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 A132271 #7 May 14 2019 14:56:24
%S A132271 1,2,3,4,5,6,7,8,9,10,22,24,26,28,30,32,34,36,38,40,63,66,69,72,75,78,
%T A132271 81,84,87,90,124,128,132,136,140,144,148,152,156,160,205,210,215,220,
%U A132271 225,230,235,240,245,250,306,312,318,324,330,336,342,348,354,360,427
%N A132271 Product{k>=0, 1+floor(n/10^k)}.
%C A132271 If n is written in base-10 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 (1+d(m)d(m-1)d(m-2)...d(2)d(1)d(0))*(1+d(m)d(m-1)d(m-2)...d(2)d(1))*(1+d(m)d(m-1)d(m-2)...d(2))*...*(1+d(m)d(m-1)d(m-2))*(1+d(m)d(m-1))*(1+d(m)).
%F A132271 The following formulas are given for a general parameter p considering the product of terms 1+floor(n/p^k) for 0<=k<=floor(log_p(n)), where p=10 for this sequence.
%F A132271 Recurrence: a(n)=(1+n)*a(floor(n/p)); a(pn)=(1+pn)*a(n); a(n*p^m)=product{1<=k<=m, 1+n*p^k}*a(n).
%F A132271 a(k*p^m-j)=(k*p^m-j+1)*k^m*p^(m(m-1)/2), for 0<k<p, 0<j<p, m>=1, a(p^m)=p^(m(m+1)/2)*product{0<=k<=m, 1+1/p^k}, m>=1.
%F A132271 a(n)=A132272(p*n)=(1+n)*A132272(n).
%F A132271 Asymptotic behavior: a(n)=O(n^((1+log_p(n))/2)); this follows from the inequalities below.
%F A132271 a(n)<=A067080(n)*product{0<=k<=floor(log_p(n)), 1+1/p^k}.
%F A132271 a(n)>=A067080(n)/product{1<=k<=floor(log_p(n)), 1-1/p^k}.
%F A132271 a(n)<c*n^((1+log_p(n))/2)=c*p^A000217(log_p(n)), where c=product{k>=0, 1+1/p^k}=2.2244691382741012... (for p=10 see constant A132325).
%F A132271 a(n)>n^((1+log_p(n))/2)=p^A000217(log_p(n)).
%F A132271 lim sup a(n)/A067080(n)=2*product{k>0, 1+1/p^k}=2.2244691382741012..., for n-->oo (for p=10 see constant A132325).
%F A132271 lim inf a(n)/A067080(n)=1/product{k>0, 1-1/p^k}=1/0.8900100999989990000001000..., for n-->oo (for p=10 see constant A132038).
%F A132271 lim inf a(n)/n^((1+log_p(n))/2)=1, for n-->oo.
%F A132271 lim sup a(n)/n^((1+log_p(n))/2)=2*product{k>0, 1+1/p^k}=2.2244691382741012..., for n-->oo (for p=10 see constant A132325).
%F A132271 lim inf a(n+1)/a(n)=2*product{k>0, 1+1/p^k}=2.2244691382741012... for n-->oo (for p=10 see constant A132325).
%e A132271 a(12)=(1+floor(12/10^0))*(1+floor(12/10^1))=13*2=26; a(21)=63 since 21=21(base-10) and so
%e A132271 a(21)=(1+21)*(1+2)(base-10)=22*3=66.
%t A132271 f[n_] := Block[{k = 0, p = 1}, While[a = Floor[n/10^k]; a > 0, p *= 1 + a; k++]; p]; Array[f, 61, 0] (* _Robert G. Wilson v_, May 10 2011 *)
%t A132271 Table[Product[1+Floor[n/10^k],{k,0,n}],{n,0,60}] (* _Harvey P. Dale_, May 14 2019 *)
%Y A132271 Cf. A132038, A132325, A132269(p=2), A132327(p=3), A132272.
%Y A132271 For the product of terms floor(n/p^k) see A098844, A067080, A132027-A132033, A132263, A132264.
%K A132271 nonn
%O A132271 0,2
%A A132271 _Hieronymus Fischer_, Aug 20 2007