A169986 -id:A169986 - cp's OEIS frontend

A014217 a(n) = floor(phi^n), where phi = (1+sqrt(5))/2 is the golden ratio.

1, 1, 2, 4, 6, 11, 17, 29, 46, 76, 122, 199, 321, 521, 842, 1364, 2206, 3571, 5777, 9349, 15126, 24476, 39602, 64079, 103681, 167761, 271442, 439204, 710646, 1149851, 1860497, 3010349, 4870846, 7881196, 12752042, 20633239, 33385281, 54018521, 87403802

Offset: 0

Clark Kimberling

a(n) = floor(lim_{k->oo} Fibonacci(k)/Fibonacci(k-n)). - Jon Perry, Jun 10 2003
For n > 1, a(n) is the maximum element in the continued fraction for A000045(n)*phi. - Benoit Cloitre, Jun 19 2005
a(n) is also the curvature (rounded down) of the circle inscribed in the n-th kite arranged in a spiral, starting with a unit circle, as shown in the illustration in the links section. - Kival Ngaokrajang, Aug 29 2013
a(n) is the n-th Lucas number (A000032) if n is odd, and a(n) is the n-th Lucas number minus 1 if n is even. (Mario Catalani's formula below expresses this fact.) This is related to the fact that the powers of phi approach the values of the Lucas numbers, the odd powers from above and the even powers from below. - Geoffrey Caveney, Apr 18 2014
a(n) is the sum of the last summands over all Arndt compositions of n (see the Checa link). - Daniel Checa, Dec 25 2023
a(n) is the number of (saturated or unsaturated) substituted N-heterocycles in chemistry (N = nitrogen). That means the number of matchings in a cycle graph when the two maximum matchings in every cycle with an even number of vertices are indistinguishable (because the corresponding resonance structures in the molecule are equivalent). - Stefan Schuster, Mar 20 2025

Alois P. Heinz, Table of n, a(n) for n = 0..4784 (first 301 terms from T. D. Noe)
Mohammad K. Azarian, Problem 123, Missouri Journal of Mathematical Sciences, Vol. 10, No. 3, Fall 1998, p. 176. Solution published in Vol. 12, No. 1, Winter 2000, pp. 61-62.
Daniel F. Checa, Arndt Compositions: Connections with Fibonacci Numbers, Statistics, and Generalizations, 2023. p. 29.
Daniel F. Checa and José L. Ramírez, Arndt Compositions: A Generating Functions Approach, Integers (2024) Vol. 24, A35. See p. 14.
G. Harman, One hundred years of normal numbers
Kival Ngaokrajang, Illustration for n = 0..7
Stefan Schuster and Tatjana Malycheva, Enumeration of saturated and unsaturated substituted N-heterocycles, arXiv:2309.02343 [q-bio.BM], 2023.
Index entries for linear recurrences with constant coefficients, signature (1,2,-1,-1).

Cf. A000032, A000045, A001622, A020956, A022319, A052952, A057146, A062114, A128440, A153382, A169985, A169986, A203976, A226328.

First differences give A181716.

a014217 n = a014217_list !! n
a014217_list = 1 : 1 : zipWith (+)
   a000035_list (zipWith (+) a014217_list $ tail a014217_list)
-- Reinhard Zumkeller, Jan 06 2012

[Floor( ((1+Sqrt(5))/2)^n ): n in [0..100]]; // Vincenzo Librandi, Apr 16 2011

A014217 := proc(n)
    option remember;
    if n <= 3 then
        op(n+1,[1,1,2,4]) ;
    else
        procname(n-1)+2*procname(n-2)-procname(n-3)-procname(n-4) ;
    end if;
end proc: # R. J. Mathar, Jun 23 2013
#
a:= n-> (<<0|1|0|0>, <0|0|1|0>, <0|0|0|1>, <-1|-1|2|1>>^n. <<1, 1, 2, 4>>)[1, 1]:
seq(a(n), n=0..40);  # Alois P. Heinz, Oct 12 2017

Table[Floor[GoldenRatio^n], {n, 0, 36}] (* Vladimir Joseph Stephan Orlovsky, Dec 12 2008 *)
LinearRecurrence[{1, 2, -1, -1}, {1, 1, 2, 4}, 40] (* Jean-François Alcover, Nov 05 2017 *)

my(x='x+O('x^44)); Vec((1-x^2+x^3)/((1+x)*(1-x)*(1-x-x^2))) \\ Joerg Arndt, Jul 10 2023

from sympy import floor, sqrt
def A014217(n): return floor(((1+sqrt(5))/2)**n) # Chai Wah Wu, Dec 17 2021

[floor(golden_ratio^n) for n in range(37)] # Danny Rorabaugh, Apr 19 2015

a(n) = a(n-1) + 2*a(n-2) - a(n-3) - a(n-4).
a(n) = a(n-1) + a(n-2) + (1-(-1)^n)/2 = a(n-1) + a(n-2) + A000035(n).
a(n) = A000032(n) - (1 + (-1)^n)/2. - Mario Catalani (mario.catalani(AT)unito.it), Jan 17 2003
G.f.: (1-x^2+x^3)/((1+x)*(1-x)*(1-x-x^2)). - R. J. Mathar, Sep 06 2008
a(2n-1) = (Fibonacci(4n+1)-2)/Fibonacci(2n+2). - Gary Detlefs, Feb 16 2011
a(n) = floor(Fibonacci(2n+3)/Fibonacci(n+3)). - Gary Detlefs, Feb 28 2011
a(2n) = Fibonacci(2*n-1) + Fibonacci(2*n+1) - 1. - Gary Detlefs, Mar 10 2011
a(n+2*k) - a(n) = A203976(k)*A000032(n+k) if k odd, a(n+2*k) - a(n) = A203976(k)*A000045(n+k) if k even, for k > 0. - Paul Curtz, Jun 05 2013
a(n) = A052952(n) - A052952(n-2) + A052952(n-3). - R. J. Mathar, Jun 13 2013
a(n+6) - a(n-6) = 40*A000045(n), case k=6 of my formula above. - Paul Curtz, Jun 13 2013
From Paul Curtz, Jun 17 2013: (Start)
a(n-3) + a(n+3) = A153382(n).
a(n-1) + a(n+2) = A022319(n). (End)
For k > 0, a(2k) = A169985(2k)-1 and a(2k+1) = A169985(2k+1) (which is equivalent to Catalani's 2003 formula). - Danny Rorabaugh, Apr 15 2015
a(n) = ((-1)^(1+n)-1)/2 + ((1-sqrt(5))/2)^n + ((1+sqrt(5))/2)^n. - Colin Barker, Nov 05 2017
a(n) = floor(2*sinh(n*arccsch(2))). - Federico Provvedi, Feb 23 2022
E.g.f.: 2*exp(x/2)*cosh(sqrt(5)*x/2) - cosh(x). - Stefano Spezia, Jul 26 2022
a(n) = floor(Fibonacci(n)*phi) + Fibonacci(n-1) = A074331(n) + A000045(n-1) = A052952(n-1) + A000045(n-1). This is the case k=1 of the formula (also found in A128440): floor(k * phi^n) = floor(Fibonacci(n)*k*phi) + Fibonacci(n-1) * k. - Chunqing Liu, Oct 03 2023

Corrected by T. D. Noe, Nov 09 2006

Edited by N. J. A. Sloane, Aug 29 2008 at the suggestion of R. J. Mathar

A169985 Round phi^n to the nearest integer.

Original entry on oeis.org

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

N. J. A. Sloane, Sep 26 2010

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

			a(4) = 7 because we have: {}, {1}, {2}, {3}, {4}, {1,3}, {2,4}. - _Geoffrey Critzer_, Sep 23 2013

Danny Rorabaugh, Table of n, a(n) for n = 0..4000
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.
Index entries for linear recurrences with constant coefficients, signature (1,1).

Cf. A000032, A000045, A000204, A001622, A014217, A169986.

Cf. A000126, A000296, A001610.

Concatenation([1,2], List([2..40], n-> Lucas(1,-1,n)[2] )); # G. C. Greubel, Jul 09 2019

[Round(Sqrt(Fibonacci(2*n) + 2*Fibonacci(2*n-1))): n in [0..40]]; // Vincenzo Librandi, Apr 16 2015

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 *)

my(x='x+O('x^40)); Vec((1+x-x^3)/(1-x-x^2)) \\ G. C. Greubel, Feb 13 2019

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

[round(golden_ratio^n) for n in range(40)] # Danny Rorabaugh, Apr 16 2015

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

A214986 Power ceiling array for the golden ratio, by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 4, 3, 1, 1, 7, 8, 5, 1, 1, 12, 21, 22, 7, 1, 1, 20, 55, 94, 48, 12, 1, 1, 33, 144, 399, 329, 134, 18, 1, 1, 54, 377, 1691, 2255, 1487, 323, 30, 1, 1, 88, 987, 7164, 15456, 16492, 5796, 872, 47, 1, 1, 143, 2584, 30348, 105937, 182900

Offset: 1

Clark Kimberling, Oct 28 2012

row 0: A000012 ... row 6: A049660
row 1: A000071 ... row 8: A049668
row 2: A001906 ... col 0: A000012
row 3: A049652 ... col 1: A169986
row 4: A004187
For x>1, define c(x,0) = 1 and c(x,n) = ceiling(x*c(x,n-1)) for n>0.  Row m of A214986 is the sequence c(r^m,n), where r = golden ratio = (1 + sqrt(5))/2.  The name of the array corresponds to the power ceiling function f(x) = limit of c(x,n)/x^n as n increases without bound; f(x) generalizes the case for x = 3/2 as described under "Power Ceilings" at MathWorld.  For a graph of f(x), see the Mathematica program at A083286.
The term "power ceiling sequence" extends to sequences generated by recurrences P(n) = ceiling(x*P(n-1)) + g(n), and "power ceiling functions" f(x) to the limit of P(n)/x^n in case x>1 and g(n)/x^n -> 0.
Suppose that h is a nonnegative integer and g(n) is a constant.  If x is a positive integer power of the golden ratio r, then f(x), in many cases, lies in the field Q(sqrt(5)).  Examples matching rows of A214986, using g(n) = 0, follow:
...
x ... P ........ f(x)
r ... A000071 .. (5 + 2*sqrt(5))/2 = 1.8944... (A010532)
r^2 . A001906 .. (5 + 3*sqrt(5))/10 = 1.7082...(A176015)
r^3 . A049652 .. (25 + 11*sqrt(5))/40 = 1.2399...
r^4 . A004187 .. (15 + 7*sqrt(5))/10 = 1.0219...
...
If k is odd, then f(r^k) = r^k((b(k) + c(k))/d(k)), where
b(k) = L(j)^2 + L(j-1)^2, where j=[(k+1)/2], L=A000032 (Lucas numbers); c(k) = (L(k)+2)*sqrt(5); d(k) = 10*F(k)*L(k), where F=A000045 (Fibonacci numbers).  If k is even, then f(r^k) = r^k/(F(k)*sqrt(5)).

			Northwest corner:
1...1....1.....1......1.......1
1...2....4.....7......12......20
1...3....8.....21.....55......144
1...5....22....94.....399.....1691
1...7....48....329....2255....15456
1...19...134...1487...16492...182900

Clark Kimberling, Antidiagonals n = 1..35, flattened
Eric Weisstein's World of Mathematics, Power Ceilings

Cf. A000045, A214978, A214984, A214987.

r = GoldenRatio;
s[x_, 0] := 1; s[x_, n_] := Ceiling[x*s[x, n - 1]];
t = TableForm[Table[s[r^m, n], {m, 0, 10}, {n, 0, 10}]  ]
u = Flatten[Table[s[r^m, n - m], {n, 0, 10}, {m, 0, n}]]

The odd-numbered rows of A214986 are even-numbered rows of A213978; the even-numbered rows of A214986 are odd-numbered rows of A214984.

A062724 a(n) = floor(tau^n) + 1, where tau = (1 + sqrt(5))/2.

Original entry on oeis.org

2, 2, 3, 5, 7, 12, 18, 30, 47, 77, 123, 200, 322, 522, 843, 1365, 2207, 3572, 5778, 9350, 15127, 24477, 39603, 64080, 103682, 167762, 271443, 439205, 710647, 1149852, 1860498, 3010350, 4870847, 7881197, 12752043, 20633240, 33385282

Offset: 0

Jason Earls, Jul 15 2001

nonn

Apart from the first term, this sequence also gives the ceiling of the powers of the golden ratio (cf. A169986). - Mohammad K. Azarian, Apr 14 2008

Harry J. Smith, Table of n, a(n) for n = 0..400

Equals A014217 + 1.

Cf. A169985, A169986.

Floor[GoldenRatio^Range[0,40]]+1 (* Harvey P. Dale, Dec 18 2019 *)

j=[]; for(n=0,60,t=(1+sqrt(5))/2; j=concat(j,floor((t^n))+1)); j

{ default(realprecision, 200); t=(1 + sqrt(5))/2; p=1; for (n=0, 400, if (n, p*=t); write("b062724.txt", n, " ", p\1 + 1) ) } \\ Harry J. Smith, Aug 09 2009

a(n) = 3*Fibonacci(n-1) + Fibonacci(n-2) + (n mod 2), n > 0. - Gary Detlefs, Dec 29 2010

A118841 Numbers k such that ceiling(phi^k) is prime.

Original entry on oeis.org

1, 2, 3, 4, 8, 16

Offset: 1

Eric W. Weisstein, May 01 2006

nonn

a(7) if it exists is greater than 250000. - Mark Rodenkirch, Feb 22 2020

Eric Weisstein's World of Mathematics, Phi-Prime

Cf. A118842, A001606.

Select[Range[20],PrimeQ[Ceiling[GoldenRatio^#]]&] (* James C. McMahon, Sep 14 2024 *)

c(n) = 3*fibonacci(n-1) + fibonacci(n-2) + (n % 2); \\ A169986(n) for n >= 1
for(n=1,10^7,if(ispseudoprime(c(n)),print1(n,", ")))
\\ Joerg Arndt, Feb 22 2020

a(7)-a(8) from Charles R Greathouse IV, Jul 30 2011
a(9)-a(11) from Charles R Greathouse IV, Aug 01 2011
Incorrect a(7)-a(11) removed by Mark Rodenkirch, Feb 22 2020

A279100 a(n) = Sum_{k=0..n} ceiling(phi^k), where phi is the golden ratio (A001622).

Original entry on oeis.org

1, 3, 6, 11, 18, 30, 48, 78, 125, 202, 325, 525, 847, 1369, 2212, 3577, 5784, 9356, 15134, 24484, 39611, 64088, 103691, 167771, 271453, 439215, 710658, 1149863, 1860510, 3010362, 4870860, 7881210, 12752057, 20633254, 33385297, 54018537, 87403819, 141422341, 228826144, 370248469, 599074596

Offset: 0

Ilya Gutkovskiy, Dec 06 2016

Partial sums of A169986.

Eric Weisstein's World of Mathematics, Golden Ratio
Index entries for linear recurrences with constant coefficients, signature (2,1,-3,0,1).

Cf. A001622, A020956, A169986.

Accumulate[Table[Ceiling[GoldenRatio^n], {n, 0, 40}]]
LinearRecurrence[{2, 1, -3, 0, 1}, {1, 3, 6, 11, 18}, 41]

G.f.: (1 + x - x^2 - x^3 - x^4)/((1 - x)^2*(1 - 2*x^2 - x^3)).
a(n) = 2*a(n-1) + a(n-2) - 3*a(n-3) + a(n-5).
a(n) = (10*n - 5*(-1)^n + 2^(1-n)*sqrt(5)*(5 + 3*sqrt(5))*(1 + sqrt(5))^n + sqrt(5)*2^(1-n)*(3*sqrt(5) - 5) *(1 - sqrt(5))^n - 35)/20.
a(n) ~ phi^(n+2).

cp's OEIS Frontend

A014217 a(n) = floor(phi^n), where phi = (1+sqrt(5))/2 is the golden ratio.

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A169985 Round phi^n to the nearest integer.

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A214986 Power ceiling array for the golden ratio, by antidiagonals.

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A062724 a(n) = floor(tau^n) + 1, where tau = (1 + sqrt(5))/2.

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A118841 Numbers k such that ceiling(phi^k) is prime.

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A279100 a(n) = Sum_{k=0..n} ceiling(phi^k), where phi is the golden ratio (A001622).

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