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 A132029 #9 Oct 16 2019 12:07:52
%S A132029 1,2,3,4,5,6,7,8,9,20,22,24,26,28,45,48,51,54,57,80,84,88,92,96,125,
%T A132029 130,135,140,145,180,186,192,198,204,245,252,259,266,273,320,328,336,
%U A132029 344,352,405,414,423,432,441,1000,1020,1040,1060,1080,1210,1232,1254,1276
%N A132029 Product{0<=k<=floor(log_5(n)), floor(n/5^k)}, n>=1.
%C A132029 If n is written in base-5 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).
%H A132029 Harvey P. Dale, <a href="/A132029/b132029.txt">Table of n, a(n) for n = 1..1000</a>
%F A132029 Recurrence: a(n)=n*a(floor(n/5)); a(n*5^m)=n^m*5^(m(m+1)/2)*a(n).
%F A132029 a(k*5^m)=k^(m+1)*5^(m(m+1)/2), for 0<k<5.
%F A132029 Asymptotic behavior: a(n)=O(n^((1+log_5(n))/2)); this follows from the inequalities below.
%F A132029 a(n)<=b(n), where b(n)=n^(1+floor(log_5(n)))/5^((1+floor(log_5(n)))*floor(log_5(n))/2); equality holds for n=k*5^m, 0<k<5, m>=0. b(n) can also be written n^(1+floor(log_5(n)))/5^A000217(floor(log_5(n))).
%F A132029 Also: a(n)<=2^((1-log_5(2))/2)*n^((1+log_5(n))/2)=1.2181246...*5^A000217(log_5(n)), equality holds for n=2*5^m, m>=0.
%F A132029 a(n)>c*b(n), where c=0.438796837203638531... (see constant A132021).
%F A132029 Also: a(n)>c*(sqrt(2)/2^log_5(sqrt(2)))*n^((1+log_5(n))/2)=0.534509224...*5^A000217(log_5(n)).
%F A132029 lim inf a(n)/b(n)=0.438796837203638531..., for n-->oo.
%F A132029 lim sup a(n)/b(n)=1, for n-->oo.
%F A132029 lim inf a(n)/n^((1+log_5(n))/2)=0.438796837203638531...*sqrt(2)/2^log_5(sqrt(2)), for n-->oo.
%F A132029 lim sup a(n)/n^((1+log_5(n))/2)=sqrt(2)/2^log_5(sqrt(2))=1.2181246..., for n-->oo.
%F A132029 lim inf a(n)/a(n+1)=0.438796837203638531... for n-->oo (see constant A132021).
%e A132029 a(26)=floor(26/5^0)*floor(26/5^1)*floor(26/5^2)=26*5*1=130; a(34)=204 since 34=114(base-5) and so a(34)=114*11*1(base-5)=34*6*1=204.
%t A132029 Table[Product[Floor[n/5^k],{k,0,Floor[Log[5,n]]}],{n,60}] (* _Harvey P. Dale_, Oct 16 2019 *)
%Y A132029 Cf. A048651, A132021, A100222, A000217.
%Y A132029 For formulas regarding a general parameter p (i.e. terms floor(n/p^k)) see A132264.
%Y A132029 For the product of terms floor(n/p^k) for p=2 to p=12 see A098844(p=2), A132027(p=3)-A132033(p=9), A067080(p=10), A132263(p=11), A132264(p=12).
%Y A132029 For the products of terms 1+floor(n/p^k) see A132269-A132272, A132327, A132328.
%K A132029 nonn
%O A132029 1,2
%A A132029 _Hieronymus Fischer_, Aug 20 2007