OFFSET
1,2
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).
LINKS
Ivan Neretin, Table of n, a(n) for n = 1..10000
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)).
EXAMPLE
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.
MATHEMATICA
Table[(f = If[# < 3, #, #*f[Quotient[#, 3]]] &)[n], {n, 51}] (* Ivan Neretin, May 29 2016 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Hieronymus Fischer, Aug 13 2007, Aug 20 2007
STATUS
approved