A067080 -id:A067080 - cp's OEIS frontend

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

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)).

A004128 a(n) = Sum_{k=1..n} floor(3*n/3^k).

Original entry on oeis.org

0, 1, 2, 4, 5, 6, 8, 9, 10, 13, 14, 15, 17, 18, 19, 21, 22, 23, 26, 27, 28, 30, 31, 32, 34, 35, 36, 40, 41, 42, 44, 45, 46, 48, 49, 50, 53, 54, 55, 57, 58, 59, 61, 62, 63, 66, 67, 68, 70, 71, 72, 74, 75, 76, 80, 81, 82, 84, 85, 86, 88, 89, 90, 93, 94, 95, 97, 98, 99, 101, 102

Offset: 0

N. J. A. Sloane

nonn

3-adic valuation of (3n)!; cf. A054861.

Denominators of expansion of (1-x)^{-1/3} are 3^a(n). Numerators are in |A067622|.

Gary W. Adamson, in "Beyond Measure, A Guided Tour Through Nature, Myth and Number", by Jay Kappraff, World Scientific, 2002, p. 356.

T. D. Noe, Table of n, a(n) for n = 0..1000
Vaclav Kotesovec, Graph - the asymptotic ratio (100000 terms)

Cf. A004117, A001511, A051064, A055457.

Cf. A054861, A067080, A098844, A132027, A005187, A054899.

a004128 n = a004128_list !! (n-1)
a004128_list = scanl (+) 0 a051064_list
-- Reinhard Zumkeller, May 23 2013

[n + Valuation(Factorial(n), 3): n in [0..70]]; // Vincenzo Librandi, Jun 12 2019

A004128 := proc(n)
    A054861(3*n) ;
end proc:
seq(A004128(n),n=0..100) ; # R. J. Mathar, Nov 04 2017

Table[Total[NestWhileList[Floor[#/3] &, n, # > 0 &]], {n, 0, 70}] (* Birkas Gyorgy, May 20 2012 *)
A004128 = Log[3, CoefficientList[ Series[1/(1+x)^(1/3), {x, 0, 100}], x] // Denominator] (* Jean-François Alcover, Feb 19 2015 *)
Flatten[{0, Accumulate[Table[IntegerExponent[3*n, 3], {n, 1, 100}]]}] (* Vaclav Kotesovec, Oct 17 2019 *)

{a(n) = my(s, t=1); while(t<=n, s += n\t; t*=3);s}; /* Michael Somos, Feb 26 2004 */

a(n) = (3*n-sumdigits(n,3))/2; \\ Christian Krause, Jun 10 2025

def A007949(n):
    c = 0
    while not (a:=divmod(n,3))[1]:
        c += 1
        n = a[0]
    return c
def A004128(n): return n+sum(A007949(i) for i in range(3,n+1)) # Chai Wah Wu, Feb 28 2025

A004128 = lambda n: A004128(n//3) + n if n > 0 else 0
[A004128(n) for n in (0..70)]  # Peter Luschny, Nov 16 2012

A051064(n) = a(n+1) - a(n). - Alford Arnold, Jul 19 2000
a(n) = n + floor(n/3) + floor(n/9) + floor(n/27) + ... = n + a(floor(n/3)) = n + A054861(n) = A054861(3n) = (3*n - A053735(n))/2. - Henry Bottomley, May 01 2001
a(n) = Sum_{k>=0} floor(n/3^k). a(n) = Sum_{k=0..floor(log_3(n))} floor(n/3^k), n >= 1. - Hieronymus Fischer, Aug 14 2007
Recurrence: a(n) = n + a(floor(n/3)); a(3n) = 3*n + a(n); a(n*3^m) = 3*n*(3^m-1)/2 + a(n). - Hieronymus Fischer, Aug 14 2007
a(k*3^m) = k*(3^(m+1)-1)/2, 0 <= k < 3, m >= 0. - Hieronymus Fischer, Aug 14 2007
Asymptotic behavior: a(n) = (3/2)*n + O(log(n)), a(n+1) - a(n) = O(log(n)); this follows from the inequalities below. - Hieronymus Fischer, Aug 14 2007
a(n) <= (3n-1)/2; equality holds for powers of 3. - Hieronymus Fischer, Aug 14 2007
a(n) >= (3n-2)/2 - floor(log_3(n)); equality holds for n = 3^m - 1, m > 0. - Hieronymus Fischer, Aug 14 2007
Lim inf (3n/2 - a(n)) = 1/2, for n->oo. - Hieronymus Fischer, Aug 14 2007
Lim sup (3n/2 - log_3(n) - a(n)) = 0, for n->oo. - Hieronymus Fischer, Aug 14 2007
Lim sup (a(n+1) - a(n) - log_3(n)) = 1, for n->oo. - Hieronymus Fischer, Aug 14 2007
G.f.: (Sum_{k>=0} x^(3^k)/(1-x^(3^k)))/(1-x). - Hieronymus Fischer, Aug 14 2007
a(n) = Sum_{k>=0} A030341(n,k)*A003462(k+1). - Philippe Deléham, Oct 21 2011
a(n) ~ 3*n/2 - log(n)/(2*log(3)). - Vaclav Kotesovec, Oct 17 2019

Current definition suggested by Jason Earls, Jul 04 2001

A132034 Decimal expansion of Product_{k>0} (1-1/6^k).

Original entry on oeis.org

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

Offset: 0

Hieronymus Fischer, Aug 14 2007

			0.805687728162164940923750...

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, A027873, A048651, A067080, A098844, A100220, A132022, A132026, A132038.

digits = 103; NProduct[1-1/6^k, {k, 1, Infinity}, NProductFactors -> 100, WorkingPrecision -> digits+20] // N[#, digits+20]& // RealDigits[#, 10, digits]& // First (* Jean-François Alcover, Feb 18 2014 *)
N[QPochhammer[1/6,1/6]] (* G. C. Greubel, Nov 30 2015 *)

prodinf(x=1, 1-(1/6)^x) \\ Altug Alkan, Dec 01 2015

Equals exp( -Sum_{n>0} sigma_1(n)/(n*6^n) ).
Equals (1/6; 1/6){infinity}, where (a;q){infinity} is the q-Pochhammer symbol. - G. C. Greubel, Nov 30 2015
From Amiram Eldar, May 09 2023: (Start)
Equals sqrt(2*Pi/log(6)) * exp(log(6)/24 - Pi^2/(6*log(6))) * Product_{k>=1} (1 - exp(-4*k*Pi^2/log(6))) (McIntosh, 1995).
Equals Sum_{n>=0} (-1)^n/A027873(n). (End)

A132037 Decimal expansion of Product_{k>0} (1-1/9^k).

Original entry on oeis.org

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

Offset: 0

Hieronymus Fischer, Aug 14 2007

			0.8765603540359642058360198...

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. A010815, A027877, A048651, A100220, A132025, A132026, A132038, A098844, A067080, A000203.

digits = 103; NProduct[1-1/9^k, {k, 1, Infinity}, NProductFactors -> 100, WorkingPrecision -> digits+3] // N[#, digits+3]& // RealDigits[#, 10, digits]& // First (* Jean-François Alcover, Feb 18 2014 *)
RealDigits[QPochhammer[1/9], 10, 100][[1]] (* Amiram Eldar, May 09 2023 *)

prodinf(k=1, 1 - 1/(9^k)) \\ Amiram Eldar, May 09 2023

Equals exp(-Sum_{n>0} sigma_1(n)/(n*9^n)) = exp(-Sum_{n>0} A000203(n)/(n*9^n)).
Equals Sum_{n>=0} A010815(n)/9^n. - R. J. Mathar, Mar 04 2009
From Amiram Eldar, May 09 2023: (Start)
Equals sqrt(Pi/log(3)) * exp(log(3)/12 - Pi^2/(12*log(3))) * Product_{k>=1} (1 - exp(-2*k*Pi^2/log(3))) (McIntosh, 1995).
Equals Sum_{n>=0} (-1)^n/A027877(n). (End)

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).

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)

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A132026 Decimal expansion of Product_{k>=0} (1 - 1/(2*10^k)).

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A132019 Decimal expansion of Product_{k>=0} 1-1/(2*3^k).

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

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A004128 a(n) = Sum_{k=1..n} floor(3*n/3^k).

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A132034 Decimal expansion of Product_{k>0} (1-1/6^k).

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A132037 Decimal expansion of Product_{k>0} (1-1/9^k).

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

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