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 11-13 of 13 results.

A091832 Pierce expansion of 1/e^2.

Original entry on oeis.org

7, 18, 19, 136, 349, 357, 1354, 6996, 7135, 9531, 11558, 15996, 17432, 52118, 151048, 427802, 821834, 877819, 972918, 1046690, 1540789, 3653077, 8200738, 9628573, 164153335, 5607624822, 86457467082, 141885251873, 151882622551
Offset: 1

Views

Author

Benoit Cloitre, Mar 09 2004

Keywords

Comments

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

Crossrefs

Cf. A006784 (Pierce expansion definition), A059194.

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[1/E^2, 7!], 15] (* G. C. Greubel, Nov 14 2016 *)
  • PARI
    default(realprecision, 100000); r=exp(2); for(n=1, 100, s=(r/(r-floor(r))); print1(floor(r), ", "); r=s) \\ Benoit Cloitre [amended by Georg Fischer, Nov 20 2020]

Formula

Let u(0) = exp(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)).
1/e^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->oo} a(n)^(1/n) = e.

Extensions

a(1)=7 inserted by Georg Fischer, Nov 20 2020

A091833 Pierce expansion of 1/zeta(2).

Original entry on oeis.org

1, 2, 4, 7, 22, 29, 51, 173, 210, 262, 417, 746, 12341, 207220, 498538, 1286415, 2351289, 3702952, 7664494, 54693034, 75971438, 269954954, 6674693008, 13449203581, 59799655308, 98912303039, 948887634688, 3557757020909, 5898230078743
Offset: 1

Views

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), A059186.

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[1/Zeta[2], 7!], 25] (* G. C. Greubel, Nov 14 2016 *)
  • PARI
    default(realprecision, 100000); r=zeta(2); for(n=1, 100, s=(r/(r-floor(r))); print1(floor(r), ", "); r=s) \\ Benoit Cloitre [amended by Georg Fischer, Nov 20 2020]

Formula

Let u(0) = Pi^2/6 and u(n+1) = u(n)/frac(u(n)) where frac(x) is the fractional part of x; then a(n) = floor(u(n)).
1/zeta(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->oo} a(n)^(1/n) = e.

Extensions

a(1)=1 inserted by Georg Fischer, Nov 20 2020

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

Views

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.
Previous Showing 11-13 of 13 results.