1,2

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).

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...*sqr(2)/2^log_3(sqr(2)), for n-->oo.

lim sup a(n)/n^((1+log_3(n))/2)=sqr(2)/2^log_3(sqr(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)).

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.

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.

Sequence in context: A039007 A050745 A077375 this_sequence A103651 A093713 A057472

Adjacent sequences: A132024 A132025 A132026 this_sequence A132028 A132029 A132030

nonn

Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Aug 13 2007, Aug 20 2007