A139030 -id:A139030 - cp's OEIS frontend

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

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

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

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

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

Seiichi Manyama, Table of n, a(n) for n = 0..1000
J. M. Borwein and R. Girgensohn, Addition theorems and binary expansions, Canadian J. Math. 47 (1995) 262-273.
E. Eckert, The group of primitive Pythagorean triangles, Mathematics Magazine 57 (1984) 22-27.
Steven R. Finch, Plouffe's Constant [Broken link]
Steven R. Finch, Plouffe's Constant [From the Wayback machine]
Simon Plouffe, The Computation of Certain Numbers Using a Ruler and Compass, J. Integer Seqs. Vol. 1 (1998), #98.1.3.
Marc Renault, The Period, Rank, and Order of the (a,b)-Fibonacci Sequence mod m, Math. Mag. 86 (2013) pp. 372-380.
Index entries for linear recurrences with constant coefficients, signature (6,-25).

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.

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))

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))
```

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

A139031 Imaginary part of (4 + 3i)^n.

Original entry on oeis.org

3, 24, 117, 336, -237, -10296, -76443, -354144, -922077, 1476984, 34867797, 242017776, 1064447283, 2465133864, -6890111163, -116749235904, -761741108157, -3175197967656, -6358056037323, 28515500892816, 387075408075603, 2383715742284424, 9392840736385317, 15549832333971936

Offset: 1

Gary W. Adamson, Apr 06 2008

Division of each term by 3 generates an integer sequence 1, 8, 39, 112, -79, -3432, -25481, -118048, -307359, 492328, ... - R. J. Mathar, Apr 08 2008

			a(3) = 117 since (4 + 3i)^3 = (-44 + 117i).
a(4) = 336 = 8*a(3) - 25*a(2) = 8*117 - 25*24.
a(3) = 117 = term (2,1) of [4,-3; 3,4]^3.

Index entries for linear recurrences with constant coefficients, signature (8,-25).

Cf. A139030, A066771, A066776.

a[n_]:=Im[(4+3I)^n];Array[a,24] (* James C. McMahon, Jun 25 2025 *)

a(n) = imag((4 + 3*I)^n); \\ Michel Marcus, Jun 25 2025

Imaginary part of (4 + 3i)^n.
Term (2,1) of [4,-3; 3,4]^n.
a(n)^2 + A139030(n)^2 = 5^(2*n).
a(n) = 8*a(n-1) - 25*a(n-2), n>2, given a(1) = 3, a(2) = 24.
(unsigned): Odd-indexed terms of A066771 interleaved with even-indexed terms of A066776.
O.g.f.: 3*x/(1-8*x+25*x^2). - R. J. Mathar, Apr 08 2008

A376283 a(n) = (40)^n * cos(nB), where B is the angle opposite side CA in a triangle ABC having sidelengths |BC|=3, |CA|=4, |AB|=5; ABC is the smallest integer-sided right triangle.

Original entry on oeis.org

1, 24, -448, -59904, -2158592, -7766016, 3080978432, 160312590336, 2765438844928, -123759079981056, -10365137990975488, -299512095597133824, 2207640196898357248, 585186082406535266304, 24556707640476321185792, 242424234892406990831616

Offset: 0

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; 17 divides a(8n+4) for n>= 0. See the Renault paper in References.

Marc Renault, "The Period, Rank, and Order of the (a,b)-Fibonacci Sequence mod m", Math. Mag. 86 (2013) 372 - 380.

Index entries for linear recurrences with constant coefficients, signature (48,-1600).

Cf. A374880, A139030, A376285, A376455.

(*Program 1*)
A[a_, b_, c_] := ArcCos[(b^2 + c^2 - a^2)/(2  b  c)];
Table[TrigExpand[(20)^n  Cos[n  A[4, 5, 3]]], {n, 0, 30}]
(*Program 2*)
LinearRecurrence[{48, -1600}, {1, 24}, 30]

a(n) = (40)^n * cos(nB), where B is the angle opposite side CA in a triangle ABC having sidelengths |BC|=3, |CA|=4, |AB|=5; ABC is the smallest integer-sided right triangle.
a(n) = 48 a(n-1) - 1600 a(n-2), where a(0) = 1, a(1) = 24.
From Stefano Spezia, Oct 03 2024: (Start)
G.f.: (1 - 24*x)/(1 - 48*x + 1600*x^2).
E.g.f.: exp(24*x)*cos(32*x). (End)

A376285 a(n) = 20^n * cos(n*A), where A is the angle opposite side BC in a triangle ABC having sidelengths |BC|=3, |CA|=4, |AB|=5; ABC is the smallest integer-sided right triangle.

Original entry on oeis.org

1, 16, 112, -2816, -134912, -3190784, -48140288, -264175616, 10802495488, 451350102016, 10122205069312, 143370521411584, 538974657445888, -40101019526365184, -1498822487822041088, -31921911799759241216, -421972182463479283712, -734345118927640592384

Offset: 0

Clark Kimberling, Oct 03 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(4*n+2) for n >= 0; 17 divides a(8*n+3) for n >= 0. See the Renault paper in References.

Marc Renault, "The Period, Rank, and Order of the (a,b)-Fibonacci Sequence mod m", Math. Mag. 86 (2013) 372 - 380.

Index entries for linear recurrences with constant coefficients, signature (32,-400).

Cf. A066771, A374880, A139030, A376455.

(*Program 1*)
A[a_, b_, c_] := ArcCos[(b^2 + c^2 - a^2)/(2  b  c)];
Table[TrigExpand[(20)^n  Cos[n  A[3, 4, 5]]], {n, 0, 30}]
(*Program 2*)
LinearRecurrence[{32, -400}, {1, 16}, 30]

a(n) = 20^n * cos(n*A), where A is the angle opposite side BC in a triangle ABC having sidelengths |BC|=3, |CA|=4, |AB|=5; ABC is the smallest integer-sided right triangle.
a(n) = 32*a(n-1) - 400*a(n-2), where a(0) = 1, a(1) = 16.
From Stefano Spezia, Oct 03 2024: (Start)
G.f.: (1 - 16*x)/(1 - 32*x + 400*x^2).
E.g.f.: exp(16*x)*cos(12*x). (End)

A376322 (1/4) times obverse convolution (2)**(2^n + 1); see Comments.

Original entry on oeis.org

1, 5, 35, 385, 7315, 256025, 17153675, 2247131425, 582007039075, 299733625123625, 307826433001962875, 631352014087025856625, 2587911905742718986305875, 21207938067561582092776645625, 347534481113131645754330891856875, 11389052480558437163015177657041650625

Offset: 0

Clark Kimberling, Sep 20 2024

nonn

See A374848 for the definition of obverse convolution and a guide to related sequences.

Cf. A000051, A000079, A007395, A010701, A374848, A375054, A139030.

s[n_] := 2; t[n_] := 2^n + 1;
u[n_] := (1/4) Product[s[k] + t[n - k], {k, 0, n}];
Table[u[n], {n, 0, 20}]
(* or *)
Table[2^(n*(n+1)/2 - 2) * QPochhammer[-3, 1/2, n+1], {n, 0, 15}] (* Vaclav Kotesovec, Sep 20 2024 *)

a(n) = a(n-1)*A062709(n) for n>=1.

a(n) = (1/4)((3)**(2^n)) = (1/4)(A010701(n)**A000079(n)) for n>=0.

A376323 (1/5) times obverse convolution (3)**(2^n + 1); see Comments.

Original entry on oeis.org

1, 6, 48, 576, 11520, 414720, 28200960, 3722526720, 967856947200, 499414184755200, 513397781928345600, 1053492248516965171200, 4319318218919557201920000, 35401132122264690826936320000, 580153753219673753271832412160000, 19012798800515148242224491811307520000

Offset: 0

Clark Kimberling, Sep 20 2024

nonn

See A374848 for the definition of obverse convolution and a guide to related sequences.

Cf. A000051, A010701, A052548, A374848, A375053, A139030.

s[n_] := 3; t[n_] := 2^n + 1;
u[n_] := (1/5) Product[s[k] + t[n - k], {k, 0, n}];
Table[u[n], {n, 0, 20}]
(* or *)
Table[2^(n*(n+1)/2) * QPochhammer[-4, 1/2, n+1]/5, {n, 0, 15}] (* Vaclav Kotesovec, Sep 20 2024 *)

a(n) = 2 a(n-1)*A052548(n-1) for n>=1.

cp's OEIS Frontend

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

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A139031 Imaginary part of (4 + 3i)^n.

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A376283 a(n) = (40)^n * cos(nB), where B is the angle opposite side CA in a triangle ABC having sidelengths |BC|=3, |CA|=4, |AB|=5; ABC is the smallest integer-sided right triangle.

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A376285 a(n) = 20^n * cos(n*A), where A is the angle opposite side BC in a triangle ABC having sidelengths |BC|=3, |CA|=4, |AB|=5; ABC is the smallest integer-sided right triangle.

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A376322 (1/4) times obverse convolution (2)**(2^n + 1); see Comments.

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Mathematica

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A376323 (1/5) times obverse convolution (3)**(2^n + 1); see Comments.

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Mathematica

Formula

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