A020956 -id:A020956 - 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

A182801 Joint-rank array of the numbers j*r^(i-1), where r = golden ratio = (1+sqrt(5))/2, i>=1, j>=1, read by antidiagonals.

Original entry on oeis.org

1, 3, 2, 5, 6, 4, 7, 9, 11, 8, 10, 13, 16, 19, 14, 12, 18, 23, 28, 32, 25, 15, 21, 31, 39, 48, 54, 42, 17, 26, 36, 52, 66, 81, 89, 71, 20, 29, 44, 61, 86, 110, 134, 147, 117, 22, 34, 49, 73, 102, 141, 181, 221, 240, 193, 24, 38, 57, 82

Offset: 1

Clark Kimberling, Dec 04 2010

Joint-rank arrays are introduced here as follows.
Suppose that R={f(i,j)} is set of positive numbers, where i and j range through countable sets I and J, respectively, such that for every n, then number f(i,j) < n is finite.  Let T(i,j) be the position of f(i,j) in the joint ranking of all the numbers in R.  The joint-rank array of R is the array T whose i-th row is T(i,j).
For A182801, f(i,j)=j*r^(i-1), where r=(1+sqrt(5))/2 and I=J={1,2,3,...}.
(row 1)=A020959; (row 2)=A020960; (row 3)=A020961.
(col 1)=A020956; (col 2)=A020957; (col 3)=A020958.
Every positive integer occurs exactly once in A182801, so that as a sequence it is a permutation of the positive integers.

			Northwest corner:
1....3....5....7...10...12...
2....6....9...13...18...21...
4...11...16...23...31...36...
8...19...28...39...52...61...

Cf. A001622, A182802, A182846, A182847, A182848, A182849, A252229.

r=GoldenRatio;
f[i_,j_]:=Sum[Floor[j*r^(i-k)],{k,1,i+Log[r,j]}];
TableForm[Table[f[i,j],{i,1,16},{j,1,16}]] (* A182801 *)

T(i,j)=Sum{floor(j*r^(i-k)): k>=1}.

A347068 Rectangular array (T(n,k)), by downward antidiagonals: T(n,k) = position of k in the ordering of {h*r^m, r = 1/(golden ratio), h >= 1, 0 <= m <= n}.

Original entry on oeis.org

2, 5, 4, 7, 10, 8, 10, 14, 18, 14, 13, 20, 26, 31, 25, 15, 26, 36, 46, 53, 42, 18, 30, 47, 63, 79, 88, 71, 20, 36, 55, 81, 107, 132, 146, 117, 23, 40, 65, 96, 136, 178, 219, 239, 193, 26, 46, 73, 112, 162, 225, 294, 359, 391, 315, 28, 52, 84, 127, 189, 269

Offset: 1

Clark Kimberling, Sep 02 2021

Row 1: A001950 (upper Wythoff sequence);
row 2: A283234;
row 3: A190508;
col 1: A020956.

			Corner:
    2,   5,   7,  10,  13,  15,  18,  20,  23, ...
    4,  10,  14,  20,  26,  30,  36,  40,  46, ...
    8,  18,  26,  36,  47,  55,  65,  73,  84, ...
   14,  31,  46,  63,  81,  96, 112, 127, 145, ...
   25,  53,  79, 107, 136, 162, 189, 215, 244, ...
   42,  88, 132, 178, 225, 269, 314, 358, 405, ...
   71, 146, 219, 294, 370, 443, 517, 590, 666, ...
   ...

Cf. A001950, A020956, A283234, A190508, A347065, A347066, A347067, A347069.

z = 1000; r = N[(-1+Sqrt[5])/2];
s[m_] := Range[z] r^m; t[0] = s[0];
t[n_] := Sort[Union[s[n], t[n - 1]]]
row[n_] := Flatten[Table[Position[t[n], N[k]], {k, 1, z}]]
TableForm[Table[row[n], {n, 1, 10}]] (* A347068, array *)
w[n_, k_] := row[n][[k]];
Table[w[n - k + 1, k], {n, 12}, {k, n, 1, -1}] // Flatten (* A347068, sequence *)

A215004 a(0) = a(1) = 1; for n>1, a(n) = a(n-1) + a(n-2) + floor(n/2).

Original entry on oeis.org

1, 1, 3, 5, 10, 17, 30, 50, 84, 138, 227, 370, 603, 979, 1589, 2575, 4172, 6755, 10936, 17700, 28646, 46356, 75013, 121380, 196405, 317797, 514215, 832025, 1346254, 2178293, 3524562, 5702870, 9227448, 14930334, 24157799, 39088150, 63245967, 102334135, 165580121

Offset: 0

Alex Ratushnyak, Jul 31 2012

If the first two terms are {0,1}, we get A020956 except for the first term.

If the first two terms are {1,2}, we get A281362.

Colin Barker, Table of n, a(n) for n = 0..1000
Jean-Luc Baril, Sergey Kirgizov, and Armen Petrossian, Dyck Paths with catastrophes modulo the positions of a given pattern, Australasian J. Comb. (2022) Vol. 84, No. 2, 398-418.
Nathan Fox, Proof of formula for a(n).
Index entries for linear recurrences with constant coefficients, signature (2,1,-3,0,1).

Cf. A020956, except for first term: same formula, seed {0,1}.

Cf. A000045, A281362.

[Fibonacci(n+3)-(2*n+5-(-1)^n)/4: n in [0..40]]; // _G. C. Greubel, Feb 01 2018

Table[((-1)^n - 2 n + 8 Fibonacci[n] + 4 LucasL[n] - 5)/4, {n, 0, 20}] (* Vladimir Reshetnikov, May 18 2016 *)
RecurrenceTable[{a[0]==a[1]==1,a[n]==a[n-1]+a[n-2]+Floor[n/2]},a,{n,40}] (* or *) LinearRecurrence[{2,1,-3,0,1},{1,1,3,5,10},40] (* Harvey P. Dale, Jul 11 2020 *)

Vec(-(x^3-x+1)/((x-1)^2*(x+1)*(x^2+x-1)) + O(x^100)) \\ Colin Barker, Sep 16 2015

a(n)=([0,1,0,0,0;0,0,1,0,0;0,0,0,1,0;0,0,0,0,1;1,0,-3,1,2]^n* [1;1;3;5;10])[1,1] \\ Charles R Greathouse IV, Jan 16 2017

prpr = prev = 1
for n in range(2,100):
    print(prpr, end=', ')
    curr = prpr+prev + n//2
    prpr = prev
    prev = curr

[fibonacci(n+3) -(n+2+(n%2))//2 for n in range(41)] # G. C. Greubel, Apr 05 2024

From Colin Barker, Sep 16 2015: (Start)
a(n) = 2*a(n-1) + a(n-2) - 3*a(n-3) + a(n-5) for n>4.
G.f.: (1-x+x^3) / ((1-x)^2*(1+x)*(1-x-x^2)). (End)
a(n) = Fibonacci(n+3) - floor((n+3)/2). - Nathan Fox, Jan 27 2017
a(n) = (-3/4 + (-1)^n/4 + (2^(-n)*((1-t)^n*(-2+t) + (1+t)^n*(2+t)))/t + (-1-n)/2) where t=sqrt(5). - Colin Barker, Feb 09 2017
From G. C. Greubel, Apr 05 2024: (Start)
a(n) =  Fibonacci(n+3) - (1/4)*(2*n + 5 - (-1)^n).
E.g.f.: 2*exp(x/2)*( cosh(sqrt(5)*x/2) + (2/sqrt(5))*sinh(sqrt(5)*x/2) ) - (1/2)*( (x+2)*cosh(x) + (x+3)*sinh(x) ). (End)

A277752 a(n) = Sum_{k=0..n} (-1)^k*floor(phi^k), where phi is the golden ratio (A001622).

Original entry on oeis.org

1, 0, 2, -2, 4, -7, 10, -19, 27, -49, 73, -126, 195, -326, 516, -848, 1358, -2213, 3564, -5785, 9341, -15135, 24467, -39612, 64069, -103692, 167750, -271454, 439192, -710659, 1149838, -1860511, 3010335, -4870861, 7881181, -12752058, 20633223, -33385298, 54018504, -87403820, 141422306

Offset: 0

Ilya Gutkovskiy, Oct 31 2016

Alternating sum of A014217.

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

Cf. A000032, A000045, A001622, A014217, A020956.

Accumulate[Table[(-1)^n Floor[GoldenRatio^n], {n, 0, 40}]]
LinearRecurrence[{0, 3, -1, -2, 1}, {1, 0, 2, -2, 4}, 41]

G.f.: (1 - x^2 - x^3)/((1 - x)^2*(1 + 2*x - x^3)).
a(n) = 3*a(n-2) - a(n-3) - 2*a(n-4) + a(n-5).
a(n) = Sum_{k=0..n} (-1)^k*floor(Fibonacci(2k+3)/Fibonacci(k+3)).
a(n) = Sum_{k=0..n} (-1)^k*(L(k) - (1 + (-1)^k)/2), where L(k) is the Lucas numbers beginning at 2 (A000032).
a(n) = 2^(-n-2)*(9*2^n - 2^(n+1)*n - (-2)^n - 2*(1 + sqrt(5))*(sqrt(5) - 1)^n + 2*(sqrt(5) - 1)*(-1-sqrt(5))^n).
a(n) ~ (-1)^n*phi^(n-1).
a(n) = (-1)^n*Lucas(n-1) - (1/4)*(2*n -9 +(-1)^n). - G. C. Greubel, Oct 31 2016

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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A182801 Joint-rank array of the numbers j*r^(i-1), where r = golden ratio = (1+sqrt(5))/2, i>=1, j>=1, read by antidiagonals.

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A347068 Rectangular array (T(n,k)), by downward antidiagonals: T(n,k) = position of k in the ordering of {h*r^m, r = 1/(golden ratio), h >= 1, 0 <= m <= n}.

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A215004 a(0) = a(1) = 1; for n>1, a(n) = a(n-1) + a(n-2) + floor(n/2).

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A277752 a(n) = Sum_{k=0..n} (-1)^k*floor(phi^k), where phi is the golden ratio (A001622).

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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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