A132264 -id:A132264 - cp's OEIS frontend

A067080 If n = ab...def in decimal notation then the left digitorial function Ld(n) = ab...defab...deab...d...ab*a.

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 90, 93, 96, 99, 102, 105, 108, 111, 114, 117, 160, 164, 168, 172, 176, 180, 184, 188, 192, 196, 250, 255, 260, 265, 270, 275, 280, 285, 290, 295, 360, 366, 372

Offset: 1

Amarnath Murthy, Jan 05 2002

This entry should probably start at n=0, just as A067079 does. But that would require a number of changes, so it can wait until the editors have more free time. - N. J. A. Sloane, Nov 29 2014

			Ld(256) = 256*25*2 =12800.
a(31)=floor(31/10^0)*floor(31/10^1)=31*3=93;
a(42)=168 since 42=42(base-10) and so a(42)=42*4(base-10)=42*4=168.

Hieronymus Fischer, Table of n, a(n) for n = 1..10000

Cf. A067079, A065039, A048651, A098844, A132019, A132026, A132038, A000217.
For formulas regarding a general parameter p (i.e. terms floor(n/p^k)) see A132264.
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), A132263(p=11), A132264(p=12).
For the products of terms 1+floor(n/p^k) see A132269-A132272, A132327, A132328.

a067080 n = if n <= 9 then n else n * a067080 (n `div` 10)
-- Reinhard Zumkeller, Nov 29 2012

Table[d = IntegerDigits[n]; rd = 1; While[ Length[d] > 0, rd = rd*FromDigits[d]; d = Drop[d, -1]]; rd, {n, 1, 75} ]
Table[Times@@NestList[Quotient[#,10]&,n,IntegerLength[n]-1],{n,70}] (* Harvey P. Dale, Dec 16 2013 *)

a(n)=my(t=n);while(n\=10,t*=n); t \\ Charles R Greathouse IV, Nov 20 2012

a(n) = Product_{k=1..length(n)} floor(n/10^(k-1)). - Vladeta Jovovic, Jan 08 2002
From Hieronymus Fischer, Aug 13 2007: (Start)
a(n) = product{0<=k<=floor(log_10(n)), floor(n/10^k)}, n>=1.
Recurrence:
a(n) = n*a(floor(n/10));
a(n*10^m) = n^m*10^(m(m+1)/2)*a(n).
a(k*10^m) = k^(m+1)*10^(m(m+1)/2), for 0a(n) <= b(n), where b(n)=n^(1+floor(log_10(n)))/10^(1/2*(1+floor(log_10(n)))*floor(log_10(n))); equality holds for n=k*10^m, m>=0, 1<=k<10. Here b(n) can also be written n^(1+floor(log_10(n)))/10^A000217(floor(log_10(n))).
Also: a(n) <= 3^((1-log_10(3))/2)*n^((1+log_10(n))/2)=1.332718...*10^A000217(log_10(n)), equality for n=3*10^m, m>=0.
a(n) > c*b(n), where c=0.472362443816572... (see constant A132026).
Also: a(n) > c*2^((1-log_10(2))/2)*n^((1+log_10(n))/2) = 0.601839...*10^A000217(log_10(n)).
lim inf a(n)/b(n) = 0.472362443816572..., for n-->oo.
lim sup a(n)/b(n) = 1, for n-->oo.
lim inf a(n)/n^((1+log_10(n))/2) = 0.472362443816572...*sqrt(2)/2^log_10(sqrt(2)), for n-->oo.
lim sup a(n)/n^((1+log_10(n))/2) = sqrt(3)/3^log_10(sqrt(3)), for n-->oo.
lim inf a(n)/a(n+1) = 0.472362443816572... for n-->oo (see constant A132026).
a(n) = O(n^((1+log_10(n))/2)). (End)

More terms from Robert G. Wilson v, Jan 07 2002

A098844 a(1)=1, a(n) = n*a(floor(n/2)).

Original entry on oeis.org

1, 2, 3, 8, 10, 18, 21, 64, 72, 100, 110, 216, 234, 294, 315, 1024, 1088, 1296, 1368, 2000, 2100, 2420, 2530, 5184, 5400, 6084, 6318, 8232, 8526, 9450, 9765, 32768, 33792, 36992, 38080, 46656, 47952, 51984, 53352, 80000, 82000, 88200, 90300

Offset: 1

Benoit Cloitre, Nov 03 2004

nonn

			a(10) = floor(10/2^0)*floor(10/2^1)*floor(10/2^2)*floor(10/2^3) = 10*5*2*1 = 100;
a(17) = 1088 since 17 = 10001(base 2) and so a(17) = 10001*1000*100*10*1(base 2) = 17*8*4*2*1 = 1088.

Hieronymus Fischer, Table of n, a(n) for n = 1..1000

Cf. A048651, A067080, A132027, A132028, A132029, A132030, A132019, A132026, A132038.
For formulas regarding a general parameter p (i.e., terms floor(n/p^k)) see A132264.
For the product of terms floor(n/p^k) for p=3 to p=12 see A132027(p=3)-A132033(p=9), A067080(p=10), A132263(p=11), A132264(p=12).
For the products of terms 1+floor(n/p^k) see A132269-A132272, A132327, A132328.

lst={};Do[p=n;s=1;While[p>1,p=IntegerPart[p/2];s*=p;];AppendTo[lst,s],{n,1,6!,2}];lst (* Vladimir Joseph Stephan Orlovsky, Jul 28 2009 *)

PARI
```
a(n)=if(n<2,1,n*a(floor(n/2)))
```

from math import prod
def A098844(n): return n*prod(n//2**k for k in range(1,n.bit_length()-1)) # Chai Wah Wu, Jun 07 2022

a(2^n) = 2^(n*(n+1)/2) = A006125(n+1).
From Hieronymus Fischer, Aug 13 2007: (Start)
a(n) = Product_{k=0..floor(log_2(n))} floor(n/2^k), n>=1.
Recurrence:
a(n*2^m) = n^m*2^(m(m+1)/2)*a(n).
a(n) <= n^((1+log_2(n))/2) = 2^A000217(log_2(n)); equality iff n is a power of 2.
a(n) >= c(n)*(n+1)^((1 + log_2(n+1))/2) for n != 2,
where c(n) = Product_{k=1..floor(log_2(n))} (1 - 1/2^k); equality holds iff n+1 is a power of 2.
a(n) > c*(n+1)^((1 + log_2(n+1))/2)
where c = 0.288788095086602421... (see constant A048651).
lim inf a(n)/n^((1+log_2(n))/2)=0.288788095086602421... for n-->oo.
lim sup a(n)/n^((1+log_2(n))/2) = 1 for n-->oo.
lim inf a(n)/a(n+1) = 0.288788095086602421... for n-->oo (see constant A048651).
a(n) = O(n^((1+log_2(n))/2)). (End)

Formula section edited by Hieronymus Fischer, Jun 13 2012

A132027 a(n) = Product_{k=0..floor(log_3(n))} floor(n/3^k), n>=1.

Original entry on oeis.org

1, 2, 3, 4, 5, 12, 14, 16, 27, 30, 33, 48, 52, 56, 75, 80, 85, 216, 228, 240, 294, 308, 322, 384, 400, 416, 729, 756, 783, 900, 930, 960, 1089, 1122, 1155, 1728, 1776, 1824, 2028, 2080, 2132, 2352, 2408, 2464, 3375, 3450, 3525, 3840, 3920, 4000, 4335

Offset: 1

Hieronymus Fischer, Aug 13 2007, Aug 20 2007

nonn

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 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).

			a(11)=floor(11/3^0)*floor(11/3^1)*floor(11/3^2)=11*3*1=33;
a(13)=52 since 13=111(base-3) and so a(13)=111*11*1(base-3)=13*4*1=52.

Ivan Neretin, Table of n, a(n) for n = 1..10000

Cf. A048651, A098844, A067080, A132019, A100220, A000217.
For formulas regarding a general parameter p (i.e. terms floor(n/p^k)) see A132264.
For the product of terms floor(n/p^k) for p=2 to p=12 see A098844(p=2), A132028(p=4)-A132033(p=9), A067080(p=10), A132263(p=11), A132264(p=12).
For the products of terms 1+floor(n/p^k) see A132269-A132272, A132327, A132328.

Table[(f = If[# < 3, #, #*f[Quotient[#, 3]]] &)[n], {n, 51}] (* Ivan Neretin, May 29 2016 *)

Recurrence: a(n)=n*a(floor(n/3)); a(n*3^m)=n^m*3^(m(m+1)/2)*a(n).
a(k*3^m)=k^(m+1)*3^(m(m+1)/2), for k=1 or 2.
a(n)<=b(n), where b(n)=n^(1+floor(log_3(n)))/3^(1/2*(1+floor(log_3(n)))*floor(log_3(n))); equality holds if n is a power of 3 or two times a power of 3.
Also: a(n)<=2^((1-log_3(2))/2)*n^((1+log_3(n))/2)=1.1364507...*3^A000217(log_3(n)), equality for n=2*3^m, m>=0.
a(n)>c*b(n), where c=0.3826631966790330232889550... (see constant A132019).
Also: a(n)>c*2^((1-log_3(2))/2)*n^((1+log_3(n))/2)=0.434877...*3^A000217(log_3(n)).
lim inf a(n)/b(n)=0.3826631966790330232889550..., for n-->oo.
lim sup a(n)/b(n)=1, for n-->oo.
lim inf a(n)/n^((1+log_3(n))/2)=0.3826631966790330232889550...*sqrt(2)/2^log_3(sqrt(2)), for n-->oo.
lim sup a(n)/n^((1+log_3(n))/2)=sqrt(2)/2^log_3(sqrt(2)), for n-->oo.
lim inf a(n)/a(n+1)=0.3826631966790330232889550... for n-->oo (see constant A132019).
a(n)=O(n^((1+log_3(n))/2)).

A132272 a(n) = Product_{k>0} (1 + floor(n/10^k)).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10

Offset: 0

Hieronymus Fischer, Aug 20 2007

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))*(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)).

a(n) = A179051(n) for n < 100. [From Reinhard Zumkeller, Jun 27 2010]

			a(121)=(1+floor(121/10^1))*(1+floor(121/10^2))=13*2=26; a(132)=28 since 132=132(base-10) and so
a(132)=(1+13)*(1+1)(base-10)=14*2=28.

R. J. Mathar, Table of n, a(n) for n = 0..500

Cf. A132038, A132270(p=2), A132271, A132325, A132328(p=3), A132271.

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

A132272 := proc(n)
    a := 1;
    for k from 1 do
        f := floor(n/10^k) ;
        if f <=0 then
            return a;
        else
            a := a*(1+f) ;
        end if;
    end do:
end proc:
seq(A132272(n),n=1..120) ; # R. J. Mathar, Jun 13 2025

Table[Product[1+Floor[n/10^k],{k,n}],{n,0,100}] (* Harvey P. Dale, Jul 20 2024 *)

The following formulas are given for a general parameter p considering the product of terms 1+floor(n/p^k) for 0

Recurrence: a(n)=(1+floor(n/p))*a(floor(n/p)); a(pn)=(1+n)*a(n); a(n*p^m)=product{0<=k

a(k*p^m-j)=k^m*p^(m(m-1)/2), for 0=1. a(p^m)=p^(m(m-1)/2)*product{0<=k

a(n)=A132271(floor(n/p))=A132271(n)/(1+n).

Asymptotic behavior: a(n)=O(n^((log_p(n)-1)/p)); this follows from the inequalities below.

a(n)<=A067080(n)/(n+1)*product{0<=k<=floor(log_p(n)), 1+1/p^k}.

a(n)>=A067080(n)/((n+1)*product{0

a(n)A000217(log_p(n))/(n+1), where c=product{k>0, 1+1/p^k}=2.2244691382741012... (for p=10 see constant A132325).

a(n)>n^((1+log_p(n))/2)/(n+1)=p^A000217(log_p(n))/(n+1).

lim sup n*a(n)/A067080(n)=2*product{k>0, 1+1/p^k}=2.2244691382741012..., for n-->oo (for p=10 see constant A132325).

lim inf n*a(n)/A067080(n)=1/product{k>0, 1-1/p^k}=1/0.8900100999989990000001000..., for n-->oo (for p=10 s. constant A132038).

lim inf a(n)/n^((1+log_p(n))/2)=1, for n-->oo.

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).

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).

A132033 Product{0<=k<=floor(log_9(n)), floor(n/9^k)}, n>=1.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 36, 38, 40, 42, 44, 46, 48, 50, 52, 81, 84, 87, 90, 93, 96, 99, 102, 105, 144, 148, 152, 156, 160, 164, 168, 172, 176, 225, 230, 235, 240, 245, 250, 255, 260, 265, 324, 330, 336, 342, 348, 354, 360, 366, 372

Offset: 1

Hieronymus Fischer, Aug 20 2007

If n is written in base-9 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).

			a(85)=floor(85/9^0)*floor(85/9^1)*floor(85/9^2)=85*9*1=765; a(88)=792 since 88=107(base-9) and so a(88)=107*10*1(base-9)=88*9*1=792.

Cf. A048651, A132025, A132037, A000217.
For formulas regarding a general parameter p (i.e. terms floor(n/p^k)) see A132264.
For the product of terms floor(n/p^k) for p=2 to p=12 see A098844(p=2), A132027(p=3)-A132032(p=8), A067080(p=10), A132263(p=11), A132264(p=12).
For the products of terms 1+floor(n/p^k) see A132269-A132272, A132327, A132328.

Table[Product[Floor[n/9^k],{k,0,Floor[Log[9,n]]}],{n,62}] (* James C. McMahon, Mar 03 2025 *)

Recurrence: a(n)=n*a(floor(n/9)); a(n*9^m)=n^m*9^(m(m+1)/2)*a(n).

a(k*9^m)=k^(m+1)*9^(m(m+1)/2), for 0

Asymptotic behavior: a(n)=O(n^((1+log_9(n))/2)); this follows from the inequalities below.

a(n)<=b(n), where b(n)=n^(1+floor(log_9(n)))/9^((1+floor(log_9(n)))*floor(log_9(n))/2); equality holds for n=k*9^m, 0=0. b(n) can also be written n^(1+floor(log_9(n)))/9^A000217(floor(log_9(n))).

Also: a(n)<=3^(1/4)*n^((1+log_9(n))/2)=1.316074013...*9^A000217(log_9(n)), equality holds for n=3*9^m, m>=0.

a(n)>c*b(n), where c=0.4689451783670236932832800... (see constant A132024).

Also: a(n)>c*2^((1-log_9(2))/2)*n^((1+log_9(n))/2)=0.4689451783670...*1.267747616...*9^A000217(log_9(n)).

lim inf a(n)/b(n)=0.4689451783670236932832800..., for n-->oo.

lim sup a(n)/b(n)=1, for n-->oo.

lim inf a(n)/n^((1+log_9(n))/2)=0.4689451783670236932832800...*sqrt(2)/2^log_9(sqrt(2)), for n-->oo.

lim sup a(n)/n^((1+log_9(n))/2)=3^(1/4)=1.316074013..., for n-->oo.

lim inf a(n)/a(n+1)=0.4689451783670236932832800... for n-->oo (see constant A132025).

A132328 a(n) = Product_{k>0} (1+floor(n/3^k)).

Original entry on oeis.org

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

Hieronymus Fischer, Aug 20 2007

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)).

			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.

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

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.

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

(* 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 *)

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

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

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).

A132263 Product{0<=k<=floor(log_11(n)), floor(n/11^k)}, n>=1.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 99, 102, 105, 108, 111, 114, 117, 120, 123, 126, 129, 176, 180, 184, 188, 192, 196, 200, 204, 208, 212, 216, 275, 280, 285, 290, 295, 300, 305, 310

Offset: 1

Hieronymus Fischer, Aug 20 2007

If n is written in base-11 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).

			a(50)=floor(50/11^0)*floor(50/11^1)=50*4=200; a(63)=315 since 63=58(base-11) and so a(63)=58*5(base-11)=63*5=315.

Cf. A048651, A132019-A132026, A132265, A132267, A000217.
For formulas regarding a general parameter p (i.e. terms floor(n/p^k)) see A132264.
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), A132264(p=12).
For the products of terms 1+floor(n/p^k) see A132269-A132272, A132327, A132328.

Recurrence: a(n)=n*a(floor(n/11)); a(n*11^m)=n^m*11^(m(m+1)/2)*a(n).

a(k*11^m)=k^(m+1)*11^(m(m+1)/2), for 0

Asymptotic behavior: a(n)=O(n^((1+log_11(n))/2)); this follows from the inequalities below.

a(n)<=b(n), where b(n)=n^(1+floor(log_11(n)))/p^((1+floor(log_11(n)))*floor(log_11(n))/2); equality holds for n=k*11^m, 0=0. b(n) can also be written n^(1+floor(log_11(n)))/11^A000217(floor(log_11(n))).

Also: a(n)<=3^((1-log_11(3))/2)*n^((1+log_11(n))/2)=1.346673852...^((1-log_11(3))/2)*11^A000217(log_11(n)), equality holds for n=3*11^m, m>=0.

a(n)>c*b(n), where c=0.4751041275076031053975644472... (see constant A132265).

Also: a(n)>c*(sqrt(2)/2^log_11(sqrt(2)))*n^((1+log_11(n))/2)=0.607848303...*11^00217(log_11(n)).

lim inf a(n)/b(n)=0.4751041275076031053975644472..., for n-->oo.

lim sup a(n)/b(n)=1, for n-->oo.

lim inf a(n)/n^((1+log_p(n))/2)=0.4751041275076031...*sqrt(2)/2^log_11(sqrt(2)), for n-->oo.

lim sup a(n)/n^((1+log_p(n))/2)=sqrt(3)/3^log_11(sqrt(3))=1.346673852..., for n-->oo.

lim inf a(n)/a(n+1)=0.4751041275076031053975644472... for n-->oo (see constant A132265).

A132269 a(n) = Product_{k>=0} (1 + floor(n/2^k)).

Original entry on oeis.org

1, 2, 6, 8, 30, 36, 56, 64, 270, 300, 396, 432, 728, 784, 960, 1024, 4590, 4860, 5700, 6000, 8316, 8712, 9936, 10368, 18200, 18928, 21168, 21952, 27840, 28800, 31744, 32768, 151470, 156060, 170100, 174960, 210900, 216600, 234000, 240000, 340956, 349272, 374616

Offset: 0

Hieronymus Fischer, Aug 20 2007

nonn

If n is written in base 2 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)).
From Gary W. Adamson, Aug 25 2016: (Start)
Given the following production matrix M =
  1, 0, 0, 0, 0, ...
  2, 0, 0, 0, 0, ...
  0, 3, 0, 0, 0, ...
  0, 4, 0, 0, 0, ...
  0, 0, 5, 0, 0, ...
  0, 0, 6, 0, 0, ...
  0, 0, 0, 7, 0, ...
  ...
the sequence is the left-shifted vector as lim_{n->infinity} M^n. (End)

			a(10) = (1 + floor(10/2^0))*(1 + floor(10/2^1))*(1 + floor(10/2^2))*(1 + floor(10/2^3)) = 11*6*3*2 = 396;
a(17) = 4860 since 17 = 10001_2 and so a(17) = (1+10001_2)*(1+1000_2)*(1+100_2)*(1+10_2)*(1+1) = 18*9*5*3*2 = 4860.

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

Cf. A048651, A081845, A132270, A132271(p=10), A132272, A132327(p=3), A132328.
For formulas regarding a general parameter p (i.e., terms 1+floor(n/p^k)) see A132271.
For the product of terms floor(n/p^k) see A098844, A067080, A132027-A132033, A132263, A132264.

[1] cat [n le 1 select 2 else (1+n)*Self(Floor(n/2)): n in [1..50]]; // Vincenzo Librandi, Aug 26 2016

f:= proc(n) option remember; (1+n)*procname(floor(n/2)) end proc:
f(0):= 1:
map(f, [$0..100]); # Robert Israel, Aug 26 2016

Table[Product[1 + Floor[2 n/2^k], {k, 2 n}], {n, 0, 42}] (* or *)
Table[Function[w, Times @@ Map[1 + FromDigits[PadRight[w, #], 2] &, Range@ Length@ w]]@ IntegerDigits[n, 2], {n, 0, 42}] (* Michael De Vlieger, Aug 26 2016 *)

Recurrence: a(n)=(1+n)*a(floor(n/2)); a(2n)=(1+2n)*a(n); a(n*2^m) = (Product_{k=1..m} (1 + n*2^k))*a(n).
a(2^m-1) = 2^(m*(m+1)/2), a(2^m) = 2^(m*(m+1)/2)*Product_{k=0..m} (1 + 1/2^k), m>=1.
a(n) = A132270(2n) = (1+n)*A132270(n).
Asymptotic behavior: a(n) = O(n^((1+log_2(n))/2)); this follows from the inequalities below.
a(n) <= A098844(n)*Product_{k=0..floor(log_2(n))} (1 + 1/2^k).
a(n) >= A098844(n)/Product_{k=1..floor(log_2(n))} (1 - 1/2^k).
a(n) < c*n^((1+log_2(n))/2) = c*2^A000217(log_2(n)), where c = Product_{k>=0} (1 + 1/2^k) = 4.7684620580627... (see constant A081845).
a(n) > n^((1+log_2(n))/2) = 2^A000217(log_2(n)),
lim sup a(n)/A098844(n) = Product_{k>=0} (1 + 1/2^k) = 4.7684620580627..., for n->oo (see constant A081845).
lim inf a(n)/A098844(n) = 1/Product_{k>=1} (1 - 1/2^k) = 1/0.288788095086602421..., for n->oo (see constant A048651).
lim inf a(n)/n^((1+log_2(n))/2) = 1, for n->oo.
lim sup a(n)/n^((1+log_2(n))/2) = Product_{k>=0} (1 + 1/2^k) = 4.7684620580627..., for n->oo (see constant A081845).
lim inf a(n+1)/a(n) = Product_{k>=0} (1 + 1/2^k) = 4.7684620580627... for n->oo (see constant A081845).
G.f. g(x) satisfies g(x) = (1+2x)*g(x^2) + 2*x^2*(1+x)*g'(x^2). - Robert Israel, Aug 26 2016

A132327 a(n) = Product{k>=0} (1 + floor(n/3^k)).

Original entry on oeis.org

1, 2, 3, 8, 10, 12, 21, 24, 27, 80, 88, 96, 130, 140, 150, 192, 204, 216, 399, 420, 441, 528, 552, 576, 675, 702, 729, 2240, 2320, 2400, 2728, 2816, 2904, 3264, 3360, 3456, 4810, 4940, 5070, 5600, 5740, 5880, 6450, 6600, 6750, 8832, 9024, 9216, 9996, 10200

Offset: 0

Hieronymus Fischer, Aug 20 2007

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)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)).

			a(12)=(1+floor(12/3^0))*(1+floor(12/3^1))*(1+floor(12/3^2))=13*5*2=130; a(20)=441 since 20=202(base-3) and so
a(20)=(1+202)*(1+20)*(1+2)(base-3)=21*7*3=441.

Cf. A100220, A132027, A132038, A132264, A132269(for p=2), A132271(for p=10).
For formulas regarding a general parameter p (i.e. terms 1+floor(n/p^k)) see A132271.
For the product of terms floor(n/p^k) see A098844, A067080, A132027-A132033, A132263, A132264.

Table[Product[1+Floor[n/3^k],{k,0,n}],{n,0,49}] (* James C. McMahon, Mar 07 2025 *)

Recurrence: a(n)=(1+n)*a(floor(n/3)); a(3n)=(1+3n)*a(n); a(n*3^m)=product{1<=k<=m, 1+n*3^k}*a(n).
a(k*3^m-j)=(k*3^m-j+1)*3^m*p^(m(m-1)/2), for 0=1, a(3^m)=3^(m(m+1)/2)*product{0<=k<=m, 1+1/3^k}, m>=1.
a(n)=A132328(3*n)=(1+n)*A132328(n).
Asymptotic behavior: a(n)=O(n^((1+log_3(n))/2)); this follows from the inequalities below.
a(n)<=A132027(n)*product{0<=k<=floor(log_3(n)), 1+1/3^k}.
a(n)>=A132027(n)/product{1<=k<=floor(log_3(n)), 1-1/3^k}.
a(n)A000217(log_3(n)), where c=product{k>=0, 1+1/p^k}=3.12986803713402307587769821345767... (see constant A132323).
a(n)>n^((1+log_3(n))/2)=3^A000217(log_3(n)).
lim sup a(n)/A132027(n)=2*product{k>0, 1+1/3^k}=3.12986803713402307587769821345767..., for n-->oo (see constant A132323).
lim inf 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).

A132271 Product{k>=0, 1+floor(n/10^k)}.

Original entry on oeis.org

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, 81, 84, 87, 90, 124, 128, 132, 136, 140, 144, 148, 152, 156, 160, 205, 210, 215, 220, 225, 230, 235, 240, 245, 250, 306, 312, 318, 324, 330, 336, 342, 348, 354, 360, 427

Offset: 0

Hieronymus Fischer, Aug 20 2007

nonn

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)).

			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
a(21)=(1+21)*(1+2)(base-10)=22*3=66.

Cf. A132038, A132325, A132269(p=2), A132327(p=3), A132272.

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

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 *)
Table[Product[1+Floor[n/10^k],{k,0,n}],{n,0,60}] (* Harvey P. Dale, May 14 2019 *)

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.
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).
a(k*p^m-j)=(k*p^m-j+1)*k^m*p^(m(m-1)/2), for 0=1, a(p^m)=p^(m(m+1)/2)*product{0<=k<=m, 1+1/p^k}, m>=1.
a(n)=A132272(p*n)=(1+n)*A132272(n).
Asymptotic behavior: a(n)=O(n^((1+log_p(n))/2)); this follows from the inequalities below.
a(n)<=A067080(n)*product{0<=k<=floor(log_p(n)), 1+1/p^k}.
a(n)>=A067080(n)/product{1<=k<=floor(log_p(n)), 1-1/p^k}.
a(n)A000217(log_p(n)), where c=product{k>=0, 1+1/p^k}=2.2244691382741012... (for p=10 see constant A132325).
a(n)>n^((1+log_p(n))/2)=p^A000217(log_p(n)).
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).
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).
lim inf a(n)/n^((1+log_p(n))/2)=1, for n-->oo.
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).
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).

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cp's OEIS Frontend

A067080 If n = ab...def in decimal notation then the left digitorial function Ld(n) = ab...def*ab...de*ab...d*...*ab*a.

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A098844 a(1)=1, a(n) = n*a(floor(n/2)).

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A132027 a(n) = Product_{k=0..floor(log_3(n))} floor(n/3^k), n>=1.

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A132272 a(n) = Product_{k>0} (1 + floor(n/10^k)).

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A132033 Product{0<=k<=floor(log_9(n)), floor(n/9^k)}, n>=1.

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A132328 a(n) = Product_{k>0} (1+floor(n/3^k)).

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A132263 Product{0<=k<=floor(log_11(n)), floor(n/11^k)}, n>=1.

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A067080 If n = ab...def in decimal notation then the left digitorial function Ld(n) = ab...defab...deab...d...ab*a.