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.

User: Sanjar Abrarov

Sanjar Abrarov's wiki page.

Sanjar Abrarov has authored 7 sequences.

A375935 Denominator of rational number tan(2^(n-1)*arctan(1/A024810(n))).

Original entry on oeis.org

1, 3, 119, 72697201, 466125047039779152001, 1298332088620573402611989909424448036756809624806401
Offset: 1

Author

Sanjar Abrarov, Sep 03 2024

Keywords

Comments

r(n) = tan(2^(n-1)*arctan(1/A024810(n))) is always a rational number such that its limit lim_{n->oo} r(n)=1.

Crossrefs

Cf. A024810, A374723 (numerators).

Programs

  • Mathematica
    a[n_] := Module[{}, z[n] = (Floor[2^(n + 1)/\[Pi]] + I)^2^(n - 1); Denominator[Im[z[n]]/Re[z[n]]]];
    Print[a[1], " ", a[2], " ", a[3], " ", a[4], " ", a[5], " ", a[6]];

Formula

a(n) = denominator(tan(2^(n-1)*arctan(1/A024810(n)))).

A374723 Numerator of rational number tan(2^(n-1)*arctan(1/A024810(n))).

Original entry on oeis.org

1, 4, 120, 74455920, 479301523749226879680, 1336367227479573478314645756081634359006026455038720
Offset: 1

Author

Sanjar Abrarov, Sep 03 2024

Keywords

Comments

r(n) = tan(2^(n-1)*arctan(1/A024810(n))) is always a rational number such that its limit lim_{n->oo} r(n)=1.

Crossrefs

Cf. A024810, A375935 (denominators).

Programs

  • Mathematica
    a[0] := 0;
    a[n_] := Sqrt[2 + a[n - 1]];
    b[n_] := Floor[a[n]/Sqrt[2 - a[n - 1]]];
    r[0, x_] := 1;
    r[1, x_] := (2*x)/(1 - x^2);
    r[n_, x_] := (2*r[n - 1, x])/(1 - r[n - 1, x]^2);
    k = 1;
    While[k <= 6, Print[k, " ", Numerator[r[k - 1, 1/b[k]]]]; k++];

Formula

a(n) = numerator(tan(2^(n-1)*arctan(1/A024810(n)))).

A338879 Denominators in a set of expansions of the single-term Machin-like formula for Pi.

Original entry on oeis.org

1, 3, 2, 5, 11, 3, 7, 11, 9, 4, 9, 37, 31, 27, 5, 11, 28, 8, 7, 19, 6, 13, 79, 69, 61, 55, 51, 7, 15, 53, 47, 21, 19, 35, 33, 8, 17, 137, 41, 37, 101, 31, 29, 83, 9, 19, 86, 78, 71, 13, 12, 56, 53, 51, 10, 21, 211, 193, 177, 163, 151, 141, 133, 127, 123, 11, 23, 127, 39, 18, 50, 31, 29, 41, 13, 25, 73, 12
Offset: 1

Author

Sanjar Abrarov, Nov 13 2020

Keywords

Comments

Numerators are A338878.
Abrarov et al. give an identity arctan(n*x) = Sum_{m=1..n} arctan(x / (1 + (m-1)*m*x^2)). At x=1/n this identity provides set of expansions of the single-term Machin-like formula for Pi in form Pi/4 = arctan(1) = Sum_{m=1..n} arctan(n/((m-1)*m + n^2)). For m = n - k + 1 at k=1..n the fractions n / ((m-1)*m + n^2) constitute the triangle with rows in ascending order:
k= 1 2 3 4 5 6
n=1: 1;
n=2: 1/3, 1/2;
n=3: 1/5, 3/11, 1/3;
n=4: 1/7, 2/11, 2/9, 1/4;
n=5: 1/9, 5/37, 5/31, 5/27, 1/5;
n=6: 1/11, 3/28, 1/8, 1/7, 3/19, 1/6;

Examples

			The triangle T(n,k) begins:
    k=  1   2   3   4   5   6
  n=1:  1;
  n=2:  3,  2;
  n=3:  5,  11, 3;
  n=4:  7,  11, 9,  4;
  n=5:  9,  37, 31, 27, 5;
  n=6:  11, 28, 8,  7,  19, 6;
For example, at n = 3 the expansion formula is Pi/4 = arctan(1/5) + arctan(3/11) + arctan(1/3) and the corresponding sequence in the denominators is 5,11,3.
		

Crossrefs

Cf. A338878 (numerators), A003881 (Pi/4).

Programs

  • Mathematica
    (*Define variable*)
    PiOver4[m_] := Sum[ArcTan[m/((k - 1)*k + m^2)], {k, 1, m}];
    (*Expansions*)
    m := 1;
    While[m <= 10,
      If[m == 1, Print["\[Pi]/4 = ArcTan[1/1]"],
        Print["\[Pi]/4 = ", PiOver4[m]]]; m = m + 1];
    (*Verification*)
    m := 1;
    While[m <= 10, Print[PiOver4[m] == Pi/4]; m = m + 1];
    (*Denominators*)
    For[n = 1, n <= 10, n++, {k := 1; sq := {};
      While[n >= k, AppendTo[sq, Denominator[n/((n - k)*(n - k
        + 1) + n^2)]]; k++]}; Print[sq]];
  • PARI
    T(n, k) = if (n>=k, denominator(n/((n - k)*(n - k + 1) + n^2)))
    matrix(10, 10, n, k, T(n, k)) \\ Michel Marcus, Nov 14 2020

Formula

T(n,k) = denominator of n / ((n-k)*(n-k+1) + n^2), for n>=1 and 1 <= k <= n.
Pi/4 = Sum_{k=1..n} arctan(A338878(n,k) / T(n,k)).

A338878 Numerators in a set of expansions of the single-term Machin-like formula for Pi.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 2, 2, 1, 1, 5, 5, 5, 1, 1, 3, 1, 1, 3, 1, 1, 7, 7, 7, 7, 7, 1, 1, 4, 4, 2, 2, 4, 4, 1, 1, 9, 3, 3, 9, 3, 3, 9, 1, 1, 5, 5, 5, 1, 1, 5, 5, 5, 1, 1, 11, 11, 11, 11, 11, 11, 11, 11, 11, 1, 1, 6, 2, 1, 3, 2, 2, 3, 1, 2, 6, 1
Offset: 1

Author

Sanjar Abrarov, Nov 13 2020

Keywords

Comments

Denominators are A338879.
Abrarov et al. (see section LINKS) give an identity arctan(n*x) = Sum_{m=1..n} arctan(x/(1 + (m-1)*m*x^2)). At x=1/n this identity provides set of expansions of the single-term Machin-like formula for Pi in form Pi/4 = arctan(1) = Sum_{m=1..n} arctan(n/((m-1)*m + n^2)). For m = n - k + 1 at k=1..n the fractions n/((m-1)*m + n^2) constitute the triangle with rows in ascending order:
k= 1 2 3 4 5 6
n=1: 1;
n=2: 1/3, 1/2;
n=3: 1/5, 3/11, 1/3;
n=4: 1/7, 2/11, 2/9, 1/4;
n=5: 1/9, 5/37, 5/31, 5/27, 1/5;
n=6: 1/11, 3/28, 1/8, 1/7, 3/19, 1/6;
Section EXAMPLE shows the corresponding triangle T(n,k) with numerators. This triangle T(n,k) possesses a sifting property for primes. In particular, in a row of kind {1,p,p,p,...,p,p,1} the integer p must be a prime and n = p except the case n = 4 when prime p = 2.

Examples

			Triangle T(n,k) begins:
     k= 1  2  3  4  5  6
  n=1:  1;
  n=2:  1, 1;
  n=3:  1, 3, 1;
  n=4:  1, 2, 2, 1;
  n=5:  1, 5, 5, 5, 1;
  n=6:  1, 3, 1, 1, 3, 1;
For example, for row n = 3 the corresponding expansion formula is Pi/4 = arctan(1/5) + arctan(3/11) + arctan(1/3) and the numerators are 1,3,1.
At n = 3, n = 4 and n = 5 the rows are {1,3,1}, {1,2,2,1} and {1,5,5,5,1} and the primes are 3, 2 and 5, respectively.
		

Crossrefs

Cf. A338879 (denominators), A003881 (Pi/4).

Programs

  • Mathematica
    (*Define variable*)
    PiOver4[m_] := Sum[ArcTan[m/((k - 1)*k + m^2)], {k, 1, m}];
    (*Expansions*)
    m := 1;
    While[m <= 10,
      If[m == 1, Print["\[Pi]/4 = ArcTan[1/1]"],
        Print["\[Pi]/4 = ", PiOver4[m]]]; m = m + 1];
    (*Verification*)
    m := 1;
    While[m <= 10, Print[PiOver4[m] == Pi/4]; m = m + 1];
    (*Numerators*)
    For[n = 1, n <= 15, n++, {k := 1; sq := {};
      While[n >= k, AppendTo[sq, n/GCD[n, k*(k - 1)]]; k++]};
        Print[sq]];
  • PARI
    T(n, k) = if (n>=k, n/gcd(n,k*(k - 1)))
    matrix(10, 10, n, k, T(n, k)) \\ Michel Marcus, Nov 14 2020

Formula

T(n,k) = numerator(n/((n-k)*(n-k+1) + n^2)) = n/gcd(n,k*(k-1)), for n>=1 and 1 <= k <= n.
Pi/4 = Sum_{k=1..n} arctan(T(n,k) / A338879(n,k)).

A282089 Decimal expansion of constant 1.287194... related to a conjectural Viète-like formula for Pi.

Original entry on oeis.org

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

Author

Sanjar Abrarov, Feb 06 2017

Keywords

Comments

Conjecture: Pi = lim_{k -> infinity} 2^{k + 1}*(1 - c_k), where the variable c_k is defined by a set of the Viète-like recurrence relations {a_1 = sqrt(2), a_k = sqrt(2 + a_{k - 1}), b_k = sqrt(2 - a_k)/a_{k + 1}, c_1 = b_1, c_k = (c_{k - 1} + b_k)/(1 - c_{k - 1}*b_k)}. From this conjecture it follows that Sum_{k >= 1} (1 - c_k) is convergent [Abrarov and Quine].

Examples

			1.287194036067924017020927807581...
		

Crossrefs

Cf. A000796.

Programs

  • Mathematica
    Clear[a, b, c]
    a[k_] := N[Nest[Sqrt[2 + #1] &, 0, k], 100]
    b[k_] := b[k] = Sqrt[2 - a[k]]/a[k + 1]
    c[1] := b[1] = b[1]
    c[k_] := c[k] = (c[k - 1] + b[k])/(1 - c[k - 1]*b[k])
    k := 90
    Print["Index k = ", k]
    m := 1
    Print["Power m = ", m]
    (* The equation (12) *)
    apprPi := 2^(k + 1)*(1 - c[k]^m)
    Print["Actual value of Pi is ", N[Pi, 30]]
    Print["At k = ", k, " the approximated value of Pi is ", N[apprPi, 30]]
    K := 300
    Print["Truncating integer K = ", K]
    Print["Computing the digits ..."]
    RealDigits[N[Sum[1 - c[k]^m, {k, 1, K}], 30]][[1]]

Formula

Sum_{k >= 1} (1 - c_k) = 1.287194... , where c_k is computed by the recurrence equations a_1 = sqrt(2), a_k = sqrt(2 + a_{k - 1}), b_k = sqrt(2 - a_k)/a_{k + 1}, c_1 = b_1 and c_k = (c_{k - 1} + b_k)/(1 - c_{k - 1}*b_k).

A278924 A sequence showing numerators in ratios tending to the constant Pi/4 = 0.785398163397448... .

Original entry on oeis.org

4, 296, 36772, 1288688, 96641548, 26576092808, 8637277012172, 1079658805128928, 91770997994914276, 43591225139846360008
Offset: 1

Author

Sanjar Abrarov, Dec 01 2016

Keywords

Comments

The ratios c(n)/d(n) rapidly tend to the constant Pi/4 = 0.785398163397448... with increasing index n: abs(Pi/4 - c(1)/d(1)) > abs(Pi/4 - c(2)/d(2)) > abs(Pi/4 - c(3)/d(3)) > abs(Pi/4 - c(4)/d(4)) ..., where c(n) = A278924(n) and d(n) = A278364(n) are even and odd positive integers, respectively.

Examples

			------------------------------------------------
n    c(n)                   d(n)
------------------------------------------------
1    4                      5
2    296                    375
3    36772                  46875
4    1288688                1640625
5    96641548               123046875
6    26576092808            33837890625
7    8637277012172          10997314453125
8    1079658805128928       1374664306640625
9    91770997994914276      116846466064453125
10   43591225139846360008   55502071380615234375
------------------------------------------------
At n = 6 the ratio c(6)/d(6) = 26576092808/33837890625 is close to Pi/4. However, at n = 10 the ratio c(10)/d(10) = 43591225139846360008/55502071380615234375 becomes more closer to Pi/4.
		

Crossrefs

Programs

  • Mathematica
    x := 1;(* argument x *)
    M := 1;(* initial value for the integer M *)
    n := 1; (* index *)
    (* Note that arctan(1) = Pi/4 *)
    atan := I*Sum[(1/(2*m - 1))*(1/(1 + 2*(I/x))^(2*m - 1) - 1/(1 - 2*(I/x))^(2*m - 1)),{m, 1, Floor[M/2] + 1}];
    sqn := {};(* initiate the sequence *)
    AppendTo[sqn,{"Index n", "Numerators", "Denominators"}];
    While[M <= 20, AppendTo[sqn,{n, Numerator[atan], Denominator[atan]}];
    {M = M + 2, n++}];
    Print[MatrixForm[sqn]]

Formula

arctan(1) = I*lim_{M -> inf}Sum_{m = 1..floor(M/2) + 1}(1/(2*m - 1))*(1/(1 + 2*I)^(2*m - 1) - 1/(1 - 2*I)^(2*m - 1))

A278364 A sequence showing denominators in ratios tending to the constant Pi/4 = 0.785398163397448... .

Original entry on oeis.org

5, 375, 46875, 1640625, 123046875, 33837890625, 10997314453125, 1374664306640625, 116846466064453125, 55502071380615234375
Offset: 1

Author

Sanjar Abrarov, Dec 04 2016

Keywords

Comments

The ratios c(n)/d(n) rapidly tend to the constant Pi/4 = 0.785398163397448... with increasing index n: abs(Pi/4 - c(1)/d(1)) > abs(Pi/4 - c(2)/d(2)) > abs(Pi/4 - c(3)/d(3)) > abs(Pi/4 - c(4)/d(4)) ..., where c(n) = A278924(n) and d(n) = A278364(n) are even and odd positive integers, respectively. All denominators d(n) are divisible by 5.

Examples

			------------------------------------------------
n    c(n)                   d(n)
------------------------------------------------
1    4                      5
2    296                    375
3    36772                  46875
4    1288688                1640625
5    96641548               123046875
6    26576092808            33837890625
7    8637277012172          10997314453125
8    1079658805128928       1374664306640625
9    91770997994914276      116846466064453125
10   43591225139846360008   55502071380615234375
------------------------------------------------
At n = 6 the ratio c(6)/d(6) = 26576092808/33837890625 is close to Pi/4. However, at n = 10 the ratio c(10)/d(10) = 43591225139846360008/55502071380615234375 becomes more closer to Pi/4.
		

Crossrefs

Programs

  • Mathematica
    x := 1; (* argument x *)
    M := 1; (* initial value for the integer M *)
    n := 1; (* index *)
    (* Note that arctan(1) = Pi/4 *)
    atan := I*Sum[(1/(2*m - 1))*(1/(1 + 2*(I/x))^(2*m - 1) - 1/(1 - 2*(I/x))^(2*m - 1)), {m, 1, Floor[M/2] + 1}];
    sqn := {}; (* initiate the sequence *)
    AppendTo[sqn, {"Index n", "Numerators", "Denominators"}];
    While[M <= 20, AppendTo[sqn, {n, Numerator[atan], Denominator[atan]}];
    {M = M + 2, n++}];
    Print[MatrixForm[sqn]]

Formula

arctan(1) = I*lim_{M -> inf}Sum_{m = 1..floor(M/2) + 1}(1/(2*m - 1))*(1/(1 + 2*I)^(2*m - 1) - 1/(1 - 2*I)^(2*m - 1))