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 15 results. Next

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

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

Views

Author

Hieronymus Fischer, Aug 20 2007

Keywords

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)).

Examples

			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.
		

Crossrefs

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.

Programs

  • 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 *)

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
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).

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

Author

Hieronymus Fischer, Aug 20 2007

Keywords

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)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)).

Examples

			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.
		

Crossrefs

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.

Programs

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

Formula

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).

A015502 a(1) = 1, a(n) = Sum_{k=1..n-1} (3^k - 1)/2 * a(k).

Original entry on oeis.org

1, 1, 5, 70, 2870, 350140, 127801100, 139814403400, 458731057555400, 4514831068460246800, 133300387296288786770000, 11806948504381482999365980000, 3137354163532752044074527571580000, 2500979519710095684958538548015855960000
Offset: 1

Keywords

Crossrefs

Sequences with the recurrence a(n) = (m^(n-1) + m-2)*a(n-1)/(m-1): A036442 (m=2), this sequence (m=3), A015503 (m=4), A015506 (m=5), A015507 (m=6), A015508 (m=7), A015509 (m=8), A015511 (m=9), A015512 (m=10), A015513 (m=11), A015515 (m=12).
Cf. A156296.

Programs

  • Magma
    [n le 2 select 1 else ((3^(n-1)+1)/2)*Self(n-1): n in [1..15]]; // Vincenzo Librandi, Nov 11 2012
    
  • Mathematica
    Flatten[{1, Table[QPochhammer[-1, 3, n]/2^(n+1), {n, 2, 15}]}] (* Vaclav Kotesovec, Mar 24 2017 *)
    a[n_, m_]:= a[n, m]= If[n<3, 1, (m^(n-1)+m-2)*a[n-1,m]/(m-1)];
    Table[a[n,3], {n,20}] (* G. C. Greubel, Apr 29 2023 *)
  • SageMath
    @CachedFunction # a = A015502
    def a(n,m): return 1 if (n<3) else (m^(n-1) + m-2)*a(n-1,m)/(m-1)
    [a(n,3) for n in range(1,31)] # G. C. Greubel, Apr 29 2023

Formula

a(n) = ((3^(n-1) + 1)/2) * a(n-1). - Vincenzo Librandi, Nov 11 2012
a(n) ~ c * 3^(n*(n-1)/2) / 2^(n+1), where c = A132323 = QPochhammer(-1, 1/3) = 3.129868... . - Vaclav Kotesovec, Mar 24 2017

A132265 Decimal expansion of Product_{k>=0} (1 - 1/(2*11^k)).

Original entry on oeis.org

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

Author

Hieronymus Fischer, Aug 20 2007

Keywords

Examples

			0.47510412750760310539756444...
		

Programs

  • Mathematica
    digits = 105; NProduct[1-1/(2*11^k), {k, 0, Infinity}, NProductFactors -> 100, WorkingPrecision -> digits+10] // N[#, digits+10]& // RealDigits[#, 10, digits]& // First (* Jean-François Alcover, Feb 18 2014 *)
    N[QPochhammer[1/2,1/11]] (* G. C. Greubel, Nov 30 2015 *)
  • PARI
    prodinf(x=0, 1 - 1/(2*11^x)) \\ Altug Alkan, Dec 01 2015

Formula

lim inf Product_{k=0..floor(log_11(n))} floor(n/11^k)*11^k/n for n-->oo.
lim inf A132263(n)*11^((1+floor(log_11(n)))*floor(log_11(n))/2)/n^(1+floor(log_11(n))) for n-->oo.
lim inf A132263(n)*11^A000217(floor(log_11(n)))/n^(1+floor(log_11(n))) for n-->oo.
(1/2)*exp(-Sum_{n>0} 11^(-n)*Sum_{k|n} 1/(k*2^k)).
lim inf A132263(n)/A132263(n+1) = 0.47510412750760310539756444... for n-->oo.
Equals (1/2; 1/11){infinity}, where (a;q){infinity} is the q-Pochhammer symbol. - G. C. Greubel, Nov 30 2015

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)

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)

A132325 Decimal expansion of Product_{k>=0} (1+1/10^k).

Original entry on oeis.org

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

Author

Hieronymus Fischer, Aug 20 2007

Keywords

Comments

Twice the constant A132326.

Examples

			2.22446913827410126425215613418881160749501...
		

Programs

  • Mathematica
    digits = 105; NProduct[1+1/10^k, {k, 1, Infinity}, NProductFactors -> 100, WorkingPrecision -> digits+3] // N[#, digits+3]& // RealDigits[#, 10, digits]& // First (* Jean-François Alcover, Feb 18 2014 *)
    2*N[QPochhammer[-1/10,1/10]] (* G. C. Greubel, Dec 02 2015 *)
  • PARI
    prodinf(x=0, 1+(1/10)^x) \\ Altug Alkan, Dec 03 2015

Formula

Equals lim sup_{n->oo} Product_{0<=k<=floor(log_10(n))} (1+1/floor(n/10^k)).
Equals lim sup_{n->oo} A132271(n)/n^((1+log_10(n))/2).
Equals lim sup_{n->oo} A132272(n)/n^((log_10(n)-1)/2).
Equals 2*exp(Sum_{n>0} 10^(-n)*Sum_{k|n} -(-1)^k/k) = 2*exp(Sum_{n>0} A000593(n)/(n*10^n)).
Equals lim sup_{n->oo} A132271(n+1)/A132271(n).
Equals 2*(-1/10; 1/10){infinity}, where (a;q){infinity} is the q-Pochhammer symbol. - G. C. Greubel, Dec 02 2015
Equals sqrt(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
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