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 1-7 of 7 results.

A066771 a(n) = 5^n*cos(2*n*arctan(1/2)) or denominator of tan(2*n*arctan(1/2)).

Original entry on oeis.org

1, 3, -7, -117, -527, -237, 11753, 76443, 164833, -922077, -9653287, -34867797, 32125393, 1064447283, 5583548873, 6890111163, -98248054847, -761741108157, -2114245277767, 6358056037323, 91004468168113, 387075408075603, 47340744250793, -9392840736385317
Offset: 0

Views

Author

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

Keywords

Comments

Let A =
[ -3/5 -(2/5)i, -(2/5)i, -(2/5)i, -(2/5)i ]
[ -(2/5)i, -3/5 +(2/5)i, -(2/5)i, (2/5)i ]
[ -(2/5)i, -(2/5)i, -3/5 +(2/5)i, (2/5)i ]
[ -(2/5)i, (2/5)i, (2/5)i, -3/5 -(2/5)i ]
be the Cayley transform of the matrix iH, where H =
[1, 1, 1, 1]
[1, -1, 1, -1]
[1, 1, -1, -1]
[1, -1, -1, 1]
is a Hadamard matrix of order 4 and i is the imaginary unit. Any diagonal entry of the matrix A^n is one of the four complex numbers (+ or -)(X/5^n)(+ or -)(Y/(5^n)i). Then a(n) is the X in [A^n](j,j), j=1,2,3,4. - _Simone Severini, Apr 28 2004
Related to the (3,4,5) Pythagorean triple. Each unsigned term represents a leg in a Pythagorean triple in which the hypotenuse = 5^n. E.g., (3 + 4i)^3 = (-117 + 44i), considered as two legs of a triangle, hypotenuse = 125 = 5^3. - Gary W. Adamson, Aug 06 2006
a(n) = 5^n*cos(nC-nA), where C is the angle opposite side AB and A is the angle opposite side BC in a triangle ABC having sidelengths |BC|=3, |CA|=4, |AB|=5 - Clark Kimberling, Oct 02 2024
If a prime p divides a term, then the indices n such that p divides a(n) comprise an arithmetic sequence; e.g., 7 divides a(4n+2) for n >= 0; 13 divides a(6n+3) for n>= 0. See the Renault paper in References. - Clark Kimberling, Oct 03 2024

References

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

Links

Crossrefs

Cf. A066770 5^n sin(2n arctan(1/2)), A000351 powers of 5 and also hypotenuse of right triangle with legs given by A066770 and A066771.
Note that A066770, A066771 and A000351 are primitive Pythagorean triples with hypotenuse 5^n. The offset of A000351 is 0, but the offset is 1 for A066770, A066771.
Cf. A093378.
Cf. A193410, A121622.
Cf. A139030.

Programs

  • Maple
    a[1] := 4/3; for n from 1 to 40 do a[n+1] := (4/3+a[n])/(1-4/3*a[n]):od: seq(abs(denom(a[n])), n=1..40);# a[n]=tan(2n arctan(1/2))
  • Mathematica
    CoefficientList[Series[(1-3x)/(1-6x+25x^2),{x,0,30}],x] (* or *) LinearRecurrence[{6,-25},{1,3},30] (* Harvey P. Dale, Jul 16 2011 *)
  • PARI
    a(n)=real((2+I)^(2*n))

Formula

G.f.: ( 1-3*x ) / ( 1-6*x+25*x^2 ).
A recursive formula for T(n) = tan(2*n*arctan(1/2)) is T(n+1) = (4/3 + T(n))/(1 - (4/3)*T(n)). Unsigned A(n) is the absolute value of the denominator of T(n).
a(n) is the real part of (2+i)^(2n) = Sum_{k=0..n} 4^(n-k)*(-1)^k*C(2n, 2k). - Benoit Cloitre, Aug 03 2002
a(n) = real part of (3 + 4i)^n. - Gary W. Adamson, Aug 06 2006
a(n) = 6*a(n-1) - 25*a(n-2). - Gary Detlefs, Jun 10 2010
a(n) = 5^n*cos(n*arccos(3/5)). - Gary Detlefs, Dec 11 2010
a(n) = (-1)^n * hypergeom([1,-n,1/2-n],[1/2,1],-4). - Gerry Martens, Jul 28 2023

A139030 Real part of (4 + 3i)^n.

Original entry on oeis.org

1, 4, 7, -44, -527, -3116, -11753, -16124, 164833, 1721764, 9653287, 34182196, 32125393, -597551756, -5583548873, -29729597084, -98248054847, -42744511676, 2114245277767, 17982575014036, 91004468168113, 278471369994004, -47340744250793, -7340510203856444, -57540563024581727
Offset: 0

Views

Author

Gary W. Adamson, Apr 06 2008

Keywords

Comments

sqrt (a(n)^2 + (A139031(n))^2) = 5^n. Example: a(3) = -44, A139031(3) = 117. Sqrt (-44^2 + 117^2) = 5^3.
If a prime p divides a term, then the indices n such that p divides a(n) comprise an arithmetic sequence; e.g., 11 divides a(6n+3) for n >= 0; 31 divides a(8n+4) for n>= 0. See the Renault paper in Links. - Clark Kimberling, Oct 02 2024

Examples

			a(5) = -3116 since (4 + 3i)^5 = (-3116 - 237i) where -237 = A139031(5).

Links

Crossrefs

Cf. A139031, A066771, A066770, A374880, A376283, A376455.

Programs

  • Maple
    a:= n-> Re((4+3*I)^n):
seq(a(n), n=0..24);  # Alois P. Heinz, Oct 15 2024
  • Mathematica
    Re[(4+3I)^Range[40]] (* or *) LinearRecurrence[{8,-25},{4,7},40] (* Harvey P. Dale, Nov 09 2011 *)
A[a_, b_, c_] := ArcCos[(b^2 + c^2 - a^2)/(2 b c)];
{a, b, c} = {3, 4, 5};
Table[TrigExpand[5^n Cos[n (A[b, c, a] - A[c, a, b])]], {n, 0, 50}] (* Clark Kimberling, Oct 02 2024 *)

Formula

Real part of (4 + 3i)^n. Term (1,1) of [4,-3; 3,4]^n. a(n), n>=2 = 8*a(n-1) - 25*a(n-2), given a(0) = 1, a(1) = 4. Odd-indexed terms of A066770 interleaved with even-indexed terms of A066771, irrespective of sign.
G.f.: (1-4*x) / ( 1-8*x+25*x^2 ). - R. J. Mathar, Feb 05 2011
a(n) = 5^n * cos(nB-nC), where B is the angle opposite side CA and C is the angle opposite side AB in a triangle ABC having sidelengths |BC|=3, |CA|=4, |AB|=5; ABC is the smallest integer-sided right triangle. - Clark Kimberling, Oct 02 2024
E.g.f.: exp(4*x)*cos(3*x). - Stefano Spezia, Oct 03 2024

Extensions

More terms from Harvey P. Dale, Nov 09 2011
a(0)=1 prepended by Alois P. Heinz, Oct 15 2024

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.

Links

Crossrefs

Cf. A067361 (17^n cos(2n arctan(1/4))).
Cf. A066770, A066771, A067358, A067359, A020888, A014498, A020892.

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.

Links

Crossrefs

Cf. A067360 (17^n sin(2n arctan(1/4))).
Cf. A066770, A066771, A067358, A067359, A020888, A014498, A020892.

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.

Links

Crossrefs

Cf. A067359 (13^n cos(2n arctan(2/3))).
Cf. A066770, A066771, A067360, A067361, A020888, A014498, A020892.

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.

Links

Crossrefs

Cf. A067358 (13^n sin(2n arctan(2/3))).
Cf. A066770, A066771, A067360, A067361, A020888, A014498, A020892.

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

A105667 1/(2k)-th of area of primitive Pythagorean triangle with hypotenuse 5^(2^n), where k is the product of all Mersenne primes not exceeding 2^(n+2) - 1.

Original entry on oeis.org

1, 2, 4216, 44834576
Offset: 0

Views

Author

Lekraj Beedassy, May 04 2005

Keywords

Links

Crossrefs

Cf. A066770; A066771; A098918.
Showing 1-7 of 7 results.

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