cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 10 results.

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

Views

Author

Benoit Cloitre, Jan 25 2003

Keywords

Examples

			2.38423102903137172414989928867839723877161951650843345769...
		

Crossrefs

Programs

  • Mathematica
    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 *)
  • PARI
    prodinf(n=1,1+2.^-n) \\ Charles R Greathouse IV, May 27 2015
    
  • PARI
    1/prodinf(n=0, 1-2^(-2*n-1)) \\ Gleb Koloskov, Apr 04 2021

Formula

(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)

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

Views

Author

Benoit Cloitre, Apr 09 2003

Keywords

Comments

Twice the product in A079555.

Examples

			4.76846205806274344829979857....
		

Crossrefs

Programs

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

Formula

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

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

Views

Author

Hieronymus Fischer, Aug 20 2007

Keywords

Comments

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).

Examples

			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.
		

Crossrefs

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.

Formula

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).

A132264 Product{0<=k<=floor(log_12(n)), floor(n/12^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, 22, 23, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 108, 111, 114, 117, 120, 123, 126, 129, 132, 135, 138, 141, 192, 196, 200, 204, 208, 212, 216, 220, 224, 228, 232, 236, 300, 305, 310, 315
Offset: 1

Author

Hieronymus Fischer, Aug 20 2007

Keywords

Comments

If n is written in base-12 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).

Examples

			a(50)=floor(50/12^0)*floor(50/12^1)=50*4=200.
a(65)=325 since 65=55(base-12) and so a(65)=55*5(base-12)=65*5=325.
		

Crossrefs

For the product of terms floor(n/p^k) for p=2 to p=11 see A098844(p=2), A132027(p=3)-A132033(p=9), A067080(p=10), A132263(p=11).
For the products of terms 1+floor(n/p^k) see A132269-A132272, A132327, A132328.

Formula

The following formulas are given for a general parameter p considering the product of terms floor(n/p^k) for 0<=k<=floor(log_p(n)), where p=12 for this sequence.
Recurrence: a(n)=n*a(floor(n/p)); a(n*p^m)=n^m*p^(m(m+1)/2)*a(n).
a(k*p^m)=k^(m+1)*p^(m(m+1)/2), for 0
Asymptotic behavior: a(n)=O(n^((1+log_p(n))/2)); this follows from the inequalities below.
a(n)<=b(n), where b(n)=n^(1+floor(log_p(n)))/p^((1+floor(log_p(n)))*floor(log_p(n))/2); equality holds for n=k*p^m, 0=0. b(n) can also be written n^(1+floor(log_p(n)))/p^A000217(floor(log_p(n))).
Also: a(n)<=q^((1-log_p(q))/2)*n^((1+log_p(n))/2)=q^((1-log_p(q))/2)*p^A000217(log_p(n)), equality holds for n=q*p^m, m>=0, where q=floor(sqrt(p)+1/2). Also, equality holds for n=(q+1)*p^m, provided p is a A002378-number (in this case we have p=q*(q+1) and so q^((1-log_p(q))/2)=(q+1)^((1-log_p(q+1))/2)).
a(n)>c*b(n), where c=product{k>0, 1-1/(2*p^k)}=0.47735217025489380... (for p=12 see constant A132265).
Also: a(n)>c*(sqrt(2)/2^log_p(sqrt(2)))*n^((1+log_p(n))/2)=0.612870619...*p^A000217(log_p(n)), (p=12).
lim inf a(n)/b(n)=product{k>0, 1-1/(2*p^k)}=0.47735217025489380198334286365820..., for n-->oo (for p=12 see constant A132265).
lim sup a(n)/b(n)=1, for n-->oo.
lim inf a(n)/n^((1+log_p(n))/2)=(sqrt(2)/2^log_p(sqrt(2)))*product{k>0, 1-1/(2*p^k)}=0.612870619..., for n-->oo, (p=12).
lim sup a(n)/n^((1+log_p(n))/2)=sqrt(q)/q^log_p(sqrt(q))=1.358593737..., for n-->oo, (p=12, q=round(sqrt(p))=3).
lim inf a(n)/a(n+1)=product{k>0, 1-1/(2*p^k)}=0.47735217025489380... for n-->oo (for p=12 see constant A132265).

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

Author

Hieronymus Fischer, Aug 20 2007

Keywords

Comments

Twice the constant A132324.

Examples

			3.12986803713402307587769821345767...
		

Programs

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

Formula

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

A132268 Decimal expansion of Product_{k>0} (1-1/12^k).

Original entry on oeis.org

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

Author

Hieronymus Fischer, Aug 20 2007

Keywords

Examples

			0.9097262689059948886363620469770...
		

Programs

  • Mathematica
    digits = 106; NProduct[1-1/12^k, {k, 1, Infinity}, NProductFactors -> 100, WorkingPrecision -> digits+3] // N[#, digits+3]& // RealDigits[#, 10, digits]& // First (* Jean-François Alcover, Feb 18 2014 *)
    N[QPochhammer[1/12,1/12]] (* G. C. Greubel, Dec 05 2015 *)
  • PARI
    prodinf(k=1, (1-1/12^k)) \\ Michel Marcus, Dec 05 2015

Formula

Equals exp(-Sum_{n>0} sigma_1(n)/(n*12^n)) = exp(-Sum_{n>0} A000203(n)/(n*12^n)).
Equals (1/12; 1/12){infinity}, where (a;q){infinity} is the q-Pochhammer symbol. - G. C. Greubel, Dec 05 2015
From Amiram Eldar, May 09 2023: (Start)
Equals sqrt(2*Pi/log(12)) * exp(log(12)/24 - Pi^2/(6*log(12))) * Product_{k>=1} (1 - exp(-4*k*Pi^2/log(12))) (McIntosh, 1995).
Equals Sum_{n>=0} (-1)^n/A027880(n). (End)

A132326 Decimal expansion of Product_{k>=1} (1+1/10^k).

Original entry on oeis.org

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

Author

Hieronymus Fischer, Aug 20 2007

Keywords

Comments

Half the constant A132325.

Examples

			1.1122345691370506321260780670944...
		

Programs

  • Mathematica
    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 *)
    N[QPochhammer[-1/10,1/10]] (* G. C. Greubel, Dec 01 2015 *)
  • PARI
    prodinf(k=1, 1+1/10^k) \\ Amiram Eldar, May 20 2023

Formula

Equals (1/2)*lim sup_{n->oo} Product_{0<=k<=floor(log_10(n))} (1+1/floor(n/10^k)).
Equals (1/2)*lim sup_{n->oo} A132271(n)/n^((1+log_10(n))/2).
Equals (1/2)*lim sup_{n->oo} A132272(n)/n^((log_10(n)-1)/2).
Equals exp(Sum_{n>0} 10^(-n)*Sum_{k|n} -(-1)^k/k) = exp(Sum_{n>0} A000593(n)/(n*10^n)).
Equals (1/2)*lim sup_{n->oo} A132271(n+1)/A132271(n).
Equals (-1/10; 1/10){infinity}, where (a;q){infinity} is the q-Pochhammer symbol. - G. C. Greubel, Dec 01 2015
Equals (sqrt(2)/2) * exp(log(10)/24 + Pi^2/(12*log(10))) * Product_{k>=1} (1 - exp(-2*(2*k-1)*Pi^2/log(10))) (McIntosh, 1995). - Amiram Eldar, May 20 2023

A132267 Decimal expansion of Product_{k>0} (1-1/11^k).

Original entry on oeis.org

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

Author

Hieronymus Fischer, Aug 20 2007

Keywords

Examples

			0.900832706809715279949862694760...
		

Programs

  • Mathematica
    digits = 105; NProduct[1-1/11^k, {k, 1, Infinity}, NProductFactors -> 100, WorkingPrecision -> digits+3] // N[#, digits+3]& // RealDigits[#, 10, digits]& // First (* Jean-François Alcover, Feb 18 2014 *)
    N[QPochhammer[1/11, 1/11], 200] (* G. C. Greubel, Dec 20 2015 *)
  • PARI
    prodinf(x=1, 1-1/11^x) \\ Altug Alkan, Dec 20 2015

Formula

Equals exp(-Sum_{n>0} sigma_1(n)/(n*11^n)) = exp(-Sum_{n>0} A000203(n)/(n*11^n)).
Equals (1/11; 1/11){infinity}, where (a;q){infinity} is the q-Pochhammer symbol. - G. C. Greubel, Dec 20 2015
From Amiram Eldar, May 09 2023: (Start)
Equals sqrt(2*Pi/log(11)) * exp(log(11)/24 - Pi^2/(6*log(11))) * Product_{k>=1} (1 - exp(-4*k*Pi^2/log(11))) (McIntosh, 1995).
Equals Sum_{n>=0} (-1)^n/A027879(n). (End)

A132324 Decimal expansion of Product_{k>=1} (1+1/3^k).

Original entry on oeis.org

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

Author

Hieronymus Fischer, Aug 20 2007

Keywords

Comments

Half the constant A132323.

Examples

			1.56493401856701153793884910...
		

Programs

  • Mathematica
    digits = 105; NProduct[1+1/3^k, {k, 1, Infinity}, NProductFactors -> 100, WorkingPrecision -> digits+5] // N[#, digits+5]& // RealDigits[#, 10, digits]& // First (* Jean-François Alcover, Feb 18 2014 *)
    N[QPochhammer[-1/3,1/3]] (* G. C. Greubel, Dec 01 2015 *)

Formula

(1/2)*lim sup Product{k=0..floor(log_3(n))} (1+1/floor(n/3^k)) for n-->oo.
(1/2)*lim sup A132327(n)/A132027(n) for n-->oo.
(1/2)*lim sup A132327(n)/n^((1+log_3(n))/2) for n-->oo.
(1/2)*lim sup A132328(n)/n^((log_3(n)-1)/2) for n-->oo.
exp(Sum_{n>0} 3^(-n)*Sum_{k|n} -(-1)^k/k) = exp(Sum_{n>0} A000593(n)/(n*3^n)).
(1/2)*lim sup A132327(n+1)/A132327(n) = 1.56493401856701153793884910... for n-->oo.
Equals (-1/3; 1/3){infinity}, where (a;q){infinity} is the q-Pochhammer symbol. - G. C. Greubel, Dec 01 2015
From Amiram Eldar, Feb 19 2022: (Start)
Equals (sqrt(2)/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).
Equals Sum_{n>=0} 1/A027871(n). (End)

A132266 Decimal expansion of Product_{k>=0} (1 - 1/(2*12^k)).

Original entry on oeis.org

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

Author

Hieronymus Fischer, Aug 20 2007

Keywords

Examples

			0.47735217025489380198334286365820...
		

Programs

  • Mathematica
    digits = 105; NProduct[1-1/(2*12^k), {k, 0, Infinity}, NProductFactors -> 100, WorkingPrecision -> digits+5] // N[#, digits+5]& // RealDigits[#, 10, digits]& // First (* Jean-François Alcover, Feb 18 2014 *)
    (1/2)*N[QPochhammer[1/24, 1/12], 200] (* G. C. Greubel, Dec 20 2015 *)
  • PARI
    prodinf(x=0, 1-1/(2*12^x)) \\ Altug Alkan, Dec 20 2015

Formula

lim inf (Product_{k=0..floor(log_12(n))} floor(n/12^k)*12^k/n) for n-->oo.
lim inf A132264(n)*12^((1+floor(log_12(n)))*floor(log_12(n))/2)/n^(1+floor(log_12(n))) for n-->oo.
lim inf A132264(n)*12^A000217(floor(log_12(n)))/n^(1+floor(log_12(n))) for n-->oo.
(1/2)*exp(-Sum_{n>0} 12^(-n)*Sum_{k|n} 1/(k*2^k)).
lim inf A132264(n)/A132264(n+1) = 0.47735217025489380... for n-->oo.
Equals (1/2)*(1/24; 1/12){infinity}, where (a;q){infinity} is the q-Pochhammer symbol. - G. C. Greubel, Dec 20 2015
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