A132328 a(n) = Product_{k>0} (1+floor(n/3^k)).
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, 80, 80, 80, 88, 88, 88, 96, 96, 96, 130, 130, 130, 140, 140, 140, 150, 150, 150, 192, 192, 192, 204, 204, 204, 216, 216, 216, 399, 399, 399, 420, 420, 420, 441, 441, 441, 528
Offset: 0
Examples
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.
Links
- Robert Israel, Table of n, a(n) for n = 0..10000
Crossrefs
Programs
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Maple
f:= proc(n) option remember; local t; t:= floor(n/3); (1+t)*procname(t) end proc: f(0):= 1: f(1):= 1: f(2):= 1: map(f, [$0..100]); # Robert Israel, Oct 20 2020
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Mathematica
(* Using definition *) Table[Product[1 + Floor[n/3^k], {k, IntegerLength[n, 3] - 1}], {n, 0, 100}] (* Using recurrence -- faster *) a[0] = 1; a[n_] := a[n] = (1 + #)*a[#] & [Floor[n/3]]; Table[a[n], {n, 0, 100}] (* Paolo Xausa, Sep 23 2024 *)
Formula
Recurrence: a(n)=(1+floor(n/3))*a(floor(n/3)); a(3n)=(1+n)*a(n); a(n*3^m)=product{0<=k
a(k*3^m-j)=k^m*3^(m(m-1)/2), for 0=1. a(3^m)=p^(m(m-1)/2)*product{0<=k
Asymptotic behavior: a(n)=O(n^((log_3(n)-1)/p)); this follows from the inequalities below.
a(n)<=A132027(n)/(n+1)*product{0<=k<=floor(log_3(n)), 1+1/3^k}.
a(n)>=A132027(n)/((n+1)*product{0
a(n)A000217(log_3(n))/(n+1), where c=product{k>0, 1+1/3^k}=3.12986803713402307587769821345767... (see constant A132323).
a(n)>n^((1+log_3(n))/2)/(n+1)=3^A000217(log_3(n))/(n+1).
lim sup n*a(n)/A132027(n)=2*product{k>0, 1+1/3^k}=3.12986803713402307587769821345767..., for n-->oo (see constant A132323).
lim inf n*a(n)/A132027(n)=1/product{k>0, 1-1/3^k}=1/0.560126077927948944969792243314140014..., for n-->oo (see constant A100220).
lim inf a(n)/n^((1+log_3(n))/2)=1, for n-->oo.
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).
lim inf a(n+1)/a(n)=2*product{k>0, 1+1/3^k}=3.12986803713402307587769821345767... for n-->oo (see constant A132323).
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