A132038 -id:A132038 - cp's OEIS frontend

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).

A079555 Decimal expansion of Product_{k>=1} (1 + 1/2^k) = 2.384231029031371...

Original entry on oeis.org

2, 3, 8, 4, 2, 3, 1, 0, 2, 9, 0, 3, 1, 3, 7, 1, 7, 2, 4, 1, 4, 9, 8, 9, 9, 2, 8, 8, 6, 7, 8, 3, 9, 7, 2, 3, 8, 7, 7, 1, 6, 1, 9, 5, 1, 6, 5, 0, 8, 4, 3, 3, 4, 5, 7, 6, 9, 2, 1, 0, 1, 5, 0, 7, 9, 8, 9, 1, 8, 1, 2, 9, 3, 0, 3, 6, 0, 3, 7, 2, 5, 5, 1, 8, 6, 5, 3, 5, 2, 1, 0, 3, 6, 5, 6, 8, 0, 5, 2, 0, 0, 0, 2, 6, 8

Offset: 1

Benoit Cloitre, Jan 25 2003

			2.38423102903137172414989928867839723877161951650843345769...

G. C. Greubel, Table of n, a(n) for n = 1..1200
Richard J. McIntosh, Some Asymptotic Formulae for q-Hypergeometric Series, Journal of the London Mathematical Society, Vol. 51, No. 1 (1995), pp. 120-136; alternative link.

Cf. A028362, A048651, A081845, A098844, A100220, A132019-A132026, A132034-A132038, A132265-A132268, A132323-A132326, A132269, A132270, A000593.

Cf. A141848. - Gleb Koloskov, Apr 04 2021

digits = 105; NProduct[(1 + 1/2^k), {k, 1, Infinity}, WorkingPrecision -> digits+10, NProductFactors -> 200] // RealDigits[#, 10, digits]& // First (* Jean-François Alcover, Feb 14 2013 *)
N[QPochhammer[-1/2,1/2]] (* G. C. Greubel, Dec 05 2015 *)
1/N[QPochhammer[1/2, 1/4]] (* Gleb Koloskov, Apr 04 2021 *)

prodinf(n=1,1+2.^-n) \\ Charles R Greathouse IV, May 27 2015

1/prodinf(n=0, 1-2^(-2*n-1)) \\ Gleb Koloskov, Apr 04 2021

(1/2)*lim sup Product_{k=0..floor(log_2(n)), (1 + 1/floor(n/2^k))} for n-->oo. - Hieronymus Fischer, Aug 20 2007
(1/2)*lim sup A132369(n)/A098844(n) for n-->oo. - Hieronymus Fischer, Aug 20 2007
(1/2)*lim sup A132269(n)/n^((1+log_2(n))/2) for n-->oo. - Hieronymus Fischer, Aug 20 2007
(1/2)*lim sup A132270(n)/n^((log_2(n)-1)/2) for n-->oo. - Hieronymus Fischer, Aug 20 2007
exp(sum{n>0, 2^(-n)*sum{k|n, -(-1)^k/k}})=exp(sum{n>0, A000593(n)/(n*2^n)}). - Hieronymus Fischer, Aug 20 2007
(1/2)*lim sup A132269(n+1)/A132269(n)=2.3842310290313717241498992886... for n-->oo. - Hieronymus Fischer, Aug 20 2007
Equals (-1/2; 1/2){infinity}, where (a;q){infinity} is the q-Pochhammer symbol. - G. C. Greubel, Dec 05 2015
2 + Sum_{k>1} 1/(Product_{i=2..k} (2^i-1)) = 2 + 1/3 + 1/(3*7) + 1/(3*7*15) + 1/(3*7*15*31) + 1/(3*7*15*31*63) + ... (conjecture). - Werner Schulte, Dec 22 2016
From Peter Bala, Dec 15 2020: (Start)
The above conjecture of Schulte follows by setting x = 1/2 and t = -1 in the identity Product_{k >= 1} (1 - t*x^k) = Sum_{n >= 0} (-1)^n*x^(n*(n+1)/2)*t^n/( Product_{k = 1..n} 1 - x^k ), due to Euler.
Constant C = 1 + Sum_{n >= 0} (1/2)^(n+1)*Product_{k = 1..n} (1 + 1/2^k).
C = 2 + Sum_{n >= 0} (1/4)^(n+1)*Product_{k = 1..n} (1 + 1/2^k).
3*C = 7 + Sum_{n >= 0} (1/8)^(n+1)*Product_{k = 1..n} (1 + 1/2^k).
3*7*C = 50 + Sum_{n >= 0} (1/16)^(n+1)*Product_{k = 1..n} (1 + 1/2^k).
3*7*15*C = 751 + Sum_{n >= 0} (1/32)^(n+1)*Product_{k = 1..n} (1 + 1/2^k).
(End)
Equals 1/(1-P), where P is the Pell constant from A141848. - Gleb Koloskov, Apr 04 2021
Equals Sum_{k>=0} A000009(k)/2^k. - Vaclav Kotesovec, Sep 15 2021
From Amiram Eldar, Feb 19 2022: (Start)
Equals (sqrt(2)/2) * exp(log(2)/24 + Pi^2/(12*log(2))) * Product_{k>=1} (1 - exp(-2*(2*k-1)*Pi^2/log(2))) (McIntosh, 1995).
Equals (1/2) * A081845.
Equals Sum_{n>=0} 1/A005329(n). (End)

A100222 Decimal expansion of Product_{k>=1} (1-1/5^k).

Original entry on oeis.org

7, 6, 0, 3, 3, 2, 7, 9, 5, 8, 7, 1, 2, 3, 2, 4, 2, 0, 1, 0, 1, 4, 8, 8, 2, 9, 6, 2, 9, 2, 6, 6, 5, 1, 5, 9, 4, 7, 4, 3, 4, 3, 9, 2, 8, 8, 7, 3, 2, 0, 5, 7, 9, 5, 1, 9, 8, 7, 7, 0, 9, 8, 4, 4, 0, 0, 8, 8, 8, 8, 5, 9, 9, 5, 3, 7, 5, 5, 2, 3, 3, 6, 5, 2, 7, 5, 1, 5, 3, 4, 0, 8, 6, 6, 1, 4, 3, 2, 3, 2, 5, 6

Offset: 0

Eric W. Weisstein, Nov 09 2004

			0.76033279587123242010148829629266515947434392887320...

G. C. Greubel, Table of n, a(n) for n = 0..1200
Richard J. McIntosh, Some Asymptotic Formulae for q-Hypergeometric Series, Journal of the London Mathematical Society, Vol. 51, No. 1 (1995), pp. 120-136; alternative link.
Eric Weisstein's World of Mathematics, Infinite Product.

Cf. A027872, A048651, A100220, A100221.

Cf. A000203, A100220, A100221, A132020, A132034, A132035, A132036, A132037, A132038, A258460.

(5^(1/24)*EllipticThetaPrime[1, 0, 1/Sqrt[5]]^(1/3))/2^(1/3)
N[QPochhammer[1/5,1/5]] (* G. C. Greubel, Dec 01 2015 *)

prodinf(k=1, 1 - 1/(5^k)) \\ Amiram Eldar, May 09 2023

Equals exp(-Sum_{k>0} sigma_1(k)/(k*5^k)) = exp(-Sum_{k>0} A000203(k)/(k*5^k)). - Hieronymus Fischer, Aug 07 2007
Equals (1/5; 1/5){infinity}, where (a;q){infinity} is the q-Pochhammer symbol. - G. C. Greubel, Dec 01 2015
From Amiram Eldar, May 09 2023: (Start)
Equals sqrt(2*Pi/log(5)) * exp(log(5)/24 - Pi^2/(6*log(5))) * Product_{k>=1} (1 - exp(-4*k*Pi^2/log(5))) (McIntosh, 1995).
Equals Sum_{n>=0} (-1)^n/A027872(n). (End)

A081845 Decimal expansion of Product_{k>=0} (1 + 1/2^k).

Original entry on oeis.org

4, 7, 6, 8, 4, 6, 2, 0, 5, 8, 0, 6, 2, 7, 4, 3, 4, 4, 8, 2, 9, 9, 7, 9, 8, 5, 7, 7, 3, 5, 6, 7, 9, 4, 4, 7, 7, 5, 4, 3, 2, 3, 9, 0, 3, 3, 0, 1, 6, 8, 6, 6, 9, 1, 5, 3, 8, 4, 2, 0, 3, 0, 1, 5, 9, 7, 8, 3, 6, 2, 5, 8, 6, 0, 7, 2, 0, 7, 4, 5, 1, 0, 3, 7, 3, 0, 7, 0, 4, 2, 0, 7, 3, 1, 3, 6, 1, 0, 4, 0, 0, 0, 5, 3, 7

Offset: 1

Benoit Cloitre, Apr 09 2003

Twice the product in A079555.

			4.76846205806274344829979857....

G. C. Greubel, Table of n, a(n) for n = 1..2000
Richard J. McIntosh, Some Asymptotic Formulae for q-Hypergeometric Series, Journal of the London Mathematical Society, Vol. 51, No. 1 (1995), pp. 120-136; citeseerx.
József Sándor, On Vandiver's arithmetical function - I, Notes on Number Theory and Discrete Mathematics, Vol. 27, No. 3 (2021), pp. 29-38. See p. 36, eq. 40.

Cf. A006125, A020696, A028361, A079555, A083864.

Cf. A000593, A048651, A100220, A098844, A132019-A132026, A132034-A132038, A132265-A132268, A132323-A132326, A132269, A132270.

digits = 105; NProduct[1 + 1/2^k, {k, 0, Infinity}, WorkingPrecision -> digits+5, NProductFactors -> digits] // RealDigits[#, 10, digits]& // First (* Jean-François Alcover, Mar 04 2013 *)
N[QPochhammer[-1, 1/2], 100] (* Vaclav Kotesovec, Dec 13 2015 *)
2*N[QPochhammer[-1/2, 1/2], 200] (* G. C. Greubel, Dec 20 2015 *)

prodinf(k=0,1/2^k,1) \\ Hugo Pfoertner, Feb 21 2020

lim sup Product_{k=0..floor(log_2(n))} (1 + 1/floor(n/2^k)) for n-->oo. - Hieronymus Fischer, Aug 20 2007
lim sup A132369(n)/A098844(n) for n-->oo. - Hieronymus Fischer, Aug 20 2007
lim sup A132269(n)/n^((1+log_2(n))/2) for n-->oo. - Hieronymus Fischer, Aug 20 2007
lim sup A132270(n)/n^((log_2(n)-1)/2) for n-->oo. - Hieronymus Fischer, Aug 20 2007
2*exp(Sum_{n>0} 2^(-n)*Sum_{k|n} -(-1)^k/k) = 2*exp(Sum_{n>0} A000593(n)/(n*2^n)). - Hieronymus Fischer, Aug 20 2007
lim sup A132269(n+1)/A132269(n) = 4.76846205806274344... for n-->oo. - Hieronymus Fischer, Aug 20 2007
Sum_{k>=1} (-1)^(k+1) * 2^k / (k*(2^k-1)) = log(A081845) = 1.562023833218500307570359922772014353168080202860122... . - Vaclav Kotesovec, Dec 13 2015
Equals 2*(-1/2; 1/2){infinity}, where (a;q){infinity} is the q-Pochhammer symbol. - G. C. Greubel, Dec 20 2015
Equals 1 + Sum_{n>=1} 2^n/((2-1)*(2^2-1)*...*(2^n-1)). - Robert FERREOL, Feb 21 2020
From Peter Bala, Jan 18 2021: (Start)
Constant C = 3*Sum_{n >= 0} (1/2)^n/Product_{k = 1..n} (2^k - 1).
Faster converging series:
C = (2*3*5)/(2^3)*Sum_{n >= 0} (1/4)^n/Product_{k = 1..n} (2^k - 1),
C = (2*3*5*9)/(2^6)*Sum_{n >= 0} (1/8)^n/Product_{k = 1..n} (2^k - 1),
C = (2*3*5*9*17)/(2^10)*Sum_{n >= 0} (1/16)^n/Product_{k = 1..n} (2^k - 1), and so on. The sequence [2,3,5,9,17,...] is A000051. (End)
From Amiram Eldar, Mar 20 2022: (Start)
Equals sqrt(2) * exp(log(2)/24 + Pi^2/(12*log(2))) * Product_{k>=1} (1 - exp(-2*(2*k-1)*Pi^2/log(2))) (McIntosh, 1995).
Equals 1/A083864. (End)
Equals lim_{n->oo} A020696(2^n)/A006125(n+1) (Sándor, 2021). - Amiram Eldar, Jun 29 2022

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).

A132036 Decimal expansion of Product_{k>0} (1 - 1/8^k).

Original entry on oeis.org

8, 5, 9, 4, 0, 5, 9, 9, 4, 4, 0, 0, 7, 0, 2, 8, 6, 6, 2, 0, 0, 7, 5, 8, 5, 8, 0, 0, 6, 4, 4, 1, 8, 8, 9, 4, 9, 0, 9, 4, 8, 4, 9, 7, 9, 5, 8, 8, 0, 4, 0, 9, 1, 7, 7, 4, 2, 4, 6, 9, 8, 8, 5, 8, 3, 1, 0, 0, 1, 3, 2, 3, 0, 2, 2, 9, 0, 2, 3, 9, 6, 5, 5, 2, 3, 6, 8, 9, 6, 5, 3, 7, 4, 9, 8, 3, 5, 3, 1, 4, 1, 3, 9

Offset: 0

Hieronymus Fischer, Aug 14 2007

			0.8594059944007028662007585800...

G. C. Greubel, Table of n, a(n) for n = 0..1200
Richard J. McIntosh, Some Asymptotic Formulae for q-Hypergeometric Series, Journal of the London Mathematical Society, Vol. 51, No. 1 (1995), pp. 120-136; alternative link.

Cf. A000203, A027876, A048651, A067080, A098844, A100220, A132024, A132026, A132038.

digits = 103; NProduct[1-1/8^k, {k, 1, Infinity}, NProductFactors -> 100, WorkingPrecision -> digits+3] // N[#, digits+3]& // RealDigits[#, 10, digits]& // First (* Jean-François Alcover, Feb 18 2014 *)
N[QPochhammer[1/8,1/8]] (* G. C. Greubel, Nov 26 2015 *)

prodinf(x=1, 1-(1/8)^x) \\ Altug Alkan, Dec 01 2015

Equals exp(-Sum_{n>0} sigma_1(n)/n*(1/8)^n) where sigma_1() is A000203().
Equals (1/8; 1/8){infinity}, where (a;q){infinity} is the q-Pochhammer symbol. - G. C. Greubel, Nov 26 2015
From Amiram Eldar, May 09 2023: (Start)
Equals sqrt(2*Pi/(3*log(2))) * exp(log(2)/8 - Pi^2/(18*log(2))) * Product_{k>=1} (1 - exp(-4*k*Pi^2/(3*log(2)))) (McIntosh, 1995).
Equals Sum_{n>=0} (-1)^n/A027876(n). (End)

A132323 Decimal expansion of Product_{k>=0} (1+1/3^k).

Original entry on oeis.org

3, 1, 2, 9, 8, 6, 8, 0, 3, 7, 1, 3, 4, 0, 2, 3, 0, 7, 5, 8, 7, 7, 6, 9, 8, 2, 1, 3, 4, 5, 7, 6, 7, 0, 8, 3, 3, 1, 3, 8, 8, 5, 1, 8, 3, 9, 7, 9, 0, 0, 7, 0, 0, 1, 8, 9, 9, 3, 4, 4, 2, 0, 5, 9, 8, 4, 6, 0, 4, 2, 2, 1, 4, 5, 1, 6, 1, 9, 3, 5, 3, 3, 8, 7, 8, 0, 7, 3, 2, 0, 7, 3, 5, 4, 5, 9, 2, 7, 7, 6, 3, 0, 5, 2, 0

Offset: 1

Hieronymus Fischer, Aug 20 2007

Twice the constant A132324.

			3.12986803713402307587769821345767...

G. C. Greubel, Table of n, a(n) for n = 1..1200
Richard J. McIntosh, Some Asymptotic Formulae for q-Hypergeometric Series, Journal of the London Mathematical Society, Vol. 51, No. 1 (1995), pp. 120-136; alternative link.

Cf. A081845, A100220, A132019-A132026, A132034-A132038, A132265-A132268, A132324-A132326, A132327, A132328, A000593.

digits = 105; NProduct[1+1/3^k, {k, 0, Infinity}, NProductFactors -> 100, WorkingPrecision -> digits+3] // N[#, digits+3]& // RealDigits[#, 10, digits]& // First (* Jean-François Alcover, Feb 18 2014 *)
2*N[QPochhammer[-1/3,1/3]] (* G. C. Greubel, Dec 01 2015 *)

prodinf(x=0, 1+(1/3)^x) \\ Altug Alkan, Dec 03 2015

Equals lim sup_{n->oo} Product_{0<=k<=floor(log_3(n))} (1+1/floor(n/3^k)).
Equals lim sup_{n->oo} A132327(n)/A132027(n).
Equals lim sup_{n->oo} A132327(n)/n^((1+log_3(n))/2).
Equals lim sup_{n->oo} A132328(n)/n^((log_3(n)-1)/2).
Equals 2*exp(Sum_{n>0} 3^(-n) * Sum{k|n} -(-1)^k/k) = 2*exp(Sum_{n>0} A000593(n)/(n*3^n)).
Equals lim sup_{n->oo} A132327(n+1)/A132327(n).
Equals 2*(-1/3; 1/3){infinity}, where (a;q){infinity} is the q-Pochhammer symbol. - G. C. Greubel, Dec 01 2015
Equals sqrt(2) * exp(log(3)/24 + Pi^2/(12*log(3))) * Product_{k>=1} (1 - exp(-2*(2*k-1)*Pi^2/log(3))) (McIntosh, 1995). - Amiram Eldar, May 25 2023

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).

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