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.

Previous Showing 11-20 of 26 results. Next

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

Cf. A048651, A132019-A132026, A132265, A132267, 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)-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

Views

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

Cf. A048651, A132019-A132026, A132265, A132267, A000217, A002378.
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

Views

Author

Hieronymus Fischer, Aug 20 2007

Keywords

Comments

Twice the constant A132324.

Examples

			3.12986803713402307587769821345767...

Links

Crossrefs

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

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

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

Views

Author

Hieronymus Fischer, Aug 20 2007

Keywords

Examples

			0.47510412750760310539756444...

Links

Crossrefs

Cf. A000217, A132019-A132026, A132034-A132038, A132263, A132266-A132268, A132323-A132326.

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

Views

Author

Hieronymus Fischer, Aug 20 2007

Keywords

Examples

			0.9097262689059948886363620469770...

Links

Crossrefs

Cf. A027880, A048651, A100220, A132019-A132026, A132034-A132038, A132265-A132267, A132323-A132326, A000203.

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

Views

Author

Hieronymus Fischer, Aug 20 2007

Keywords

Comments

Half the constant A132325.

Examples

			1.1122345691370506321260780670944...

Links

Crossrefs

Cf. A000593, A067080, A079555, A132019-A132026, A132034-A132038, A132265-A132268, A132271, A132272, A132325.

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

A132020 Decimal expansion of Product_{k>=0} (1 - 1/(2*4^k)).

Original entry on oeis.org

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

Views

Author

Hieronymus Fischer, Aug 14 2007

Keywords

Comments

This is the limiting probability that a large random symmetric binary matrix is nonsingular (cf. A086812, A048651). In other words, equals Lim_{n->oo} A086812(n)/A006125(n+1).- H. Tracy Hall, Sep 07 2024

Examples

			0.41942244179510759770995610770297425223395323439266674908044991663177...

Links

Crossrefs

Cf. A004171, A048651, A098844, A067080, A132019, A132026, A132028, A100221, A000217, A081845.

Programs

  • Maple
    evalf(1+sum((-1)^n*2^(n*(n-1)/2)/product(2^k-1, k=1..n), n=1..infinity), 120); # Robert FERREOL, Feb 23 2020
  • Mathematica
    RealDigits[ Product[1 - 1/(2*4^i), {i, 0, 175}], 10, 111][[1]] (* Robert G. Wilson v, May 25 2011 *)
RealDigits[QPochhammer[1/2, 1/4], 10, 105][[1]] (* Jean-François Alcover, Nov 18 2015 *)
  • PARI
    prodinf(k=0,1-1.>>(2*k+1)) \\ Charles R Greathouse IV, Nov 16 2012

Formula

Equals lim inf_{n->oo} Product_{k=0..floor(log_4(n))} floor(n/4^k)*4^k/n.
Equals lim inf_{n->oo} A132028(n)/n^(1+floor(log_4(n)))*4^((1/2)*(1+floor(log_4(n)))*floor(log_4(n))).
Equals lim inf_{n->oo} A132028(n)/n^(1+floor(log_4(n)))*4^A000217(floor(log_4(n))).
Equals (1/2)*exp(-Sum_{n>0} (4^(-n)*(Sum_{k|n} 1/(k*2^k)))).
Equals lim inf_{n->oo} A132028(n)/A132028(n+1).
Equals Product_{k>0} (1-1/(2^k+1)). - Robert G. Wilson v, May 25 2011
From Robert FERREOL, Feb 23 2020: (Start)
Equals Product_{k>0} (1 + 1/2^k)^(-1) = 2/A081845.
Equals 1 + Sum_{n>=1} (-1)^n*2^(n*(n-1)/2)/((2-1)*(2^2-1)*...*(2^n-1)). (End)
From Peter Bala, Jan 15 2021: (Start)
Constant C = Sum_{n >= 0} 2^n/Product_{k = 1..n} (1 - 4^k).
Faster converging series:
2*C = (1/2)*Sum_{n >= 0} 2^(-n)/Product_{k = 1..n} (1 - 4^k);
(2^4)*C = 7*Sum_{n >= 0} 2^(-3*n)/Product_{k = 1..n} (1 - 4^k);
(2^9)*C = 7*31*Sum_{n >= 0} 2^(-5*n)/Product_{k = 1..n} (1 - 4^k), and so on.
Slower converging series:
C = -Sum_{n >= 0} 2^(3*n)/Product_{k = 1..n} (1 - 4^k);
7*C = Sum_{n >= 0} 2^(5*n)/Product_{k = 1..n} (1 - 4^k);
7*31*C = -Sum_{n >= 0} 2^(7*n)/Product_{k = 1..n} (1 - 4^k), and so on. (End)
Equals Product_{n>=0} (1 - 1/A004171(n)). - Amiram Eldar, May 09 2023

Extensions

Name corrected by Charles R Greathouse IV, Nov 16 2012

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

Views

Author

Hieronymus Fischer, Aug 20 2007

Keywords

Examples

			0.900832706809715279949862694760...

Links

Crossrefs

Cf. A000203, A027879, A048651, A100220, A132019-A132026, A132034-A132038, A132265-A132268, A132323, A132324, A132325, A132326.

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

Views

Author

Hieronymus Fischer, Aug 20 2007

Keywords

Comments

Half the constant A132323.

Examples

			1.56493401856701153793884910...

Links

Crossrefs

Cf. A027871, A079555, A100220, A132019-A132026, A132034-A132038, A132265-A132268, A132323-A132326, A132327, A132328, A000593.

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

Views

Author

Hieronymus Fischer, Aug 20 2007

Keywords

Comments

Twice the constant A132326.

Examples

			2.22446913827410126425215613418881160749501...

Links

Crossrefs

Cf. A081845, A067080, A100220, A132019-A132026, A132034-A132038, A132323-A132326, A132271, A132272, A000593.

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
Previous Showing 11-20 of 26 results. Next

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.