A121622 -id:A121622 - 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

A188983 Odd numbers y such that x^2 + y^2 = 13^n with x and y coprime.

Original entry on oeis.org

1, 3, 5, 9, 119, 597, 2035, 4449, 239, 56403, 341525, 1315911, 3455641, 3627003, 23161315, 186118929, 815616479, 2474152797, 4241902555, 6712571031, 95420159401, 485257533003, 1671083125805, 3718150825791, 584824319281, 44827014819597, 276564805068235, 1076637637754649, 2864483360640839, 3190610873034597, 18094618450123325

Offset: 0

T. D. Noe, Apr 14 2011

nonn

The x values are in A188982.

This is also the absolute value of the real part of (3+2i)^n, where i = sqrt(-1). The signed version is A121622.

Vincenzo Librandi, Table of n, a(n) for n = 0..183

Cf. A121622, A158936, A188982.

[Integers()!Abs(Real((3+2*Sqrt(-1))^n)): n in [0..30]]; // Bruno Berselli, May 26 2011

Table[s = Select[PowersRepresentations[13^n, 2, 2], CoprimeQ @@ # &, 1][[1]]; Select[s, OddQ][[1]], {n, 0, 30}]

A193410 Expansion of (1-3x)/(1-6x+18*x^2).

Original entry on oeis.org

1, 3, 0, -54, -324, -972, 0, 17496, 104976, 314928, 0, -5668704, -34012224, -102036672, 0, 1836660096, 11019960576, 33059881728, 0, -595077871104, -3570467226624, -10711401679872, 0, 192805230237696, 1156831381426176, 3470494144278528

Offset: 0

Bruno Berselli, Aug 04 2011

Also real parts of 3^n*(1+i)^n, where i=sqrt(-1).

If |a(n)| > 0 then it is in A130505.

Bruno Berselli, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,-18).

Cf. A066771, A121622, A130505.

m:=26; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((1-3*x)/(1-6*x+18*x^2))); /* or */ &cat[[r,3*r,0,-54*r] where r is (-324)^n: n in [0..6]];

I:=[1, 3]; [n le 2 select I[n] else 6*Self(n-1)-18*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Mar 26 2013

CoefficientList[Series[(1 - 3 x)/(1 - 6 x + 18 x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Mar 26 2013 *)
LinearRecurrence[{6,-18},{1,3},40] (* Harvey P. Dale, Jul 27 2021 *)

makelist(coeff(taylor((1-3*x)/(1-6*x+18*x^2), x, 0, n), x, n), n, 0, 25);

PARI
```
Vec((1-3*x)/(1-6*x+18*x^2) +O(x^26))
```

G.f.: (1-3*x)/(1-6*x+18*x^2).
a(n) = 3^n*A146559(n) = (1/2)*((3+3*i)^n+(3-3*i)^n), where i=sqrt(-1).
a(n) = 6*a(n-1)-18*a(n-2) for n>1.
a(n) = (3*sqrt(2))^n*cos(pi*n/4).
a(4k+2) = 0, a(4k+1) = 3*a(4k) = 18*a(4k-1) = 3*(-324)^k.
G.f.: W(0)/2, where W(k) = 1 + 1/(1 - x*(3*k+3)/(x*(3*k+6) + 1/W(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 28 2013

A348652 For any nonnegative number n with base-13 expansion Sum_{k >= 0} d_k13^k, a(n) is the real part of Sum_{k >= 0} g(d_k)(3+2i)^k where g(0) = 0, and g(1+u+3v) = (1+ui)i^v for any u = 0..2 and v = 0..3 (where i denotes the imaginary unit); see A348653 for the imaginary part.

Original entry on oeis.org

0, 1, 1, 1, 0, -1, -2, -1, -1, -1, 0, 1, 2, 3, 4, 4, 4, 3, 2, 1, 2, 2, 2, 3, 4, 5, 1, 2, 2, 2, 1, 0, -1, 0, 0, 0, 1, 2, 3, -1, 0, 0, 0, -1, -2, -3, -2, -2, -2, -1, 0, 1, -2, -1, -1, -1, -2, -3, -4, -3, -3, -3, -2, -1, 0, -5, -4, -4, -4, -5, -6, -7, -6, -6, -6

Offset: 0

Rémy Sigrist, Oct 27 2021

The function f defines a bijection from the nonnegative integers to the Gaussian integers.
The following diagram depicts g(d) for d = 0..12:
                      |
                      |    +
                      |     3
                      |
            +    +    +    +
             6    5   |4    2
                      |
         --------+----+----+-------
                  7   |0    1
                      |
                 +    +    +    +
                  8   |10   11   12
                      |
                 +    |
                  9   |

Rémy Sigrist, Table of n, a(n) for n = 0..2197
Stephen K. Lucas, Base 2 + i with digit set {0, +/-1, +/-i}, ResearchGate (October 2021).
Rémy Sigrist, Colored representation of f for n = 0..13^5-1 in the complex plane (the hue is function of n)

See A316657 for a similar sequence.

Cf. A121622, A348653.

g(d) = { if (d==0, 0, (1+I*((d-1)%3))*I^((d-1)\3)) }
a(n) = real(subst(Pol([g(d)|d<-digits(n, 13)]), 'x, 3+2*I))

a(13^k) = A121622(k).

A121621 Real part of (2 + 3i)^n.

Original entry on oeis.org

1, 2, -5, -46, -119, 122, 2035, 6554, -239, -86158, -341525, -246046, 3455641, 17021162, 23161315, -128629846, -815616479, -1590277918, 4241902555, 37641223154, 95420159401, -107655263398, -1671083125805, -5284814079046, 584824319281

Offset: 0

Gary W. Adamson and Nick Williams, Aug 10 2006

A121622 is the companion sequence generated from (3 + 2i).

			a(5) = 122 since (2 + 3i)^5 = (122 - 597i).
a(5) = 122 = 4*(-119) - 13*(-46) = 4*a(4) - 13*a(3).

Nathaniel Johnston, Table of n, a(n) for n = 0..500
Beata Bajorska-Harapińska, Barbara Smoleń, Roman Wituła, On Quaternion Equivalents for Quasi-Fibonacci Numbers, Shortly Quaternaccis, Advances in Applied Clifford Algebras (2019) Vol. 29, 54.
Index entries for linear recurrences with constant coefficients, signature (4,-13)

Cf. A121622.

A121621:=proc(n)global a:if(n=0)then a[0]:=1:elif(n=1)then a[1]:=2:else a[n]:=4*a[n-1]-13*a[n-2]:fi:return a[n]:end:
seq(A121621(n),n=0..20); # Nathaniel Johnston, Apr 15 2011

f[n_] := Re[(2 + 3I)^n]; Table[f[n], {n, 0, 24}] (* Robert G. Wilson v *)

Re(2 + 3i)^n = a(n) = 4*a(n-1) - 13*a(n-2).

G.f.: ( 1-2*x ) / ( 1 - 4*x + 13*x^2 ). - R. J. Mathar, Mar 03 2013

More terms from Robert G. Wilson v, Aug 17 2006

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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A188983 Odd numbers y such that x^2 + y^2 = 13^n with x and y coprime.

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A193410 Expansion of (1-3*x)/(1-6*x+18*x^2).

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A348652 For any nonnegative number n with base-13 expansion Sum_{k >= 0} d_k*13^k, a(n) is the real part of Sum_{k >= 0} g(d_k)*(3+2*i)^k where g(0) = 0, and g(1+u+3*v) = (1+u*i)*i^v for any u = 0..2 and v = 0..3 (where i denotes the imaginary unit); see A348653 for the imaginary part.

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A121621 Real part of (2 + 3i)^n.

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

A193410 Expansion of (1-3x)/(1-6x+18*x^2).

A348652 For any nonnegative number n with base-13 expansion Sum_{k >= 0} d_k13^k, a(n) is the real part of Sum_{k >= 0} g(d_k)(3+2i)^k where g(0) = 0, and g(1+u+3v) = (1+ui)i^v for any u = 0..2 and v = 0..3 (where i denotes the imaginary unit); see A348653 for the imaginary part.