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A169985
Round phi^n to the nearest integer.
25
1, 2, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, 843, 1364, 2207, 3571, 5778, 9349, 15127, 24476, 39603, 64079, 103682, 167761, 271443, 439204, 710647, 1149851, 1860498, 3010349, 4870847, 7881196, 12752043, 20633239, 33385282, 54018521, 87403803
OFFSET
0,2
COMMENTS
Phi = (1+sqrt(5))/2, see A001622.
a(n) is the number of subsets of {1,2,...,n} with no two consecutive elements where n and 1 are considered to be consecutive. - Geoffrey Critzer, Sep 23 2013
Equals the Lucas sequence beginning at 1 (A000204) with 2 inserted between 1 and 3.
The Lucas sequence beginning at 2 (A000032) can be written as L(n) = phi^n + (-1/phi)^n. Since |(-1/phi)^n|<1/2 for n>1, this sequence is {L(n)} (with the first two terms switched). As a consequence, for n>1: a(n) is obtained by rounding phi^n up for even n and down for odd n; a(n) is also the nearest integer to 1/|phi^n - a(n)|. - Danny Rorabaugh, Apr 15 2015
LINKS
John Machacek and George D. Nasr, Transversal and Paving Positroids, arXiv:2401.02053 [math.CO], 2024. See p. 23.
Shaoxiong (Steven) Yuan, Generalized Identities of Certain Continued Fractions, arXiv:1907.12459 [math.NT], 2019.
FORMULA
O.g.f.: (1 + x - x^3)/(1 - x - x^2). - Geoffrey Critzer, Sep 23 2013
a(n) = round(sqrt(F(2n) + 2*F(2n-1))), for n >= 0, allowing F(-1) = 1. Also phi^n -> sqrt(F(2n) + 2*F(2n-1)), within < 0.02% by n = 4, therefore converging rapidly. - Richard R. Forberg, Jun 23 2014
For k > 0, a(2k) = A169986(2k) and a(2k+1) = A014217(2k+1). - Danny Rorabaugh, Apr 15 2015
For n > 1, a(n) = A001610(n - 1) + 1. - Gus Wiseman, Feb 12 2019
a(n) = A000032(n) for n>=2. - G. C. Greubel, Jul 09 2019
EXAMPLE
a(4) = 7 because we have: {}, {1}, {2}, {3}, {4}, {1,3}, {2,4}. - Geoffrey Critzer, Sep 23 2013
MATHEMATICA
nn=34; CoefficientList[Series[(1+x-x^3)/(1-x-x^2), {x, 0, nn}], x] (* Geoffrey Critzer, Sep 23 2013 *)
Round[GoldenRatio^Range[0, 40]] (* Harvey P. Dale, Jul 13 2014 *)
Table[If[n<=1, n+1, LucasL[n]], {n, 0, 40}] (* G. C. Greubel, Jul 09 2019 *)
PROG
(Magma) [Round(Sqrt(Fibonacci(2*n) + 2*Fibonacci(2*n-1))): n in [0..40]]; // Vincenzo Librandi, Apr 16 2015
(Sage) [round(golden_ratio^n) for n in range(40)] # Danny Rorabaugh, Apr 16 2015
(PARI) my(x='x+O('x^40)); Vec((1+x-x^3)/(1-x-x^2)) \\ G. C. Greubel, Feb 13 2019
(GAP) Concatenation([1, 2], List([2..40], n-> Lucas(1, -1, n)[2] )); # G. C. Greubel, Jul 09 2019
(Python)
from gmpy2 import isqrt, fib2
def A169985(n): return int((m:=isqrt(k:=(lambda x:(x[1]<<1)+x[0])(fib2(n<<1))))+(k-m*(m+1)>=1)) # Chai Wah Wu, Jun 19 2024
KEYWORD
nonn,easy
AUTHOR
N. J. A. Sloane, Sep 26 2010
STATUS
approved