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 11-18 of 18 results.

A195710 Decimal expansion of arccos(-sqrt(2/5)).

Original entry on oeis.org

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

Views

Author

Clark Kimberling, Sep 23 2011

Keywords

Examples

			arccos(-sqrt(2/5)) = 2.25551552979717...
		

Crossrefs

Programs

  • Magma
    [Arccos(-Sqrt(2/5))]; // G. C. Greubel, Nov 18 2017
  • Mathematica
    r = Sqrt[1/5]; s = Sqrt[2/5];
    N[ArcSin[r], 100]
    RealDigits[%]  (* A073000 *)
    N[ArcCos[r], 100]
    RealDigits[%]  (* A105199 *)
    N[ArcTan[r], 100]
    RealDigits[%]  (* A188595 *)
    N[ArcCos[-r], 100]
    RealDigits[%]  (* A137218 *)
    N[ArcSin[s], 100]
    RealDigits[%]  (* A195701 *)
    N[ArcCos[s], 100]
    RealDigits[%]  (* A195708 *)
    N[ArcTan[s], 100]
    RealDigits[%]  (* A195709 *)
    N[ArcCos[-s], 100]
    RealDigits[%]  (* A195710 *)
    RealDigits[ArcCos[-Sqrt[(2/5)]],10,120][[1]] (* Harvey P. Dale, Apr 06 2023 *)
  • PARI
    acos(-sqrt(2/5)) \\ G. C. Greubel, Nov 18 2017
    

Formula

Equals Pi - arcsin(sqrt(3/5)) = Pi - arctan(sqrt(3/2)). - Amiram Eldar, Jul 08 2023

A254381 a(n) = 3^n*(2*n + 1)!/n!.

Original entry on oeis.org

1, 18, 540, 22680, 1224720, 80831520, 6304858560, 567437270400, 57878601580800, 6598160580211200, 831368233106611200, 114728816168712345600, 17209322425306851840000, 2787910232899709998080000, 485096380524549539665920000, 90227926777566214377861120000
Offset: 0

Views

Author

Peter Bala, Feb 04 2015

Keywords

Crossrefs

Programs

  • Maple
    seq(3^n*(2*n + 1)!/n!, n = 0..13);
  • Mathematica
    Table[3^n(2n + 1)!/n!, {n, 0, 19}] (* Alonso del Arte, Feb 04 2015 *)

Formula

E.g.f.: 1/(1 - 12*x)^(3/2) = 1 + 18*x + 540*x^2/2! + 22680*x^3/3! + ....
Recurrence equation: a(n) = 6*(2*n + 1)*a(n-1) with a(0) = 1.
2nd order recurrence equation: a(n) = 8*(n + 1)*a(n-1) + 12*(2*n - 1)^2*a(n-2) with a(0) = 1, a(1) = 18.
Define a sequence b(n) := a(n)*sum {k = 0..n} (-1)^k/((2*k + 1)*3^k) beginning [1, 16, 492, 20544, 1111056, 73299456, 5718022848, ...]. It is not difficult to check that b(n) also satisfies the previous 2nd order recurrence equation (and so is an integer sequence). Using this observation we obtain the continued fraction expansion Pi/(2*sqrt(3)) = Sum {k >= 0} (-1)^k/( (2*k + 1)*3^k ) = 1 - 2/(18 + 12*3^2/(24 + 12*5^2/(32 + ... + 12*(2*n - 1)^2/((8*n + 8) + ... )))). Cf. A254619 and A254620.

A086203 Decimal expansion of arctan(1/2)/Pi.

Original entry on oeis.org

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

Views

Author

Eric W. Weisstein, Jul 12 2003

Keywords

Comments

This constant is a transcendental number (Margolius, 2002; Smith, 2003). - Amiram Eldar, Aug 13 2020

Examples

			0.14758...
		

References

  • Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003. See pp. 430-433.

Crossrefs

Programs

  • Mathematica
    RealDigits[ArcTan[1/2]/Pi, 10, 100][[1]] (* Amiram Eldar, Aug 13 2020 *)
  • PARI
    atan(1/2)/Pi \\ Michel Marcus, Jul 29 2015

A105533 Decimal expansion of arctan(1/7).

Original entry on oeis.org

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

Views

Author

Bryan Jacobs (bryanjj(AT)gmail.com), Apr 12 2005

Keywords

Examples

			0.1418970546041639228128516171...
		

Crossrefs

Programs

Formula

2*A073000 - arctan(1/7) = 2*A105531 + arctan(1/7) = Pi/4.
5*arctan(1/7) + 2*arctan(3/79) = Pi/4. - Frank Ellermann, Mar 01 2020
Equals arcsin(1/(5*sqrt(2))) = arccos(7/(5*sqrt(2))). - Amiram Eldar, Jul 11 2023

A254620 a(n) = 9^n*(2*n + 1)!/n!.

Original entry on oeis.org

1, 54, 4860, 612360, 99202320, 19642059360, 4596241890240, 1240985310364800, 379741504971628800, 129871594700297049600, 49091462796712284748800, 20323865597838885886003200, 9145739519027498648701440000, 4444829406247364343268899840000
Offset: 0

Views

Author

Peter Bala, Feb 03 2015

Keywords

Crossrefs

Programs

  • Maple
    seq(9^n*(2*n + 1)!/n!, n = 0..14);
  • Mathematica
    Table[9^n (2n+1)!/n!,{n,0,20}] (* Harvey P. Dale, Aug 13 2019 *)

Formula

E.g.f.: 1/(1 - 36*x)^(3/2) = 1 + 54*x + 4860*x^2/2! + 612360*x^3/3! + ....
Recurrence equation: a(n) = 18*(2*n + 1)*a(n-1) with a(0) = 1.
2nd order recurrence equation: a(n) = (40*n + 16)*a(n-1) - 36*(2*n - 1)^2*a(n-2) with a(0) = 1, a(1) = 54.
Define a sequence b(n) := a(n)*sum {k = 0..n} 1/((2*k + 1)*9^k) beginning [1, 56, 5052, 636672, 103142544, 20422253952, 4778808090048, ...]. It is not difficult to check that b(n) also satisfies the previous 2nd order recurrence equation (and so is an integer sequence). Using this observation we obtain the continued fraction expansion log(2) = 2/3*Sum {k >= 0} 1/((2*k + 1)*9^k) = 2/3*(1 + 2/(54 - 36*3^2/(96 - 36*5^2/(136 - ... - 36*(2*n - 1)^2/((40*n + 16) - ... ))))).
Alternative 2nd order recurrence equation: a(n) = (32*n + 20)*a(n-1) + 36*(2*n - 1)^2*a(n-2) with a(0) = 1, a(1) = 54.
Define now a sequence c(n) := a(n)*sum {k = 0..n} (-1)^k/((2*k + 1)*9^k) beginning [1, 52, 4692, 591072, 95755344, 18959527872, 4436530187328, ...], which, along with a(n), satisfies the alternative 2nd order recurrence equation. From this observation we find the continued fraction expansion arctan(1/3) = 1/3*Sum {k >= 0} (-1)^k/((2*k + 1)*9^k) = 1/3*(1 - 2/(54 + 36*3^2/(84 + 36*5^2/(116 + ... + 36*(2*n - 1)^2/((32*n + 20) + ... ))))). Cf. A254381 and A254619.

A344906 Decimal expansion of Sum_{k>=0} arctan(1/2^k).

Original entry on oeis.org

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

Views

Author

Daniel Hoyt, Jun 01 2021

Keywords

Comments

This number can be interpreted geometrically as the angle in radians of a fan made of stacked right triangles, with the length to height ratio doubling each successive triangle as seen in the illustration.
Since this angle exceeds Pi/2, the set of rotation angles used in the CORDIC algorithm covers an angle range sufficient to compute sine and cosine for any angle between 0 and Pi/2. This means the algorithm can converge to any angle in that range through appropriate combinations of these basic rotations. - Daniel Hoyt, Oct 25 2024

Examples

			1.743286620472340003...
		

Crossrefs

Programs

  • Maple
    Digits:= 140:
    evalf(sum(arccot(2^k), k=0..infinity));  # Alois P. Heinz, Jun 02 2021
  • PARI
    suminf(k=0, atan(1/2^k))
    
  • PARI
    sumalt(k=1, ((-1)^(k+1))*2^(2*k-1)/((2^(2*k-1)-1)*(2*k-1)))

Formula

Equals Sum_{k>=1} (-1)^(k+1)*2^(2*k-1)/((2^(2*k-1)-1)*(2*k-1)).

A288443 a(n) = (2n + 1)*2^(2n + 1); numbers k such that v(k)*2^v(k) = k, where v(n) = A007814(n) is 2-adic valuation of n.

Original entry on oeis.org

2, 24, 160, 896, 4608, 22528, 106496, 491520, 2228224, 9961472, 44040192, 192937984, 838860800, 3623878656, 15569256448, 66571993088, 283467841536, 1202590842880, 5085241278464, 21440476741632, 90159953477632, 378231999954944, 1583296743997440, 6614661952700416, 27584547717644288, 114841790497947648
Offset: 0

Views

Author

Juri-Stepan Gerasimov, Jun 24 2017

Keywords

Crossrefs

Odd bisection of A036289.

Programs

Formula

a(n) = (2n + 1)*2^(2n + 1).
a(n) = A036289(2n + 1).
a(n) = A098713(n) + 1.
a(n) = 2*A058962(n). - Joerg Arndt, Jun 25 2017
From Amiram Eldar, Jul 03 2020: (Start)
Sum_{n>=0} 1/a(n) = arctanh(1/2) = log(3)/2 (A156057).
Sum_{n>=0} (-1)^n/a(n) = arctan(1/2) (A073000). (End)

A384806 Simple continued fraction expansion of arctan(1/2)/Pi.

Original entry on oeis.org

0, 6, 1, 3, 2, 5, 1, 6, 5, 3, 1, 1, 2, 1, 1, 2, 3, 1, 2, 3, 2, 2, 2, 2, 3, 2, 1, 1, 3, 1, 3, 2, 1, 1, 3, 1, 17, 1, 5, 1, 2, 1, 2, 1, 1, 1, 1, 4, 7, 11, 1, 2, 1, 583, 1, 2, 1, 1, 2, 22, 7, 3, 23, 2, 6, 1, 1, 1, 1, 1, 1, 1, 3, 1, 3, 1, 1, 1, 1, 5, 1, 1, 2, 3, 17, 6, 2, 70, 1, 6
Offset: 0

Views

Author

Stefano Spezia, Jun 10 2025

Keywords

Examples

			arctan(1/2)/Pi = 1/(6 + 1/(1 + 1/(3 + 1/(2 + 1/(5 + ... ))))).
		

Crossrefs

Cf. A000796, A073000, A086203 (decimal expansion).

Programs

  • Mathematica
    ContinuedFraction[ArcTan[1/2]/Pi,90]
Previous Showing 11-18 of 18 results.