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 28 results. Next

A048651 Decimal expansion of Product_{k >= 1} (1 - 1/2^k).

Original entry on oeis.org

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

Views

Author

Keywords

Comments

This is the limiting probability that a large random binary matrix is nonsingular (cf. A002884).
This constant is very close to 2^(13/24) * sqrt(Pi/log(2)) / exp(Pi^2/(6*log(2))) = 0.288788095086602421278899775042039398383022429351580356839... - Vaclav Kotesovec, Aug 21 2018
This constant is irrational (see Penn link). - Paolo Xausa, Dec 09 2024

Examples

			(1/2)*(3/4)*(7/8)*(15/16)*... = 0.288788095086602421278899721929230780088911904840685784114741...
		

References

  • Steven R. Finch, Mathematical Constants, Cambridge, 2003, pp. 318, 354-361.

Crossrefs

Programs

  • Mathematica
    RealDigits[ Product[1 - 1/2^i, {i, 100}], 10, 111][[1]] (* Robert G. Wilson v, May 25 2011 *)
    RealDigits[QPochhammer[1/2], 10, 100][[1]] (* Jean-François Alcover, Nov 18 2015 *)
  • PARI
    default(realprecision, 20080); x=prodinf(k=1, -1/2^k, 1); x*=10; for (n=0, 20000, d=floor(x); x=(x-d)*10; write("b048651.txt", n, " ", d)); \\ Harry J. Smith, May 07 2009

Formula

exp(-Sum_{k>0} sigma_1(k)/k*2^(-k)) = exp(-Sum_{k>0} A000203(k)/k*2^(-k)). - Hieronymus Fischer, Jul 28 2007
From Hieronymus Fischer, Aug 13 2007: (Start)
Equals lim inf Product_{k=0..floor(log_2(n))} floor(n/2^k)*2^k/n for n->oo.
Equals lim inf A098844(n)/n^(1+floor(log_2(n)))*2^(1/2*(1+floor(log_2(n)))*floor(log_2(n))) for n->oo.
Equals lim inf A098844(n)/n^(1+floor(log_2(n)))*2^A000217(floor(log_2(n))) for n->oo.
Equals lim inf A098844(n)/(n+1)^((1+log_2(n+1))/2) for n->oo.
Equals (1/2)*exp(-Sum_{n>0} 2^(-n)*Sum_{k|n} 1/(k*2^k)). (End)
Limit of A177510(n)/A000079(n-1) as n->infinity (conjecture). - Mats Granvik, Mar 27 2011
Product_{k >= 1} (1-1/2^k) = (1/2; 1/2){infinity}, where (a;q){infinity} is the q-Pochhammer symbol. - G. C. Greubel, Nov 27 2015
exp(Sum_{n>=1}(1/n/(1 - 2^n))) (according to Mathematica). - Mats Granvik, Sep 07 2016
(Sum_{k>0} (4^k-1)/(Product_{i=1..k} ((4^i-1)*(2*4^i-1))))*2 = 2/7 + 2/(3*7*31) + 2/(3*7*15*31*127)+2/(3*7*15*31*63*127*511) + ... (conjecture). - Werner Schulte, Dec 22 2016
Equals Sum_{k=-oo..oo} (-1)^k/2^((3*k+1)*k/2) (by Euler's pentagonal number theorem). - Amiram Eldar, Aug 13 2020
From Peter Bala, Dec 15 2020: (Start)
Constant C = Sum_{n >= 0} (-1)^n/( Product_{k = 1..n} (2^k - 1) ). The above conjectural result by Schulte follows by adding terms of this series in pairs.
C = (1/2)*Sum_{n >= 0} (-1/2)^n/( Product_{k = 1..n} (2^k - 1) ).
C = (3/8)*Sum_{n >= 0} (-1/4)^n/( Product_{k = 1..n} (2^k - 1) ).
1/C = Sum_{n >= 0} 2^(n*(n-1)/2)/( Product_{k = 1..n} (2^k - 1) ).
C = 1 - Sum_{n >= 0} (1/2)^(n+1)*Product_{k = 1..n} (1 - 1/2^k).
This latter identity generalizes as:
C = Sum_{n >= 0} (1/4)^(n+1)*Product_{k = 1..n} (1 - 1/2^k),
3*C = 1 - Sum_{n >= 0} (1/8)^(n+1)*Product_{k = 1..n} (1 - 1/2^k),
3*7*C = 6 + Sum_{n >= 0} (1/16)^(n+1)*Product_{k = 1..n} (1 - 1/2^k),
3*7*15*C = 91 - Sum_{n >= 0} (1/32)^(n+1)*Product_{k = 1..n} (1 - 1/2^k),
and so on, where the sequence [1, 0, 1, 6, 91, ...] is A005327.
(End)
From Amiram Eldar, Feb 19 2022: (Start)
Equals sqrt(2*Pi/log(2)) * exp(log(2)/24 - Pi^2/(6*log(2))) * Product_{k>=1} (1 - exp(-4*k*Pi^2/log(2))) (McIntosh, 1995).
Equals Sum_{n>=0} (-1)^n/A005329(n).
Equals exp(-A335764). (End)
Equals 1/A065446. - Hugo Pfoertner, Nov 23 2024

Extensions

Corrected by Hieronymus Fischer, Jul 28 2007

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

Original entry on oeis.org

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

Views

Author

Amarnath Murthy, Jan 05 2002

Keywords

Comments

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

Examples

			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.
		

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), A132263(p=11), A132264(p=12).
For the products of terms 1+floor(n/p^k) see A132269-A132272, A132327, A132328.

Programs

  • Haskell
    a067080 n = if n <= 9 then n else n * a067080 (n `div` 10)
    -- Reinhard Zumkeller, Nov 29 2012
  • Mathematica
    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 *)
  • PARI
    a(n)=my(t=n);while(n\=10,t*=n); t \\ Charles R Greathouse IV, Nov 20 2012
    

Formula

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 0
a(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)

Extensions

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

Author

Benoit Cloitre, Nov 03 2004

Keywords

Examples

			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.
		

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=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.

Programs

  • Mathematica
    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)))
    
  • Python
    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

Formula

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)

Extensions

Formula section edited by Hieronymus Fischer, Jun 13 2012

A132038 Decimal expansion of Product_{k>0} (1-1/10^k).

Original entry on oeis.org

8, 9, 0, 0, 1, 0, 0, 9, 9, 9, 9, 8, 9, 9, 9, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 9, 9, 9, 9, 9, 9, 9, 9, 8, 9, 9, 9, 9, 9, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 8, 9, 9, 9, 9, 9, 9, 9, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 9, 9, 9
Offset: 0

Author

Hieronymus Fischer, Aug 14 2007

Keywords

Examples

			0.8900100999989990000001000...
		

Programs

  • Mathematica
    digits = 105; Clear[p]; p[n_] := p[n] = RealDigits[Product[1-1/10^k , {k, 1, n}], 10, digits] // First; p[10]; p[n=20]; While[p[n] != p[n/2], n = 2*n]; p[n] (* Jean-François Alcover, Feb 17 2014 *)
    RealDigits[QPochhammer[1/10], 10, 105][[1]] (* Jean-François Alcover, Nov 18 2015 *)
    N[QPochhammer[1/10,1/10]] (* G. C. Greubel, Nov 30 2015 *)
  • PARI
    prodinf(x=1,-.1^x,1) \\ Charles R Greathouse IV, Nov 16 2013

Formula

Equals exp( -Sum_{n>0} sigma_1(n)/(n*10^n) ).
Equals (1/10; 1/10){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(10)) * exp(log(10)/24 - Pi^2/(6*log(10))) * Product_{k>=1} (1 - exp(-4*k*Pi^2/log(10))) (McIntosh, 1995).
Equals Sum_{n>=0} (-1)^n/A027878(n). (End)

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

Original entry on oeis.org

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

Author

Hieronymus Fischer, Aug 13 2007

Keywords

Examples

			0.3826631966790330232889550...
		

Programs

  • Mathematica
    digits = 103; NProduct[1-1/(2*3^k), {k, 0, Infinity}, NProductFactors -> 100, WorkingPrecision -> digits+3] // N[#, digits+3]& // RealDigits[#, 10, digits]& // First (* Jean-François Alcover, Feb 18 2014 *)
    RealDigits[QPochhammer[1/2, 1/3], 10, 120][[1]] (* Amiram Eldar, May 08 2023 *)

Formula

Equals lim inf_{n->oo} Product_{k=0..floor(log_3(n))} floor(n/3^k)*3^k/n.
Equals lim inf_{n->oo} A132027(n)/n^(1+floor(log_3(n)))*3^(1/2*(1+floor(log_3(n)))*floor(log_3(n))).
Equals lim inf_{n->oo} A132027(n)/n^(1+floor(log_3(n)))*3^A000217(floor(log_3(n))).
Equals (1/2)*exp(-Sum_{n>0} 3^(-n)*Sum_{k|n} 1/(k*2^k)).
Equals lim inf_{n->oo} A132027(n)/A132027(n+1).
Equals Product_{n>=1} (1 - 1/A025192(n)). - Amiram Eldar, May 08 2023

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

Author

Hieronymus Fischer, Aug 14 2007

Keywords

Examples

			0.805687728162164940923750...
		

Programs

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

Formula

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

Author

Hieronymus Fischer, Aug 14 2007

Keywords

Examples

			0.8765603540359642058360198...
		

Programs

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

Formula

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)

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

Author

Benoit Cloitre, Jan 25 2003

Keywords

Examples

			2.38423102903137172414989928867839723877161951650843345769...
		

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

Author

Benoit Cloitre, Apr 09 2003

Keywords

Comments

Twice the product in A079555.

Examples

			4.76846205806274344829979857....
		

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

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

Author

Hieronymus Fischer, Aug 14 2007

Keywords

Examples

			0.8594059944007028662007585800...
		

Programs

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

Formula

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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