A132264 - OEIS

%I #8 Oct 28 2014 05:09:01

%S 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,48,50,52,

%T 54,56,58,60,62,64,66,68,70,108,111,114,117,120,123,126,129,132,135,

%U 138,141,192,196,200,204,208,212,216,220,224,228,232,236,300,305,310,315

%N Product{0<=k<=floor(log_12(n)), floor(n/12^k)}, n>=1.

%C If n is written in base-12 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).

%F The following formulas are given for a general parameter p considering the product of terms floor(n/p^k) for 0<=k<=floor(log_p(n)), where p=12 for this sequence.

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

%F a(k*p^m)=k^(m+1)*p^(m(m+1)/2), for 0<k<p.

%F Asymptotic behavior: a(n)=O(n^((1+log_p(n))/2)); this follows from the inequalities below.

%F a(n)<=b(n), where b(n)=n^(1+floor(log_p(n)))/p^((1+floor(log_p(n)))*floor(log_p(n))/2); equality holds for n=k*p^m, 0<k<p, m>=0. b(n) can also be written n^(1+floor(log_p(n)))/p^A000217(floor(log_p(n))).

%F Also: a(n)<=q^((1-log_p(q))/2)*n^((1+log_p(n))/2)=q^((1-log_p(q))/2)*p^A000217(log_p(n)), equality holds for n=q*p^m, m>=0, where q=floor(sqrt(p)+1/2). Also, equality holds for n=(q+1)*p^m, provided p is a A002378-number (in this case we have p=q*(q+1) and so q^((1-log_p(q))/2)=(q+1)^((1-log_p(q+1))/2)).

%F a(n)>c*b(n), where c=product{k>0, 1-1/(2*p^k)}=0.47735217025489380... (for p=12 see constant A132265).

%F Also: a(n)>c*(sqrt(2)/2^log_p(sqrt(2)))*n^((1+log_p(n))/2)=0.612870619...*p^A000217(log_p(n)), (p=12).

%F lim inf a(n)/b(n)=product{k>0, 1-1/(2*p^k)}=0.47735217025489380198334286365820..., for n-->oo (for p=12 see constant A132265).

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

%F lim inf a(n)/n^((1+log_p(n))/2)=(sqrt(2)/2^log_p(sqrt(2)))*product{k>0, 1-1/(2*p^k)}=0.612870619..., for n-->oo, (p=12).

%F lim sup a(n)/n^((1+log_p(n))/2)=sqrt(q)/q^log_p(sqrt(q))=1.358593737..., for n-->oo, (p=12, q=round(sqrt(p))=3).

%F lim inf a(n)/a(n+1)=product{k>0, 1-1/(2*p^k)}=0.47735217025489380... for n-->oo (for p=12 see constant A132265).

%e a(50)=floor(50/12^0)*floor(50/12^1)=50*4=200.

%e a(65)=325 since 65=55(base-12) and so a(65)=55*5(base-12)=65*5=325.

%Y Cf. A048651, A132019-A132026, A132265, A132267, A000217, A002378.

%Y For the product of terms floor(n/p^k) for p=2 to p=11 see A098844(p=2), A132027(p=3)-A132033(p=9), A067080(p=10), A132263(p=11).

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

%K nonn,base

%O 1,2

%A _Hieronymus Fischer_, Aug 20 2007