This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A132264 #8 Oct 28 2014 05:09:01
%S A132264 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 A132264 54,56,58,60,62,64,66,68,70,108,111,114,117,120,123,126,129,132,135,
%U A132264 138,141,192,196,200,204,208,212,216,220,224,228,232,236,300,305,310,315
%N A132264 Product{0<=k<=floor(log_12(n)), floor(n/12^k)}, n>=1.
%C A132264 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 A132264 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 A132264 Recurrence: a(n)=n*a(floor(n/p)); a(n*p^m)=n^m*p^(m(m+1)/2)*a(n).
%F A132264 a(k*p^m)=k^(m+1)*p^(m(m+1)/2), for 0<k<p.
%F A132264 Asymptotic behavior: a(n)=O(n^((1+log_p(n))/2)); this follows from the inequalities below.
%F A132264 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 A132264 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 A132264 a(n)>c*b(n), where c=product{k>0, 1-1/(2*p^k)}=0.47735217025489380... (for p=12 see constant A132265).
%F A132264 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 A132264 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 A132264 lim sup a(n)/b(n)=1, for n-->oo.
%F A132264 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 A132264 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 A132264 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 A132264 a(50)=floor(50/12^0)*floor(50/12^1)=50*4=200.
%e A132264 a(65)=325 since 65=55(base-12) and so a(65)=55*5(base-12)=65*5=325.
%Y A132264 Cf. A048651, A132019-A132026, A132265, A132267, A000217, A002378.
%Y A132264 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 A132264 For the products of terms 1+floor(n/p^k) see A132269-A132272, A132327, A132328.
%K A132264 nonn,base
%O A132264 1,2
%A A132264 _Hieronymus Fischer_, Aug 20 2007