A173300 -id:A173300 - cp's OEIS frontend

A173989 a(n) is the 2-adic valuation of A173300(n).

0, 0, 1, 1, 2, 1, 3, 3, 4, 3, 5, 5, 6, 5, 7, 7, 8, 7, 9, 9, 10, 9, 11, 11, 12, 11, 13, 13, 14, 13, 15, 15, 16, 15, 17, 17, 18, 17, 19, 19, 20, 19, 21, 21, 22, 21, 23, 23, 24, 23, 25, 25, 26, 25, 27, 27, 28, 27, 29, 29, 30, 29, 31, 31, 32, 31, 33, 33, 34, 33, 35, 35, 36, 35, 37, 37, 38, 37

Offset: 1

J. Lowell, Mar 04 2010

Conjecture: always follows the pattern A, A, A+1, A, where A is an odd number.

Hugo Pfoertner, Table of n, a(n) for n = 1..1000

Cf. A102302, A173300.

From R. J. Mathar, Mar 20 2010: (Start)
A173300 := proc(n) local x,y ; x := (1+sqrt(3))/2 ; y := (1-sqrt(3))/2 ; denom(expand(x^n+y^n)) ; end proc:
A173989 := proc(n) log[2](A173300(n)) ; end proc: seq(A173989(n),n=3..100) ; (End)

Log2[Denominator[Map[First, NestList[{Last[#], Last[#] + First[#]/2} &, {1, 2}, 100]]]] (* Paolo Xausa, Feb 01 2024, after Nick Hobson in A173300 *)

\\ using Max Alekseyev's function in A173300
A173300(n) = denominator(2*polcoeff( lift( Mod((1+x)/2, x^2-3)^n ), 0))
for(k=1,74,print1(logint(A173300(k),2),", ")) \\ Hugo Pfoertner, Oct 10 2018

a(n) = log(A173300(n))/log(2).
Apparently a(n) = A102302(n) for n >= 7. - Hugo Pfoertner, Oct 10 2018
Conjectures from Colin Barker, Oct 10 2018: (Start)
G.f.: x^3*(1 + x^2 - x^3 + x^4) / ((1 - x)^2*(1 + x)*(1 + x^2)).
a(n) = a(n-1) + a(n-4) - a(n-5) for n > 7.
(End)
Apparently a(n) = A116921(n) for n>=3. - R. J. Mathar, Aug 29 2025

More terms from R. J. Mathar and Max Alekseyev, Mar 20 2010

A173299 Numerators of fractions x^n + y^n, where x + y = 1 and x^2 + y^2 = 2.

Original entry on oeis.org

1, 2, 5, 7, 19, 13, 71, 97, 265, 181, 989, 1351, 3691, 2521, 13775, 18817, 51409, 35113, 191861, 262087, 716035, 489061, 2672279, 3650401, 9973081, 6811741, 37220045, 50843527, 138907099, 94875313, 518408351, 708158977, 1934726305, 1321442641

Offset: 1

J. Lowell, Feb 15 2010

x and y are given by -A152422 and 1-A152422. - R. J. Mathar, Mar 01 2010

Letting f(n) = x^n + y^n, recurrence relation f(n) = f(n - 1) + f(n - 2)/2 implies a(n) / A173300(n) = A026150(n) / 2^(n - 1). - Nick Hobson, Jan 30 2024

			a(3) = 5 because x^3 + y^3 is 2.5 and 2.5 is 5/2.

Vincenzo Librandi, Table of n, a(n) for n = 1..200

Cf. A173300 (denominators).

Z:=PolynomialRing(Integers()); N:=NumberField(2*x^2-2*x-1); S:=[ r^n+(1-r)^n: n in [1..34] ]; [ Numerator(RationalField()!S[j]): j in [1..#S] ]; // Klaus Brockhaus, Mar 02 2010

A173299 := proc(n) local x,y ; x := (1+sqrt(3))/2 ; y := (1-sqrt(3))/2 ; expand(x^n+y^n) ; numer(%) ; end proc: # R. J. Mathar, Mar 01 2010

Module[{x=(1-Sqrt[3])/2,y},y=1-x;Table[x^n+y^n,{n,40}]]//Simplify// Numerator (* Harvey P. Dale, Aug 24 2019 *)

{ a(n) = numerator( 2 * polcoeff( lift( Mod((1+x)/2,x^2-3)^n ), 0) ) }

from fractions import Fraction
def a173299_gen(a, b):
    while True:
        yield a.numerator
        b, a = b + Fraction(a, 2), b
g = a173299_gen(1, 2)
print([next(g) for  in range(34)])  # _Nick Hobson, Feb 20 2024

a(n) = numerator of ((1 + sqrt(3))/2)^n + ((1 - sqrt(3))/2)^n.

Formula, more terms, and PARI script from Max Alekseyev, Feb 24 2010

More terms from Klaus Brockhaus and R. J. Mathar, Mar 01 2010

A305491 a(n) = numerator(r(n)) where r(n) = (((1/2)(sqrt(3) + 1))^n - ((1/2)(sqrt(3) - 1))^n * cos(Pi*n))/sqrt(3).

Original entry on oeis.org

0, 1, 1, 3, 2, 11, 15, 41, 7, 153, 209, 571, 195, 2131, 2911, 7953, 679, 29681, 40545, 110771, 37829, 413403, 564719, 1542841, 263445, 5757961, 7865521, 21489003, 7338631, 80198051, 109552575, 299303201, 12776743, 1117014753, 1525870529, 4168755811, 1423656585

Offset: 0

Peter Luschny, Jun 02 2018

nonn

Let f(x, y) = ((y+1)^x - (y-1)^x * cos(Pi*x))/(y * 2^x). Then f(n, sqrt(3)) are the rational numbers a(n)/A060723(n) and f(n, sqrt(5)) the Fibonacci numbers A000045(n).
From Paul Curtz, Dec 05 2018: (Start)
The binomial inverse of the rational sequence r(n) starts 0, 1, -1, 3/2, -2, 11/4, -15/4, 41/8, -7, 153/16, -209/16, ... and is up to signs equal to r(n). The difference table starts:
      0,     1,    1,  3/2,    2, 11/4,  15/4,   41/8, ...
      1,     0,  1/2,  1/2,  3/4,    1,  11/8,   15/8, ...
     -1,   1/2,    0,  1/4,  1/4,  3/8,   1/2,  11/16, ...
    3/2,  -1/2,  1/4,    0,  1/8,  1/8,  3/16,    1/4, ...
    ...
Let s(n) = 2*r(n+1) - r(n) then s(n) = 1, 2, 5/2, 7/2, 19/4, 13/2, ... = A173299(n)/A173300(n) for n >= 1. (End)

Cf. A060723 (denominators), A060755, A000045, A305492.

Cf. A173299/A173300.

Table[Numerator[Simplify[((1/2 (Sqrt[3] + 1))^x - (1/2 (Sqrt[3] - 1))^x Cos[Pi  x])/Sqrt[3]]], {x, 0, 36}]

A recurrence for r(n) is given in A060723.

A274845 a(0)=1, a(1)=0, a(4n+2) = a(4n+3) = a(4n+5) = (4^(n+1) +(-1)^n)/5, a(4n+4) = (24^(n+1) -3(-1)^n)/5.

Original entry on oeis.org

1, 0, 1, 1, 1, 1, 3, 3, 7, 3, 13, 13, 25, 13, 51, 51, 103, 51, 205, 205, 409, 205, 819, 819, 1639, 819, 3277, 3277, 6553, 3277, 13107, 13107, 26215, 13107, 52429, 52429, 104857, 52429, 209715, 209715, 419431, 209715, 838861, 838861, 1677721, 838861, 3355443

Offset: 0

Paul Curtz, Jul 08 2016

Antidiagonals of the array in A274613 written as a triangle:
1,
0, 1/2,
0, 1/2, 1/4,
0,   0, 1/2, 1/8,
0,   0, 1/4, 3/8, 1/16,
... .
a(n) is the numerators of the antidiagonal sums i.e. 1, 0, 1/2, 1/2, 1/4, 1/2, 3/8, 3/8, 7/16, 3/8, 13/32, 13/32, 25/64, 13/32, ... = a(n)/b(n).
The denominators b(n) are A173300(n).
a(0)+a(1) = 1, a(4n+2) +a(4n+3) +a(4n+4) +a(4n+5) = 4, 16, 64, 256, ... .

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 3, 0, 0, 0, 4).

Cf. A000302, A102900, A015521, A173300, A274613.

LinearRecurrence[{0,0,0,3,0,0,0,4}, {1,0,1,1,1,1,3,3}, 50] (* G. C. Greubel, Jul 08 2016 *)

Vec((1+x^2+x^3-2*x^4+x^5)/((1-2*x^2)*(1+2*x^2)*(1+x^4)) + O(x^60)) \\ Colin Barker, Jul 22 2016

a(4n) = A102900(n), a(4n+1) = A015521(n), a(4n+2) = a(4n+3) = A015521(n+1).
a(n) = 3*a(n-4) + 4*a(n-8). - G. C. Greubel, Jul 08 2016
G.f.: (1+x^2+x^3-2*x^4+x^5) / ((1-2*x^2)*(1+2*x^2)*(1+x^4)). - Colin Barker, Jul 22 2016

More terms from Colin Barker, Jul 22 2016

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A173989 a(n) is the 2-adic valuation of A173300(n).

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A173299 Numerators of fractions x^n + y^n, where x + y = 1 and x^2 + y^2 = 2.

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A305491 a(n) = numerator(r(n)) where r(n) = (((1/2)*(sqrt(3) + 1))^n - ((1/2)*(sqrt(3) - 1))^n * cos(Pi*n))/sqrt(3).

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A274845 a(0)=1, a(1)=0, a(4n+2) = a(4n+3) = a(4n+5) = (4^(n+1) +(-1)^n)/5, a(4n+4) = (2*4^(n+1) -3*(-1)^n)/5.

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A305491 a(n) = numerator(r(n)) where r(n) = (((1/2)(sqrt(3) + 1))^n - ((1/2)(sqrt(3) - 1))^n * cos(Pi*n))/sqrt(3).

A274845 a(0)=1, a(1)=0, a(4n+2) = a(4n+3) = a(4n+5) = (4^(n+1) +(-1)^n)/5, a(4n+4) = (24^(n+1) -3(-1)^n)/5.