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 1-2 of 2 results.

A220395 A modified Engel expansion of log(2).

Original entry on oeis.org

2, 3, 8, 6, 2, 4, 93, 60, 2, 2, 2, 2, 3, 12, 10, 2, 2, 14, 52, 6, 5, 8, 2, 2, 5, 8, 2, 2, 3, 4, 14, 273, 40, 2, 3, 4, 4, 12, 27, 16, 14, 26, 4, 6, 4, 6, 2, 3, 12, 10, 4, 6, 14, 65, 12, 8, 6, 2, 7, 90, 294, 40, 2, 2, 32, 155, 8, 7, 12, 2, 2, 2, 2, 4, 6, 3, 10
Offset: 1

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Author

Peter Bala, Dec 13 2012

Keywords

Comments

See A220393 for the definition of the modified Engel expansion of a positive real number. For further details see the Bala link.

Crossrefs

Formula

Let h(x) = x*(floor(1/x) + (floor(1/x))^2) - floor(1/x). Let x = log(2). Then a(1) = 1 + floor(1/x) and, for n >= 1, a(n+1) = floor(1/h^(n-1)(x))*(1 + floor(1/h^(n)(x))).
Put P(n) = Product_{k = 1..n} a(k). Then we have the Egyptian fraction series expansion log(2) = Sum_{n>=1} 1/P(n) = 1/2 + 1/(2*3) + 1/(2*3*8) + 1/(2*3*8*6) + 1/(2*3*8*6*2) + .... The error made in truncating this series to n terms is less than the n-th term.

A091846 Pierce expansion of log(2).

Original entry on oeis.org

1, 3, 12, 21, 51, 57, 73, 85, 96, 1388, 4117, 5268, 9842, 11850, 16192, 19667, 29713, 76283, 460550, 1333597, 1462506, 9400189, 13097390, 30254851, 190193800, 201892756, 431766247, 942050077, 6204785761, 16684400052, 23762490104
Offset: 1

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Author

Benoit Cloitre, Mar 09 2004

Keywords

Comments

If u(0)=exp(1/m) m integer>=1 and u(n+1)=u(n)/frac(u(n)) then floor(u(n))=m*n.

Crossrefs

Cf. A006784 (Pierce expansion definition), A059180.

Programs

  • Mathematica
    PierceExp[A_, n_] := Join[Array[1 &, Floor[A]], First@Transpose@ NestList[{Floor[1/Expand[1 - #[[1]] #[[2]]]], Expand[1 - #[[1]] #[[2]]]} &, {Floor[1/(A - Floor[A])], A - Floor[A]}, n - 1]]; PierceExp[N[Log[2], 7!], 25] (* G. C. Greubel, Nov 14 2016 *)
  • PARI
    r=1/log(2);for(n=1,30,r=r/(r-floor(r));print1(floor(r),","))

Formula

Let u(0)=1/log(2) and u(n+1)=u(n)/frac(u(n)) where frac(x) is the fractional part of x, then a(n)=floor(u(n)).
log(2) = 1/a(1) - 1/(a(1)*a(2)) + 1/(a(1)*a(2)*a(3)) - 1/(a(1)*a(2)*a(3)*a(4)) +- ...
limit n-->infinity a(n)^(1/n) = e.
Showing 1-2 of 2 results.