close
login
A098317
Decimal expansion of phi^3 = 2 + sqrt(5).
48
4, 2, 3, 6, 0, 6, 7, 9, 7, 7, 4, 9, 9, 7, 8, 9, 6, 9, 6, 4, 0, 9, 1, 7, 3, 6, 6, 8, 7, 3, 1, 2, 7, 6, 2, 3, 5, 4, 4, 0, 6, 1, 8, 3, 5, 9, 6, 1, 1, 5, 2, 5, 7, 2, 4, 2, 7, 0, 8, 9, 7, 2, 4, 5, 4, 1, 0, 5, 2, 0, 9, 2, 5, 6, 3, 7, 8, 0, 4, 8, 9, 9, 4, 1, 4, 4, 1, 4, 4, 0, 8, 3, 7, 8, 7, 8, 2, 2, 7, 4, 9, 6
OFFSET
1,1
COMMENTS
This sequence is also the decimal expansion of ((1+sqrt(5))/2)^3. - Mohammad K. Azarian, Apr 14 2008
This is the length/width ratio of a 4-extension rectangle; see A188640 for definitions. - Clark Kimberling, Apr 10 2011
Its continued fraction is [4, 4, ...] (see A010709). - Robert G. Wilson v, Apr 10 2011
REFERENCES
Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, pages 138-139.
Alexey Stakhov, The mathematics of harmony: from Euclid to contemporary mathematics and computer science, World Scientific, Singapore, 2009, p. 657.
LINKS
Kunle Adegoke, Some Infinite Product Identities Involving Fibonacci and Lucas Numbers, The Fibonacci Quarterly, Vol. 55, No. 4 (2017), pp. 343-351.
Wikipedia, Metallic mean.
FORMULA
2 plus the constant in A002163. - R. J. Mathar, Sep 02 2008
Equals 3 + 4*sin(Pi/10) = 1 + 4*cos(Pi/5) = 1 + 4*sin(3*Pi/10) = 3 + 4*cos(2*Pi/5) = 1 + csc(Pi/10). - Arkadiusz Wesolowski, Mar 11 2012
Equals lim_{n -> infinity} F(n+3)/F(n) = lim_{n -> infinity} (1 + 2*F(n+1)/F(n)) = 2 + sqrt(5), with F(n) = A000045(n). - Arkadiusz Wesolowski, Mar 11 2012
Equals exp(arcsinh(2)), since arcsinh(x) = log(x+sqrt(x^2+1)). - Stanislav Sykora, Nov 01 2013
Equals Sum_{n>=1} n/phi^n = phi/(phi-1)^2 = phi^3. - Richard R. Forberg, Jun 29 2014
Equals 1 + 2*phi, with phi = A001622, an integer in the quadratic number field Q(sqrt(5)). - Wolfdieter Lang, Dec 10 2022
c^n = A001076(n-1) + c * A001076(n); where c = 2 + sqrt(5). - Gary W. Adamson, Oct 09 2023
Equals lim_{n -> infinity} = S(n, 2*(-1 + 2*phi))/S(n-1, 2*(-1 + 2*phi)), with the S-Chebyshev polynomials (see A049310). See also the above limit formula with Fibonacci numbers. - Wolfdieter Lang, Nov 15 2023
From Amiram Eldar, Jan 03 2026: (Start)
Formulas from Adegoke (2017):
Equals Product_{k>=1} (sqrt(5)*Fibonacci(2*k)+1)/(sqrt(5)*Fibonacci(2*k)-1), where Fibonacci(n) = A000045(n).
Equals Product_{k>=1} (Lucas(2*k)-(-1)^k*sqrt(5))/(Lucas(2*k)+(-1)^k*sqrt(5)), where Lucas(n) = A000032(n). (End)
EXAMPLE
4.23606797749978969640917366873127623544061835961152572427...
MATHEMATICA
RealDigits[N[2+Sqrt[5], 200]][[1]] (* Vladimir Joseph Stephan Orlovsky, Feb 21 2011 *)
PROG
(PARI) sqrt(5)+2 \\ Charles R Greathouse IV, Mar 11 2012
(Magma) SetDefaultRealField(RealField(100)); 2+Sqrt(5); // G. C. Greubel, Jun 30 2019
(SageMath) numerical_approx(2+sqrt(5), digits=100) # G. C. Greubel, Jun 30 2019
KEYWORD
nonn,cons,easy
AUTHOR
Eric W. Weisstein, Sep 02 2004
EXTENSIONS
Title expanded to include observation from Mohammad K. Azarian by Charles R Greathouse IV, Mar 11 2012
STATUS
approved