(f = If[# < 3, #, #*f[Quotient[#, 3]]] &)[n], {n, 51}] (* Ivan Neretin, May 29 2016 *)

Formula

Recurrence: a(n)=n*a(floor(n/3)); a(n*3^m)=n^m*3^(m(m+1)/2)*a(n).

a(k*3^m)=k^(m+1)*3^(m(m+1)/2), for k=1 or 2.

a(n)<=b(n), where b(n)=n^(1+floor(log_3(n)))/3^(1/2*(1+floor(log_3(n)))*floor(log_3(n))); equality holds if n is a power of 3 or two times a power of 3.

Also: a(n)<=2^((1-log_3(2))/2)*n^((1+log_3(n))/2)=1.1364507...*3^A000217(log_3(n)), equality for n=2*3^m, m>=0.

a(n)>c*b(n), where c=0.3826631966790330232889550... (see constant A132019).

Also: a(n)>c*2^((1-log_3(2))/2)*n^((1+log_3(n))/2)=0.434877...*3^A000217(log_3(n)).

lim inf a(n)/b(n)=0.3826631966790330232889550..., for n-->oo.

lim sup a(n)/b(n)=1, for n-->oo.

lim inf a(n)/n^((1+log_3(n))/2)=0.3826631966790330232889550...*sqrt(2)/2^log_3(sqrt(2)), for n-->oo.

lim sup a(n)/n^((1+log_3(n))/2)=sqrt(2)/2^log_3(sqrt(2)), for n-->oo.

lim inf a(n)/a(n+1)=0.3826631966790330232889550... for n-->oo (see constant A132019).

a(n)=O(n^((1+log_3(n))/2)).