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.

A228929 Optimal ascending continued fraction expansion of Pi - 3.

Original entry on oeis.org

7, -113, 4739, -46804, 134370, -614063, 1669512, -15474114, -86232481, 1080357006, -8574121305, -24144614592, 133884333083, -2239330253016, -6347915250018, 14541933941298, -42301908155404, -298013673554972, 5177473084279656, -46709468571434452, 1201667304102142095, -68508286025632748778, 850084640720511629243, -2458418086834560217354
Offset: 1

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Author

Giovanni Artico, Sep 08 2013

Keywords

Comments

Definition of the expansion: for a positive real number x, there is always a unique sequence of signed integers with increasing absolute value |a(i)|>|a(i-1)| such that x =floor(x)+ 1/a(1) + 1/a(1)/a(2) + 1/a(1)/a(2)/a(3) + 1/a(1)/a(2)/a(3)/a(4) ... or equivalently x=floor(x)+1/a(1)*(1+1/a(2)*(1+1/a(3)*(1+1/a(4)*(1+...)))) giving the fastest converging series with this representation. This formula can be represented as a regular ascending continued fraction. The expansion is similar to Engel and Pierce expansions, but the sign of the terms is not predefined and determined by the algorithm for optimizing the convergence.
For x rational number the expansion has a finite number of terms, for x irrational an infinite number. Empirically the sequence doesn't show any evident regularity except in some interesting cases.

Examples

			Pi = 3 + 1/7*(1 - 1/113*(1 + 1/4739*(1 - 1/46804*(1 + 1/134370*(1 - 1/614063*(1 + 1/1669512*(1 + ...))))))).

Links

Crossrefs

Cf. A006784, A015884, A228930, A228931, A228932, A228933, A228934.

Programs

  • Derive
    ArticoExp(x, n) := VECTOR(ROUND(1, ABS(k))*SIGN(k), k, ITERATES(ROUND(1, ABS(u))*ABS(u) - 1, u, MOD(x), n))
Precision:=Mixed
PrecisionDigits:=10000
ArticoExp(PI,20)
  • Maple
    # Slow procedure valid for every number
ArticoExp := proc (n, q::posint)::list; local L, i, z; Digits := 50000; L := []; z := n-floor(n); for i to q+1 do if z = 0 then break end if; L := [op(L), round(1/abs(z))*sign(evalf(z))]; z := abs(z)*round(1/abs(z))-1 end do; return L end proc
# Fast procedure, not suited for rational numbers
ArticoExp := proc (n, q::posint)::list; local L, i, z; Digits := 50000; L := []; z := frac(evalf(n)); for i to q+1 do if z = 0 then break end if; L := [op(L), round(1/abs(z))*sign(z)]; z := abs(z)*round(1/abs(z))-1 end do; return L end proc
# List the first 20 terms of the expansion of Pi
ArticoExp(Pi,20)
  • Mathematica
    ArticoExp[x_, n_] :=  Round[1/#] & /@ NestList[Round[1/Abs[#]]*Abs[#] - 1 &, FractionalPart[x], n]; Block[{$MaxExtraPrecision = 50000}, ArticoExp[Pi, 20]]

Formula

Given a positive real number x, let z(0)=x-floor(x) and z(k+1)=abs(z(k))*round(1/abs(z(k)))-1 ; then a(n)=sign(z(n))*round(1/abs(z(n))) for n>0.

A228933 Optimal ascending continued fraction expansion of phi-1=1/phi=(sqrt(5)-1)/2 .

Original entry on oeis.org

2, 4, -18, 322, 103682, 10749957122, 115561578124838522882, 13354478338703157414450712387359637585922, 178342091698891843163466683840822101223162205277179656650156983624835803932590082
Offset: 1

Views

Author

Giovanni Artico, Sep 10 2013

Keywords

Comments

See A228929 for the definition of "optimal ascending continued fraction".
Conjecture: The golden ratio (phi) expansion exhibits from the fourth term the recurrence relation a(n) = a(n-1)^2 - 2 described in A228931.

Examples

			phi = 1+1/2*(1+1/4*(1-1/18*(1+1/322*(1+1/103682*(1+1/10749957122*(1+...))))))

Crossrefs

Cf. A001622, A081459, A094214, A228929, A228931, A228932.

Programs

  • Maple
    ArticoExp := proc (n, q::posint)::list; local L, i, z; Digits := 50000; L := []; z := frac(evalf(n)); for i to q+1 do if z = 0 then break end if; L := [op(L), round(1/abs(z))*sign(z)]; z := abs(z)*round(1/abs(z))-1 end do; return L end proc
# List the first 8 terms of the expansion of 1/phi
ArticoExp((sqrt(5)-1)/2,8)
  • Mathematica
    Flatten[{2, 4, RecurrenceTable[{a[n] == a[n-1]^2 - 2, a[3] == -18}, a, {n, 3, 10}]}] (* Vaclav Kotesovec, Sep 20 2013 *)

Formula

a(n) = a(n-1)^2 - 2 for n>3.
For n>3, a(n) = (sqrt(5)+2)^(2^(n-2)) + (sqrt(5)-2)^(2^(n-2)). - Vaclav Kotesovec, Sep 20 2013
a(n) = 2*A081459(n-1) for n>3. - Amiram Eldar, Apr 07 2023

A228934 Optimal ascending continued fraction expansion of sqrt(44) - 6.

Original entry on oeis.org

2, 4, 15, -99, -199, -800, -79201, -316808, -12545596801, -50182387208, -314783998186522867201, -1259135992746091468808, -198177931028585663493396958369763763148801, -792711724114342653973587833479055052595208
Offset: 1

Views

Author

Giovanni Artico, Sep 11 2013

Keywords

Comments

See A228929 for the definition of "optimal ascending continued fraction".
This is the first number whose expansion exhibits (in the first 20 terms) a different recurrence relation from that described in A228931.
Conjecture: The terms of the expansion of sqrt(x) are all negative starting from a(4) and satisfy these recurrence relations for n>=3: a(2n) = 4*a(2n-1) - 4 and a(2n+1) = -2*a(2n-1)^2 + 1.
Numbers (in the range 1..1000) that exhibit this recurrence starting from some n are 44, 125, 154, 160, 176, 207, 208, 280, 352, 384, 459, 468, 500, 608, 616, 640, 665, 686, 704, 768, 800, 832, 864, 874, 875, 924.

Examples

			sqrt(44) = 6 + 1/2*(1 + 1/4*(1 + 1/15*(1 - 1/99*(1 - 1/199*(1 - 1/800*(1 - 1/79201*(1 - 1/316808*(1 - 1/12545596801*(1 - ...))))))))).

Links

Crossrefs

Cf. A228929, A228931, A228932.

Programs

  • Maple
    ArticoExp := proc (n, q::posint)::list; local L, i, z; Digits := 50000; L := []; z := frac(evalf(n)); for i to q+1 do if z = 0 then break end if; L := [op(L), round(1/abs(z))*sign(z)]; z := abs(z)*round(1/abs(z))-1 end do; return L end proc
# List the first 20 terms of the expansion of sqrt(44)-6
ArticoExp(sqrt(44),20)
  • Mathematica
    ArticoExp[x_, n_] := Round[1/#] & /@ NestList[Round[1/Abs[#]]*Abs[#] - 1 &, FractionalPart[x], n]; Block[{$MaxExtraPrecision = 50000}, ArticoExp[Sqrt[44] - 6, 20]] (* G. C. Greubel, Dec 26 2016 *)

Formula

a(2n) = 4*a(2n-1) - 4 and a(2n+1) = -2*a(2n-1)^2 + 1 for n >= 3.

Extensions

Minor typos corrected by Giovanni Artico, Sep 24 2013
Showing 1-3 of 3 results.

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