A050140 a(n) = 2*floor(n*phi)-n, where phi = (1+sqrt(5))/2.

1, 4, 5, 8, 11, 12, 15, 16, 19, 22, 23, 26, 29, 30, 33, 34, 37, 40, 41, 44, 45, 48, 51, 52, 55, 58, 59, 62, 63, 66, 69, 70, 73, 76, 77, 80, 81, 84, 87, 88, 91, 92, 95, 98, 99, 102, 105, 106, 109, 110, 113, 116, 117, 120, 121, 124, 127, 128, 131, 134, 135, 138, 139

Offset: 1

Author

Clark Kimberling

Keywords

• nonn

Comments

Old name was a(n) = last number in repeating block in continued fraction for n*phi.

References

• Alfred Brousseau, Fibonacci and Related Number Theoretic Tables, The Fibonacci Association, 1972, 101-103.

Links

• Clark Kimberling, Table of n, a(n) for n = 1..1000
• J.-P. Allouche and F. M. Dekking, Generalized Beatty sequences and complementary triples, arXiv:1809.03424 [math.NT], 2018.

Crossrefs

Cf. A000201, A001622, A005206, A050141, A005614, A001350.

Programs

Magma

[-n + 2*Floor(n*(1+Sqrt(5))/2): n in [1..50]]; // G. C. Greubel, Oct 15 2017

Mathematica

Table[-n+2Floor[n*GoldenRatio],{n,1,100}]

PARI

for(n=1,50, print1(-n + 2*floor(n*(1+sqrt(5))/2), ", ")) \\ G. C. Greubel, Oct 15 2017

Python

def A050140(n): return (n+isqrt(5*n**2)&-2)-n # Chai Wah Wu, Aug 25 2022

Formula

a(n) = -n + 2*floor(n*phi) = A283233(n)-n.

a(n) = floor(n*phi) + floor(n*sigma) where phi = (sqrt(5)+1)/2 and sigma = (sqrt(5)-1)/2.

a(n) = last number in repeating block in continued fraction for n*phi.

Extensions

Formula and more terms from Vladeta Jovovic, Nov 23 2001

Name changed by Michel Dekking, Dec 27 2017