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.

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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...
		

Crossrefs

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

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

Views

Author

Hieronymus Fischer, Aug 20 2007

Keywords

Examples

			0.47735217025489380198334286365820...
		

Crossrefs

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

A270743 Runlength sequence of the zero-one sequence A270742.

Original entry on oeis.org

1, 1, 2, 1, 2, 4, 1, 3, 6, 1, 4, 8, 1, 5, 10, 1, 6, 12, 1, 7, 14, 1, 8, 16, 1, 9, 18, 1, 10, 20, 1, 11, 22, 1, 12, 24, 1, 13, 26, 1, 14, 28, 1, 15, 30, 1, 16, 32, 1, 17, 34, 1, 18, 36, 1, 19, 38, 1, 20, 40, 1, 21, 42, 1, 22, 44, 1, 23, 46, 1, 24, 48, 1, 25
Offset: 1

Views

Author

Clark Kimberling, Apr 04 2016

Keywords

Comments

This is the runlength sequence for C = (1/2)*(3/4)*(7/8)*(15/16)... in each base (2,3,4,...); for base 10, we have C = 0.8900100999989990000001000..., as in A132038.

Examples

			(zero-one sequence at A270742) = (0, 1, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0,...), with first 6 runs 0, 1, 00, 1, 00, 1111 of lengths 1,1,2,1,2,4.
		

Crossrefs

Programs

  • Mathematica
    d = Join[{0}, RealDigits[QPochhammer[1/2], 2, 200000][[1]]];
    u = Map[Length, Split[d]]

Formula

a(3n) = 2n for n > 0, a(3n+1) = 1 for n >= 0, a(3n+2) = n + 1 for n >= 0.
Previous Showing 31-33 of 33 results.