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 1-5 of 5 results.

A067360 a(n) = 17^n sin(2n arctan(1/4)) or numerator of tan(2n arctan(1/4)).

Original entry on oeis.org

8, 240, 4888, 77280, 905768, 4839120, -116593352, -4896306240, -113193708472, -1980778750800, -26710380775592, -228866364286560, 853309115549288, 91741652745294480, 2505643247965090168, 48655959795562600320, 735547895204966951048
Offset: 1

Views

Author

Barbara Haas Margolius (b.margolius(AT)csuohio.edu), Jan 17 2002

Keywords

Comments

Note that A067360(n), A067361(n) and 17^n are primitive Pythagorean triples with hypotenuse 17^n.

References

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

Crossrefs

Cf. A067361 (17^n cos(2n arctan(1/4))).

Programs

  • Maple
    a[1] := 8/15; for n from 1 to 40 do a[n+1] := (8/15+a[n])/(1-8/15*a[n]):od: seq(abs(numer(a[n])), n=1..40);# a[n]=tan(2n arctan(1/4))
  • Mathematica
    Table[Tan[2n ArcTan[1/4]] // TrigToExp // Simplify // Numerator, {n, 1, 17} ] (* Jean-François Alcover, Jul 25 2017 *)

Formula

a(n) = 17^n sin(2n arctan(1/4)). A recursive formula for T(n) = tan(2n arctan(1/4)) is T(n+1)=(8/15+T(n))/(1-8/15*T(n)). Unsigned a(n) is the absolute value of numerator of T(n).
Conjectures from Colin Barker, Jul 25 2017: (Start)
G.f.: 8*x / (1 - 30*x + 289*x^2).
a(n) = i*((15 - 8*i)^n - (15 + 8*i)^n)/2 where i=sqrt(-1).
a(n) = 30*a(n-1) - 289*a(n-2) for n>2.
(End)

A067361 a(n) = 17^n*cos(2*n*arctan(1/4)) or denominator of tan(2*n*arctan(1/4)).

Original entry on oeis.org

15, 161, 495, -31679, -1093425, -23647519, -393425745, -4968639359, -35359140465, 375162560801, 21473668418415, 535788072480961, 9867752001506895, 141189807098209121, 1383913884510780975, 713562283940993281, -378544244105385903345
Offset: 1

Views

Author

Barbara Haas Margolius, (b.margolius(AT)csuohio.edu), Jan 17 2002

Keywords

Comments

Note that A067360(n), A067361(n) and 17^n are primitive Pythagorean triples with hypotenuse 17^n.

References

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

Crossrefs

Cf. A067360 (17^n sin(2n arctan(1/4))).

Programs

  • Maple
    a[1] := 8/15; for n from 1 to 40 do a[n+1] := (8/15+a[n])/(1-8/15*a[n]):od: seq(abs(denom(a[n])), n=1..40);# a[n]=tan(2n arctan(1/4))
  • Mathematica
    Table[t = Tan[2 n ArcTan[1/4]] // TrigToExp // Simplify; Sign[t] * Denominator[t], {n, 1, 17}] (* Jean-François Alcover, Jul 25 2017 *)

Formula

a(n) = 17^n*cos(2*n*arctan(1/4)).
A recursive formula for T(n) = tan(2*n*arctan(1/4)) is T(n+1) = (8/15+T(n))/(1-8/15*T(n)). Unsigned a(n) is the absolute value of denominator of T(n). [And a(n) = 17^n*cos(n*arctan(8/15)). - Peter Luschny, Sep 29 2019]
From Colin Barker, Jul 25 2017: (Start)
G.f.: x*(15 - 289*x) / (1 - 30*x + 289*x^2).
a(n) = ((15 - 8*i)^n + (15 + 8*i)^n)/2 where i=sqrt(-1).
a(n) = 30*a(n-1) - 289*a(n-2) for n>2. (End)
a(n) = Re((8 + 15*i)^n) = Re((4 + i)^(2*n)) = (1/2)*V(2*n,P = 8,Q = 17), where V(n,P,Q) denotes the Lucas sequence of the second kind and i=sqrt(-1). - Peter Bala, Sep 24 2019

A067358 Imaginary part of (5+12i)^n.

Original entry on oeis.org

0, 12, 120, -828, -28560, -145668, 3369960, 58317492, 13651680, -9719139348, -99498527400, 647549275812, 23290743888720, 123471611274972, -2701419604443960, -47880898349909868, -22269070348069440, 7869181117654073292, 82455284065364468280, -505338768229893703548
Offset: 0

Views

Author

Barbara Haas Margolius, (b.margolius(AT)csuohio.edu), Jan 17 2002

Keywords

Comments

Also 13^n sin(2n arctan(2/3)) or numerator of tan(2n arctan(2/3)).
Note that a(n), A067359(n) and 13^n are primitive Pythagorean triples with hypotenuse 13^n.

References

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

Crossrefs

Cf. A067359 (13^n cos(2n arctan(2/3))).

Programs

  • Maple
    a[1] := 12/5; for n from 1 to 40 do a[n+1] := (12/5+a[n])/(1-12/5*a[n]):od: seq(abs(numer(a[n])), n=1..40);# a[n]=tan(2n arctan(2/3))
  • Mathematica
    Im[(5 + 12*I)^Range[0, 24]] (* or *)
    LinearRecurrence[{10, -169}, {0, 12}, 25] (* Paolo Xausa, Apr 22 2024 *)
  • PARI
    a(n)=imag((5+12*I)^n)

Formula

G.f.: 12*x/(1-10*x+169*x^2). a(n)=10*a(n-1)-169*a(n-2). - Michael Somos, Jun 27 2002

Extensions

Better description from Michael Somos, Jun 27 2002

A067359 Real part of (5 + 12i)^n.

Original entry on oeis.org

1, 5, -119, -2035, -239, 341525, 3455641, -23161315, -815616479, -4241902555, 95420159401, 1671083125805, 584824319281, -276564805068235, -2864483360640839, 18094618450123325, 665043872449535041, 3592448206424508485, -76467932379726337079, -1371803070683005304755
Offset: 1

Views

Author

Barbara Haas Margolius, (b.margolius(AT)csuohio.edu), Jan 17 2002

Keywords

Comments

Also 13^n*cos(2*n*arctan(2/3)) or denominator of tan(2*n*arctan(2/3)).
Note that A067358(n), a(n) and 13^n are primitive Pythagorean triples with hypotenuse 13^n.

References

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

Crossrefs

Cf. A067358 (13^n sin(2n arctan(2/3))).

Programs

  • Maple
    a[1] := 12/5; for n from 1 to 40 do a[n+1] := (12/5+a[n])/(1-12/5*a[n]):od: seq(abs(denom(a[n])), n=1..40);# a[n]=tan(2n arctan(2/3))
  • Mathematica
    Table[Re[(5+12I)^n],{n,0,20}] (* Harvey P. Dale, Aug 24 2014 *)
  • PARI
    a(n)=real((5+12*I)^n)

Formula

From Michael Somos, Jun 27 2002: (Start)
G.f.: (1-5*x)/(1-10*x+169*x^2).
a(n) = 10*a(n-1) - 169*a(n-2). (End)

Extensions

Better description from Michael Somos, Jun 27 2002

A020891 Ordered set of c + a - b as (a,b,c) runs through all primitive Pythagorean triples with a

Original entry on oeis.org

4, 6, 8, 10, 10, 12, 14, 14, 16, 18, 18, 20, 22, 22, 24, 26, 26, 28, 28, 30, 30, 32, 34, 34, 36, 36, 38, 38, 40, 42, 42, 42, 44, 44, 46, 46, 48, 48, 50, 50, 52, 52, 54, 54, 56, 58, 58, 60, 60, 60, 62, 62, 64, 66, 66, 66, 66, 68, 68, 70, 70, 72, 74, 74, 76, 76, 78, 78, 78, 78, 80, 82, 82
Offset: 1

Views

Author

Keywords

Crossrefs

Cf. A020892.

Programs

  • Mathematica
    pyth[nn_]:=Module[{tr={}}, Do[If[CoprimeQ[m,n]&&Mod[m+n,2]==1,AppendTo[tr,{m^2-n^2,2 m n,m^2+n^2}] ], {m,2,nn},{n,1,m-1}];tr]; Take[#[[3]]+#[[1]]-#[[2]]&/@(Sort/@pyth[50])//Sort,80] (* Harvey P. Dale, Jan 24 2025 *)

Formula

a(n) = 2*A020892(n).

Extensions

Extended and corrected by David W. Wilson
Offset corrected to 1 by Ray Chandler, Jan 23 2020
Showing 1-5 of 5 results.