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 A132328 #24 Sep 23 2024 11:32:18
%S A132328 1,1,1,2,2,2,3,3,3,8,8,8,10,10,10,12,12,12,21,21,21,24,24,24,27,27,27,
%T A132328 80,80,80,88,88,88,96,96,96,130,130,130,140,140,140,150,150,150,192,
%U A132328 192,192,204,204,204,216,216,216,399,399,399,420,420,420,441,441,441,528
%N A132328 a(n) = Product_{k>0} (1+floor(n/3^k)).
%C A132328 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 (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)).
%H A132328 Robert Israel, <a href="/A132328/b132328.txt">Table of n, a(n) for n = 0..10000</a>
%F A132328 Recurrence: a(n)=(1+floor(n/3))*a(floor(n/3)); a(3n)=(1+n)*a(n); a(n*3^m)=product{0<=k<m, 1+n*3^k}*a(n).
%F A132328 a(k*3^m-j)=k^m*3^(m(m-1)/2), for 0<k<3, 0<j<3, m>=1. a(3^m)=p^(m(m-1)/2)*product{0<=k<m, 1+1/3^k}
%F A132328 a(n)=A132327(floor(n/3))=A132327(n)/(1+n).
%F A132328 Asymptotic behavior: a(n)=O(n^((log_3(n)-1)/p)); this follows from the inequalities below.
%F A132328 a(n)<=A132027(n)/(n+1)*product{0<=k<=floor(log_3(n)), 1+1/3^k}.
%F A132328 a(n)>=A132027(n)/((n+1)*product{0<k<=floor(log_3(n)), 1-1/3^k}).
%F A132328 a(n)<c*n^((1+log_3(n))/2)/(n+1)=c*2^A000217(log_3(n))/(n+1), where c=product{k>0, 1+1/3^k}=3.12986803713402307587769821345767... (see constant A132323).
%F A132328 a(n)>n^((1+log_3(n))/2)/(n+1)=3^A000217(log_3(n))/(n+1).
%F A132328 lim sup n*a(n)/A132027(n)=2*product{k>0, 1+1/3^k}=3.12986803713402307587769821345767..., for n-->oo (see constant A132323).
%F A132328 lim inf n*a(n)/A132027(n)=1/product{k>0, 1-1/3^k}=1/0.560126077927948944969792243314140014..., for n-->oo (see constant A100220).
%F A132328 lim inf a(n)/n^((1+log_3(n))/2)=1, for n-->oo.
%F A132328 lim sup a(n)/n^((1+log_3(n))/2)=2*product{k>0, 1+1/3^k}=3.12986803713402307587769821345767..., for n-->oo (see constant A132323).
%F A132328 lim inf a(n+1)/a(n)=2*product{k>0, 1+1/3^k}=3.12986803713402307587769821345767... for n-->oo (see constant A132323).
%e A132328 a(12)=(1+floor(12/3^1))*(1+floor(12/3^2))=5*2=10; a(19)=21 since 19=201(base-3) and so a(19)=(1+20)*(1+2)(base-3)=7*3=21.
%p A132328 f:= proc(n) option remember; local t;
%p A132328 t:= floor(n/3);
%p A132328 (1+t)*procname(t)
%p A132328 end proc:
%p A132328 f(0):= 1: f(1):= 1: f(2):= 1:
%p A132328 map(f, [$0..100]); # _Robert Israel_, Oct 20 2020
%t A132328 (* Using definition *)
%t A132328 Table[Product[1 + Floor[n/3^k], {k, IntegerLength[n, 3] - 1}], {n, 0, 100}]
%t A132328 (* Using recurrence -- faster *)
%t A132328 a[0] = 1; a[n_] := a[n] = (1 + #)*a[#] & [Floor[n/3]];
%t A132328 Table[a[n], {n, 0, 100}] (* _Paolo Xausa_, Sep 23 2024 *)
%Y A132328 Cf. A100220, A132323, A132027, A132038, A132270(p=2), A132272(p=10).
%Y A132328 For formulas regarding a general parameter p (i.e. terms 1+floor(n/p^k)) see A132272.
%Y A132328 For the product of terms floor(n/p^k) see A098844, A067080, A132027-A132033, A132263, A132264.
%K A132328 nonn,base
%O A132328 0,4
%A A132328 _Hieronymus Fischer_, Aug 20 2007