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 1-3 of 3 results.

A219508 Pierce expansion of 4*(3 - 2*sqrt(2)).

Original entry on oeis.org

1, 3, 16, 17, 72, 577, 2312, 665857, 2663432, 886731088897, 3546924355592, 1572584048032918633353217, 6290336192131674533412872, 4946041176255201878775086487573351061418968498177
Offset: 0

Views

Author

Peter Bala, Nov 23 2012

Keywords

Comments

For x in the open interval (0,1) define the map f(x) = 1 - x*floor(1/x). The n-th term (n >= 0) in the Pierce expansion of x is given by floor(1/f^(n)(x)), where f^(n)(x) denotes the n-th iterate of the map f, with the convention that f^(0)(x) = x.
Paradis et al. have determined the Pierce expansion of the quadratic irrationality 2*(p - 1)*(p - sqrt(p^2 - 1)), p a positive integer greater than or equal to 3. The present sequence is the case p = 3. For other cases see A219509 (p = 5), A219510 (p = 7) and A219511 (p = 9).
Compare this Pierce expansion for 4*(3 - 2*sqrt(2)), with terms determined by quadratic recurrences, with the Pierce expansion of 3 - 2*sqrt(2) given in A006275, where the terms are determined by cubic recurrences.

Links

Crossrefs

Cf. A001601, A006275, A219509 (p = 5), A219510 (p = 7), A219511 (p = 9).

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[4*(3 - 2*Sqrt[2]) , 7!], 10] (* G. C. Greubel, Nov 14 2016 *)
  • PARI
    r=(3 + 2*sqrt(2))/4; for(n=1, 10, print(floor(r), ", "); r=r/(r-floor(r))) \\ G. C. Greubel, Nov 15 2016

Formula

a(2*n) = 2*{(1 + sqrt(2))^(2^n) + (1 - sqrt(2))^(2^n) + 2} for n >= 1.
a(2*n-1) = 1/2*{(1 + sqrt(2))^(2^n) + (1 - sqrt(2))^(2^n)} for n >= 1.
Recurrence equations: a(0) = 1, a(1) = 3 and for n >= 1, a(2*n) = 4*(a(2*n-1) + 1) and a(2*n+1) = 2*{a(2*n-1)}^2 - 1.
It follows that a(2*n) = 8*a(2*n-3)^2 for n >=2.
4*(3 - 2*sqrt(2)) = sum {n >= 0} 1/product {k = 0..n} a(k) = 1 - 1/3 + 1/(3*16) - 1/(3*16*17) + 1/(3*16*17*72) - ....
a(2*n) = 8*A001601(n-1)^2 for n >= 2
a(2*n-1) = A001601(n) for n >= 1.

A219509 Pierce expansion of 40 - 16*sqrt(6).

Original entry on oeis.org

1, 5, 24, 49, 200, 4801, 19208, 46099201, 184396808, 4250272665676801, 17001090662707208, 36129635465198759610694779187201, 144518541860795038442779116748808, 2610701117696295981568349760414651575095962187244375364404428801
Offset: 0

Views

Author

Peter Bala, Nov 23 2012

Keywords

Comments

Paradis et al. have determined the Pierce expansion of the quadratic irrationality 2*(p - 1)*(p - sqrt(p^2 - 1)), p a positive integer greater than or equal to 3. This is the case p = 5. For other cases see A219508 (p = 3), A219510 (p = 7) and A219511 (p = 9)

Links

Crossrefs

Cf. A084765, A219508 (p = 3), A219510 (p = 7), A219511 (p = 9).

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[4*(10 - 4*Sqrt[6]) , 7!], 10] (* G. C. Greubel, Nov 14 2016 *)
  • PARI
    r=(5 + 2*sqrt(6))/8; for(n=1, 10, print(floor(r), ", "); r=r/(r-floor(r))) \\ G. C. Greubel, Nov 15 2016

Formula

a(2*n+2) = 2*{(5 + 2*sqrt(6))^(2^n) + (5 - 2*sqrt(6))^(2^n) + 2} for n >= 0.
a(2*n+1) = 1/2*{(5 + 2*sqrt(6))^(2^n) + (5 - 2*sqrt(6))^(2^n)} for n >= 0.
Recurrence equations: a(0) = 1, a(1) = 5 and for n >= 1, a(2*n) = 4*(a(2*n-1) + 1) and a(2*n+1) = 2*(a(2*n-1))^2 - 1.
40 - 16*sqrt(6) = sum {n >= 0} 1/product {k = 0..n} a(k) = 1 - 1/5 + 1/(5*24) - 1/(5*24*49) + 1/(5*24*49*200) - ....
a(2*n) = 8*A084765(n-1)^2 for n >= 2.
a(2*n+1) = A084765(n+1) for n >= 0.

A219511 Pierce expansion of 144 - 64*sqrt(5).

Original entry on oeis.org

1, 9, 40, 161, 648, 51841, 207368, 5374978561, 21499914248, 57780789062419261441, 231123156249677045768, 6677239169351578707225356193679818792961, 26708956677406314828901424774719275171848
Offset: 0

Views

Author

Peter Bala, Nov 23 2012

Keywords

Comments

Paradis et al. have determined the Pierce expansion of the quadratic irrationality 2*(p - 1)*(p - sqrt(p^2 - 1)), p a positive integer greater than or equal to 3. This is the case p = 9. For other cases see A219508 (p = 3), A219509 (p = 5) and A219510 (p = 7).

Links

Crossrefs

Cf. A081459, A219508 (p = 3), A219509 (p = 5), A219510 (p = 7).

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[144 - 64*Sqrt[5] , 7!], 10] (* G. C. Greubel, Nov 15 2016 *)
  • PARI
    r=(9 + 4*sqrt(5))/16; for(n=1, 10, print(floor(r), ", "); r=r/(r-floor(r))) \\ G. C. Greubel, Nov 15 2016

Formula

a(2*n) = 2*{(2 + sqrt(5))^(2^n) + (2 - sqrt(5))^(2^n) + 2} for n >= 1.
a(2*n-1) = 1/2*{(2 + sqrt(5))^(2^n) + (2 - sqrt(5))^(2^n)} for n >= 1.
Recurrence equations: a(0) = 1, a(1) = 9 and for n >= 1, a(2*n) = 4*(a(2*n-1) + 1) and a(2*n+1) = 2*(a(2*n-1))^2 - 1.
144 - 64*sqrt(5) = 1 - 1/9 + 1/(9*40) - 1/(9*40*161) + 1/(9*40*161*648) - ....
a(2*n) = 8*A081459(n)^2 for n >= 2.
a(2*n+1) = A081459(n+2) for n >= 0.
Showing 1-3 of 3 results.

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