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

0,4

Comments

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)).

Links

Robert Israel, Table of n, a(n) for n = 0..10000

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<m, 1+n*3^k}*a(n).

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}

a(n)=A132327(floor(n/3))=A132327(n)/(1+n).

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<k<=floor(log_3(n)), 1-1/3^k}).

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).

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).

Example

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.

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

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 *)

Crossrefs

Cf. A100220, A132323, A132027, A132038, A132270(p=2), A132272(p=10).

For formulas regarding a general parameter p (i.e. terms 1+floor(n/p^k)) see A132272.

For the product of terms floor(n/p^k) see A098844, A067080, A132027-A132033, A132263, A132264.

Sequence in context: A029058 A046026 A139801 * A064822 A238337 A104484

Adjacent sequences: A132325 A132326 A132327 * A132329 A132330 A132331

Keyword

nonn,base

Author

Hieronymus Fischer, Aug 20 2007

Status

approved