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 1-4 of 4 results.

A078889 Decimal expansion of Sum {n>=0} 1/8^(2^n).

Original entry on oeis.org

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

Views

Author

Robert G. Wilson v, Dec 11 2002

Keywords

Examples

			0.140869200229648328104...

Links

Crossrefs

Cf. A165426.
Similar sums: A007404, A078885, A078585, A078886, A078887, A078888, A078890, A036987.

Programs

  • Mathematica
    RealDigits[ N[ Sum[1/8^(2^n), {n, 0, Infinity}], 110]][[1]]
  • PARI
    suminf(n=0, 1/8^(2^n)) \\ Michel Marcus, Nov 11 2020

Formula

Equals -Sum_{k>=1} mu(2*k)/(8^k - 1), where mu is the Möbius function (A008683). - Amiram Eldar, Jul 12 2020

A167182 a(0)=1, a(1)=2; for n>=2, a(n) = 2^A042950(n-2).

Original entry on oeis.org

1, 2, 4, 8, 64, 4096, 16777216, 281474976710656, 79228162514264337593543950336, 6277101735386680763835789423207666416102355444464034512896
Offset: 0

Views

Author

Giovanni Teofilatto, Oct 29 2009

Keywords

Comments

Term a(13) has 925 decimal digits; a(14) has 1850 decimal digits. - Michael De Vlieger, Jan 07 2015

Links

Crossrefs

Cf. A042950, A098011, A165426, A172489, A173260, A175129.

Programs

  • Mathematica
    Join[{2,4},NestList[#^2&,8,10]] (* Harvey P. Dale, Nov 30 2019 *)
Table[2^Ceiling[3 2^(i - 3)], {i, 8}] (* Trevor Cappallo, Apr 21 2021 *)

Formula

a(n) = (a(n-1))^2 for n > 3.
a(n) = 2^A098011(n+1). - R. J. Mathar, Apr 22 2010

Extensions

Definition corrected by R. J. Mathar, Apr 22 2010
More terms from Vincenzo Librandi, Apr 25 2010
Entry revised by N. J. A. Sloane, Jun 20 2021

A225160 Denominators of the sequence of fractions f(n) defined recursively by f(1) = 8/1; f(n+1) is chosen so that the sum and the product of the first n terms of the sequence are equal.

Original entry on oeis.org

1, 7, 57, 3697, 15302113, 258902783918017, 73384158961115901868286873473, 5848244449673109813614947741525727934929692392922517757697
Offset: 1

Views

Author

Martin Renner, Apr 30 2013

Keywords

Comments

Numerators of the sequence of fractions f(n) is A165426(n+1), hence sum(A165426(i+1)/a(i),i=1..n) = product(A165426(i+1)/a(i),i=1..n) = A165426(n+2)/A225167(n) = A167182(n+2)/A225167(n).

Examples

			f(n) = 8, 8/7, 64/57, 4096/3697, ...
8 + 8/7 = 8 * 8/7 = 64/7; 8 + 8/7 + 64/57 = 8 * 8/7 * 64/57 = 4096/399; ...

Crossrefs

Cf. A100441, A165426, A167182, A225167.

Programs

  • Maple
    b:=n->8^(2^(n-2)); # n > 1
b(1):=8;
p:=proc(n) option remember; p(n-1)*a(n-1); end;
p(1):=1;
a:=proc(n) option remember; b(n)-p(n); end;
a(1):=1;
seq(a(i),i=1..9);

Formula

a(n) = 8^(2^(n-2)) - product(a(i),i=1..n-1), n > 1 and a(1) = 1.
a(n) = 8^(2^(n-2)) - p(n) with a(1) = 1 and p(n) = p(n-1)*a(n-1) with p(1) = 1.

A225167 Denominators of the sequence s(n) of the sum resp. product of fractions f(n) defined recursively by f(1) = 8/1; f(n+1) is chosen so that the sum and the product of the first n terms of the sequence are equal.

Original entry on oeis.org

1, 7, 399, 1475103, 22572192792639, 5844003553148435725257076863, 428857285713570950220841681681938481172663051541516755199
Offset: 1

Views

Author

Martin Renner, Apr 30 2013

Keywords

Comments

Numerators of the sequence s(n) of the sum resp. product of fractions f(n) is A165426(n+2), hence sum(A165426(i+1)/A225160(i),i=1..n) = product(A165426(i+1)/A225160(i),i=1..n) = A165426(n+2)/a(n) = A167182(n+2)/a(n).

Examples

			f(n) = 8, 8/7, 64/57, 4096/3697, ...
8 + 8/7 = 8 * 8/7 = 64/7; 8 + 8/7 + 64/57 = 8 * 8/7 * 64/57 = 4096/399; ...
s(n) = 1/b(n) = 8, 64/7, 4096/399, ...

Crossrefs

Cf. A076628, A165426, A167182, A225160.

Programs

  • Maple
    b:=proc(n) option remember; b(n-1)-b(n-1)^2; end:
b(1):=1/8;
a:=n->8^(2^(n-1))*b(n);
seq(a(i),i=1..8);

Formula

a(n) = 8^(2^(n-1))*b(n) where b(n)=b(n-1)-b(n-1)^2 with b(1)=1/8.
Showing 1-4 of 4 results.

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