A001622 -id:A001622 - cp's OEIS frontend

A060144 a(n) = floor(n/(1+tau)), or equivalently floor(n/(tau)^2), where tau is the golden ratio (A001622).

0, 0, 0, 1, 1, 1, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 6, 6, 6, 7, 7, 8, 8, 8, 9, 9, 9, 10, 10, 11, 11, 11, 12, 12, 12, 13, 13, 14, 14, 14, 15, 15, 16, 16, 16, 17, 17, 17, 18, 18, 19, 19, 19, 20, 20, 21, 21, 21, 22, 22, 22, 23, 23, 24, 24, 24, 25, 25, 25, 26, 26, 27, 27, 27, 28, 28, 29, 29

Offset: 0

Clark Kimberling, Mar 05 2001

nonn

The RUNS transform of this sequence appears to be yet another version of the Fibonacci word (cf. A001468, A001950, A076662). - N. J. A. Sloane, Mar 29 2025

Harry J. Smith, Table of n, a(n) for n = 0..1000
Martin Griffiths, A formula for an infinite family of Fibonacci-word sequences, Fib. Q., 56 (2018), 75-80.
D. R. Hofstadter, Eta-Lore [With permission]
D. R. Hofstadter, Pi-Mu Sequences [With permission]
D. R. Hofstadter and N. J. A. Sloane, Correspondence, 1977 and 1991 (On page 4 of DRH letter, v[n] = A006336, a[n] = A060144[n+1]). - _N. J. A. Sloane_, Oct 25 2014
Johan Kok, Integer sequences with conjectured relation with certain graph parameters of the family of linear Jaco graphs, arXiv:2507.16500 [math.CO], 2025. See p. 4.

Cf. A001622, A005206, A006336.

A296184 Decimal expansion of 2 + phi, with the golden section phi from A001622.

Original entry on oeis.org

3, 6, 1, 8, 0, 3, 3, 9, 8, 8, 7, 4, 9, 8, 9, 4, 8, 4, 8, 2, 0, 4, 5, 8, 6, 8, 3, 4, 3, 6, 5, 6, 3, 8, 1, 1, 7, 7, 2, 0, 3, 0, 9, 1, 7, 9, 8, 0, 5, 7, 6, 2, 8, 6, 2, 1, 3, 5, 4, 4, 8, 6, 2, 2, 7, 0, 5, 2, 6, 0, 4, 6, 2, 8, 1, 8, 9

Offset: 1

Wolfdieter Lang, Jan 08 2018

In a regular pentagon, inscribed in a unit circle this equals twice the largest distance between a vertex and a midpoint of a side.
This is an integer in the quadratic number field Q(sqrt(5)).
Only the first digit differs from A001622.

			3.618033988749894848204586834365638117720309179805762862135448622705260462...

Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 5.25, p. 417.

Sumit Kumar Jha, Two complementary relations for the Rogers-Ramanujan continued fraction, arXiv:2112.12081 [math.NT], 2021.
Index entries for algebraic numbers, degree 2.

Cf. A001622, A094214, A104457, A176055, A020837.

2 + 2*cos(2*Pi/n): A104457 (n = 5), A116425 (n = 7), A332438 (n = 9), A019973 (n = 12).

First@ RealDigits[2 + GoldenRatio, 10, 77] (* Michael De Vlieger, Jan 13 2018 *)

(5 + sqrt(5))/2 \\ Altug Alkan, Mar 19 2018

Equals 2 + A001622 = 1 + A104457 = 3 + A094214.
From Christian Katzmann, Mar 19 2018: (Start)
Equals Sum_{n>=0} (15*(2*n)!+40*n!^2)/(2*n!^2*3^(2*n+2)).
Equals 5/2 + Sum_{n>=0} 5*(2*n)!/(2*n!^2*3^(2*n+1)). (End)
Constant c = 2 + 2*cos(2*Pi/10). The linear fractional transformation z -> c - c/z has order 10, that is, z = c - c/(c - c/(c - c/(c - c/(c - c/(c - c/(c - c/(c - c/(c - c/(c - c/(z)))))))))). - Peter Bala, May 09 2024

A004956 a(n) = ceiling(n*phi), where phi is the golden ratio, A001622.

Original entry on oeis.org

0, 2, 4, 5, 7, 9, 10, 12, 13, 15, 17, 18, 20, 22, 23, 25, 26, 28, 30, 31, 33, 34, 36, 38, 39, 41, 43, 44, 46, 47, 49, 51, 52, 54, 56, 57, 59, 60, 62, 64, 65, 67, 68, 70, 72, 73, 75, 77, 78, 80, 81, 83, 85, 86, 88, 89

Offset: 0

N. J. A. Sloane

a(0)=0, a(1)=2; for n > 1, a(n) = a(n-1) + 2 if n is already in the sequence, a(n) = a(n-1) + 1 otherwise.
Integer solutions to the equation x = ceiling(phi*floor(x/phi)). - Benoit Cloitre, Feb 14 2004
From Benoit Cloitre, Mar 05 2007: (Start)
The following is an alternative way to obtain this sequence. NP means "term not in parentheses". Write down the natural numbers and mark the least NP, which is 1:
1* 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 ...
Take the first NP (which is 1) and parenthesize it; mark the least NP (which is 2):
(1*) 2* 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 ...
Take the 2nd NP (which is 3) and parenthesize it; mark the next NP (which is 4):
(1*) 2* (3) 4* 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 ...
Take the 4th NP (which is 6) and parenthesize it; mark the next NP (which is 5):
(1*) 2* (3) 4* 5* (6) 7 8 9 10 11 12 13 14 15 16 17 18 19 ...
Continuing in this way we obtain
(1*) 2* (3) 4* 5* (6) 7* (8) 9* 10* (11) 12* 13* (14) 15* (16) 17* (18) 19* ...
The starred entries (after the first) give the sequence. (End)
From Rick L. Shepherd, Dec 05 2009: (Start)
An equivalent statement of the sieving process described by Benoit Cloitre on Mar 05 2007:
Begin with the natural numbers N. Repeatedly perform these two steps:
i) Let k = N's least remaining term not yet used in Step ii).
ii) Remove the k-th remaining term from N.
The remaining terms of N are the (positive) terms shared by this sequence and A026351.
The terms removed from N (the complement) are A026352's terms (see also A004957).
The PARI program performs this sieving process and prints the positive terms of this sequence. (End)
In the Fokkink-Joshi paper, apart from a(0) = 0, this sequence is the Cloitre (0,2,2,1)-hiccup sequence, i.e., a(1) = 2; for m < n, a(n) = a(n-1)+2 if a(m) = n, else a(n) = a(n-1)+1. - Michael De Vlieger, Jul 30 2025

Christian Krause, Table of n, a(n) for n = 0..10000
Jon Asier Bárcena-Petisco, Luis Martínez, María Merino, Juan Manuel Montoya, and Antonio Vera-López, Fibonacci-like partitions and their associated piecewise-defined permutations, arXiv:2503.19696 [math.CO], 2025. See p. 3.
Benoit Cloitre, A study of a family of self-referential sequences, arXiv:2506.18103 [math.GM], 2025. See p. 7.
Benoit Cloitre, N. J. A. Sloane and M. J. Vandermast, Numerical analogues of Aronson's sequence, J. Integer Seqs., Vol. 6 (2003), #03.2.2.
Benoit Cloitre, N. J. A. Sloane and M. J. Vandermast, Numerical analogues of Aronson's sequence, arXiv:math/0305308 [math.NT], 2003.
Robbert Fokkink and Gandhar Joshi, On Cloitre's hiccup sequences, arXiv:2507.16956 [math.CO], 2025. See p. 3.

Cf. A001622, A004957, A026352 (Cloitre (1,1,2,3)-hiccup sequence).

Essentially same as A026351.

Ceiling[Range[0,100]GoldenRatio] (* Paolo Xausa, Oct 28 2023 *)
(* Second program: *)
cloitreH[j_, x_, y_, z_, w_ : 120] := Module[{c, k}, c[] := False; k = x; c[x] = True; {x}~Join~Reap[Do[If[c[n - j], k += y, k += z]; c[k] = True; Sow[k], {n, 2, w}] ][[-1, 1]] ]; {0}~Join~cloitreH[0, 2, 2, 1] (* _Michael De Vlieger, Jul 30 2025 *)

{/* paws = Print Absolute values of all elements in vector v With same Sign as sn */
paws(v,sn) = for(m=1,matsize(v)[2], if(sign(v[m])==sign(sn),\
print1(abs(v[m]),",")))}
{/* Sieve through lim numbers; make values negative to signify "removed" */
lim=100; v=vector(lim,i,i); i=0; j=0; c=1;
while(i0, k=v[i]; c=c--;
while(c0, c++)); v[j]=-v[j])); paws(v,1)\
/* Changing "1" to "-1" in paws() above prints out the terms of A026352 instead */} \\ Rick L. Shepherd, Dec 05 2009

a(n) = ceil(n*(1 + sqrt(5))/2); \\ Michel Marcus, Apr 13 2021

A004942 a(n) = round(n*phi^7), where phi is the golden ratio, A001622.

Original entry on oeis.org

0, 29, 58, 87, 116, 145, 174, 203, 232, 261, 290, 319, 348, 377, 406, 436, 465, 494, 523, 552, 581, 610, 639, 668, 697, 726, 755, 784, 813, 842, 871, 900, 929, 958, 987, 1016, 1045, 1074, 1103, 1132, 1161, 1190, 1219, 1248, 1278, 1307, 1336, 1365, 1394, 1423, 1452, 1481, 1510, 1539, 1568, 1597, 1626, 1655, 1684

Offset: 0

N. J. A. Sloane

nonn

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Cf. A001622, A004939, A004940, A004941, A004943.

Table[Round[n GoldenRatio^7], {n, 0, 100}] (* Vincenzo Librandi, Oct 21 2014 *)

ph=(1+sqrt(5))/2; a(n)=round(n*ph^7);
vector(66,n,a(n-1)) \\ Joerg Arndt, Oct 21 2014

More terms from Joerg Arndt, Oct 21 2014

A004957 a(n) = ceiling(n*phi^2), where phi is the golden ratio, A001622.

Original entry on oeis.org

0, 3, 6, 8, 11, 14, 16, 19, 21, 24, 27, 29, 32, 35, 37, 40, 42, 45, 48, 50, 53, 55, 58, 61, 63, 66, 69, 71, 74, 76, 79, 82, 84, 87, 90, 92, 95, 97, 100, 103, 105, 108, 110, 113, 116, 118, 121, 124, 126, 129, 131, 134

Offset: 0

N. J. A. Sloane

nonn

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000

Essentially same as A026352.

Cf. A001622.

import Data.List (findIndices)
a004957 n = a004957_list !! n
a004957_list = findIndices even a060142_list
-- Reinhard Zumkeller, Nov 26 2012

A060142(a(n)) = 4 * A060142(n). - Reinhard Zumkeller, Nov 26 2012

A265796 Numerators of lower primes-only best approximates (POBAs) to the golden ratio, tau (A001622); see Comments.

Original entry on oeis.org

3, 11, 37, 163, 173, 241, 571, 1231, 1571, 2351, 3571, 25463, 69247

Offset: 1

Clark Kimberling, Dec 29 2015

Suppose that x > 0. A fraction p/q of primes is a lower primes-only best approximate, and we write "p/q is in L(x)", if u/v < p/q < x < p'/q for all primes u and v such that v < q, where p' is least prime > p.
Let q(1) be the least prime q such that u/q < x for some prime u, and let p(1) be the greatest such u. The sequence L(x) follows inductively: for n > 1, let q(n) is the least prime q such that p(n)/q(n) < p/q < x for some prime p. Let q(n+1) = q and let p(n+1) be the greatest prime p such that p(n)/q(n) < p/q < x.
For a guide to POBAs, lower POBAs, and upper POBAs, see A265759.

			The lower POBAs to tau start with 3/2, 11/7, 37/23, 163/101, 173/107, 241/149. For example, if p and q are primes and q > 101, and p/q < tau, then 163/101 is closer to tau than p/q is.

Cf. A000040, A001622, A265759, A265797, A265798, A265799, A265800, A265801.

x = GoldenRatio; z = 1000; p[k_] := p[k] = Prime[k];
t = Table[Max[Table[NextPrime[x*p[k], -1]/p[k], {k, 1, n}]], {n, 1, z}];
d = DeleteDuplicates[t]; tL = Select[d, # > 0 &] (* lower POBA *)
t = Table[Min[Table[NextPrime[x*p[k]]/p[k], {k, 1, n}]], {n, 1, z}];
d = DeleteDuplicates[t]; tU = Select[d, # > 0 &] (* upper POBA *)
v = Sort[Union[tL, tU], Abs[#1 - x] > Abs[#2 - x] &];
b = Denominator[v]; s = Select[Range[Length[b]], b[[#]] == Min[Drop[b, # - 1]] &];
y = Table[v[[s[[n]]]], {n, 1, Length[s]}] (* POBA, A265800/A265801 *)
Numerator[tL]   (* A265796 *)
Denominator[tL] (* A265797 *)
Numerator[tU]   (* A265798 *)
Denominator[tU] (* A265799 *)
Numerator[y]    (* A265800 *)
Denominator[y]  (* A265801 *)

a(12)-a(13) from Robert Price, Apr 06 2019

A265797 Denominator of lower primes-only best approximates (POBAs) to the golden ratio, tau (A001622); see Comments.

Original entry on oeis.org

2, 7, 23, 101, 107, 149, 353, 761, 971, 1453, 2207, 15737, 42797

Offset: 1

Clark Kimberling, Dec 29 2015

Suppose that x > 0. A fraction p/q of primes is a lower primes-only best approximate, and we write "p/q is in L(x)", if u/v < p/q < x < p'/q for all primes u and v such that v < q, where p' is least prime > p.
Let q(1) be the least prime q such that u/q < x for some prime u, and let p(1) be the greatest such u. The sequence L(x) follows inductively: for n > 1, let q(n) is the least prime q such that p(n)/q(n) < p/q < x for some prime p. Let q(n+1) = q and let p(n+1) be the greatest prime p such that p(n)/q(n) < p/q < x.
For a guide to POBAs, lower POBAs, and upper POBAs, see A265759.

			The lower POBAs to tau start with 3/2, 11/7, 37/23, 163/101, 173/107, 241/149. For example, if p and q are primes and q > 101, and p/q < tau, then 163/101 is closer to tau than p/q is.

Cf. A000040, A001622, A265759, A265796, A265798, A265799, A265800, A265801.

x = GoldenRatio; z = 1000; p[k_] := p[k] = Prime[k];
t = Table[Max[Table[NextPrime[x*p[k], -1]/p[k], {k, 1, n}]], {n, 1, z}];
d = DeleteDuplicates[t]; tL = Select[d, # > 0 &] (*lower POBA*)
t = Table[Min[Table[NextPrime[x*p[k]]/p[k], {k, 1, n}]], {n, 1, z}];
d = DeleteDuplicates[t]; tU = Select[d, # > 0 &] (*upper POBA*)
v = Sort[Union[tL, tU], Abs[#1 - x] > Abs[#2 - x] &];
b = Denominator[v]; s = Select[Range[Length[b]], b[[#]] == Min[Drop[b, # - 1]] &];
y = Table[v[[s[[n]]]], {n, 1, Length[s]}] (*POBA,A265800/A265801*)
Numerator[tL]   (*A265796*)
Denominator[tL] (*A265797*)
Numerator[tU]   (*A265798*)
Denominator[tU] (*A265799*)
Numerator[y]    (*A265800*)
Denominator[y]  (*A265801*)

a(12)-a(13) from Robert Price, Apr 06 2019

A265798 Numerators of upper primes-only best approximates (POBAs) to the golden ratio, tau (A001622); see Comments.

Original entry on oeis.org

5, 5, 31, 47, 157, 911, 1021, 1487, 4283, 5147, 8629, 11069, 15017, 20939, 22447, 24709, 38239, 80803

Offset: 1

Clark Kimberling, Dec 29 2015

Suppose that x > 0. A fraction p/q of primes is an upper primes-only best approximate, and we write "p/q is in U(x)", if p'/q < x < p/q < u/v for all primes u and v such that v < q, where p' is greatest prime < p in case p >= 3.
Let q(1) = 2 and let p(1) be the least prime >= x. The sequence U(x) follows inductively: for n >= 1, let q(n) is the least prime q such that x < p/q < p(n)/q(n) for some prime p. Let q(n+1) = q and let p(n+1) be the least prime p such that x < p/q < p(n)/q(n).
For a guide to POBAs, lower POBAs, and upper POBAs, see A265759.

			The upper POBAs to tau start with 5/2, 5/3, 31/19, 47/29, 157/97, 911/563, 1021/631. For example, if p and q are primes and q > 97, and p/q > tau, then 157/97 is closer to tau than p/q is.

Cf. A000040, A001622, A265759, A265796, A265797, A265799, A265800, A265801.

x = GoldenRatio; z = 1000; p[k_] := p[k] = Prime[k];
t = Table[Max[Table[NextPrime[x*p[k], -1]/p[k], {k, 1, n}]], {n, 1, z}];
d = DeleteDuplicates[t]; tL = Select[d, # > 0 &] (* lower POBA *)
t = Table[Min[Table[NextPrime[x*p[k]]/p[k], {k, 1, n}]], {n, 1, z}];
d = DeleteDuplicates[t]; tU = Select[d, # > 0 &] (* upper POBA *)
v = Sort[Union[tL, tU], Abs[#1 - x] > Abs[#2 - x] &];
b = Denominator[v]; s = Select[Range[Length[b]], b[[#]] == Min[Drop[b, # - 1]] &];
y = Table[v[[s[[n]]]], {n, 1, Length[s]}] (* POBA, A265800/A265801 *)
Numerator[tL]   (* A265796 *)
Denominator[tL] (* A265797 *)
Numerator[tU]   (* A265798 *)
Denominator[tU] (* A265799 *)
Numerator[y]    (* A265800 *)
Denominator[y]  (* A265801 *)

a(13)-a(18) from Robert Price, Apr 06 2019

A265799 Denominators of upper primes-only best approximates (POBAs) to the golden ratio, tau (A001622); see Comments.

Original entry on oeis.org

2, 3, 19, 29, 97, 563, 631, 919, 2647, 3181, 5333, 6841, 9281, 12941, 13873, 15271, 23633, 49939

Offset: 1

Clark Kimberling, Dec 29 2015

Suppose that x > 0. A fraction p/q of primes is an upper primes-only best approximate, and we write "p/q is in U(x)", if p'/q < x < p/q < u/v for all primes u and v such that v < q, where p' is greatest prime < p in case p >= 3.
Let q(1) = 2 and let p(1) be the least prime >= x. The sequence U(x) follows inductively: for n >= 1, let q(n) is the least prime q such that x < p/q < p(n)/q(n) for some prime p. Let q(n+1) = q and let p(n+1) be the least prime p such that x < p/q < p(n)/q(n).
For a guide to POBAs, lower POBAs, and upper POBAs, see A265759.

			The upper POBAs to tau start with 5/2, 5/3, 31/19, 47/29, 157/97, 911/563, 1021/631. For example, if p and q are primes and q > 97, and p/q > tau, then 157/97 is closer to tau than p/q is.

Cf. A000040, A001622, A265759, A265796, A265797, A265798, A265800, A265801.

x = GoldenRatio; z = 1000; p[k_] := p[k] = Prime[k];
t = Table[Max[Table[NextPrime[x*p[k], -1]/p[k], {k, 1, n}]], {n, 1, z}];
d = DeleteDuplicates[t]; tL = Select[d, # > 0 &] (* lower POBA *)
t = Table[Min[Table[NextPrime[x*p[k]]/p[k], {k, 1, n}]], {n, 1, z}];
d = DeleteDuplicates[t]; tU = Select[d, # > 0 &] (* upper POBA *)
v = Sort[Union[tL, tU], Abs[#1 - x] > Abs[#2 - x] &];
b = Denominator[v]; s = Select[Range[Length[b]], b[[#]] == Min[Drop[b, # - 1]] &];
y = Table[v[[s[[n]]]], {n, 1, Length[s]}] (* POBA, A265800/A265801 *)
Numerator[tL]   (* A265796 *)
Denominator[tL] (* A265797 *)
Numerator[tU]   (* A265798 *)
Denominator[tU] (* A265799 *)
Numerator[y]    (* A265800 *)
Denominator[y]  (* A265801 *)

a(13)-a(18) from Robert Price, Apr 06 2019

A019587 The left budding sequence: number of i such that 0 < i <= n and 0 < {phii} <= {phin}, where {} denotes the fractional part and phi = A001622.

Original entry on oeis.org

1, 1, 3, 2, 1, 5, 3, 8, 5, 2, 9, 5, 1, 10, 5, 15, 9, 3, 15, 8, 21, 13, 5, 20, 11, 2, 19, 9, 27, 16, 5, 25, 13, 1, 23, 10, 33, 19, 5, 30, 15, 41, 25, 9, 37, 20, 3, 33, 15, 46, 27, 8, 41, 21, 55, 34, 13, 49, 27, 5, 43, 20, 59, 35, 11, 52, 27, 2, 45, 19, 63, 36, 9, 55, 27, 74, 45, 16, 65

Offset: 1

N. J. A. Sloane and J. H. Conway

			{r} = 0.61...; {2r} = 0.23...; {3r} = 0.85...; {4r} = 0.47...; so that a(4) = 2.

J. H. Conway, personal communication.

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
Hamoon Mousavi et al., Walnut, An automatic theorem prover for words, version 7.0.
N. J. A. Sloane, Classic Sequences

Cf. A001622, A019588, A194733, A193738.

a019587 n = length $ filter (<= nTau) $
            map (snd . properFraction . (* tau) . fromInteger) [1..n]
   where (_, nTau) = properFraction (tau * fromInteger n)
         tau = (1 + sqrt 5) / 2
-- Reinhard Zumkeller, Jan 28 2012

Digits := 100;
A019587 := proc(n::posint)
    local a,k,phi,kfrac,nfrac ;
    phi := (1+sqrt(5))/2 ;
    a :=0 ;
    nfrac := n*phi-floor(n*phi) ;
    for k from 1 to n do
        kfrac := k*phi-floor(k*phi) ;
        if evalf(kfrac-nfrac)  <= 0 then
            a := a+1 ;
        end if;
    end do:
    a ;
end proc:
seq(A019587(n),n=1..100) ; # R. J. Mathar, Aug 13 2021

r = GoldenRatio; p[x_] := FractionalPart[x];
u[n_, k_] := If[p[k*r] <= p[n*r], 1, 0]
v[n_, k_] := If[p[k*r] > p[n*r], 1, 0]
s[n_] := Sum[u[n, k], {k, 1, n}]
t[n_] := Sum[v[n, k], {k, 1, n}]
Table[s[n], {n, 1, 100}]  (* A019587 *)
Table[t[n], {n, 1, 100}]  (* A194733 *)
(* Clark Kimberling, Sep 02 2011 *)

a(n) + A194733(n) = n.

The theorem prover Walnut (see link) can compute the following "linear representation" for a(n). Let v = [1,0,0,0,0,0,0,0,0,0,0,0], w = [0,1,1,3,2,1,5,3,8,5,2,9]^T, mu(0) =[[1,0,0,0,0,0,0,0,0,0,0,0], [0,0,1,0,0,0,0,0,0,0,0,0], [0,0,0,1,0,0,0,0,0,0,0,0], [0,0,0,0,0,1,0,0,0,0,0,0], [0,0,0,0,0,0,0,1,0,0,0,0], [0,0,0,0,0,0,0,0,1,0,0,0], [0,0,0,0,0,0,0,0,0,0,1,0], [0,0,0,0,0,0,0,0,0,0,0,1], [0,0,0,0,0,0,0,-1,0,0,2,0], [0,0,-2,0,0,1,0,2,0,0,0,0], [0,0,-1,-1,0,1,0,1,1,0,-1,1], [0,0,-1,0,0,0,0,0,0,0,2,0]], mu(1) = [[0,1,0,0,0,0,0,0,0,0,0,0], [0,0,0,0,0,0,0,0,0,0,0,0], [0,0,0,0,1,0,0,0,0,0,0,0], [0,0,0,0,0,0,1,0,0,0,0,0], [0,0,0,0,0,0,0,0,0,0,0,0], [0,0,0,0,0,0,0,0,0,1,0,0], [0,0,0,0,0,0,0,0,0,0,0,0], [0,0,0,0,0,0,-1,0,0,2,0,0], [0,-2,0,0,1,0,2,0,0,0,0,0],[0,0,0,0,0,0,0,0,0,0,0,0], [0,-1,0,0,0,0,0,0,0,2,0,0], [0,-4,0,0,2,0,4,0,0,-1,0,0]]. Then a(n) = v.mu(x).w, where x is the Zeckendorf (or Fibonacci) representation of n. This gives an algorithm for a(n) that runs in polynomial time in log n. - Jeffrey Shallit, Aug 09 2025

Extended by Ray Chandler, Apr 18 2009

cp's OEIS Frontend

A060144 a(n) = floor(n/(1+tau)), or equivalently floor(n/(tau)^2), where tau is the golden ratio (A001622).

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A296184 Decimal expansion of 2 + phi, with the golden section phi from A001622.

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A004956 a(n) = ceiling(n*phi), where phi is the golden ratio, A001622.

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A004942 a(n) = round(n*phi^7), where phi is the golden ratio, A001622.

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A004957 a(n) = ceiling(n*phi^2), where phi is the golden ratio, A001622.

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A265796 Numerators of lower primes-only best approximates (POBAs) to the golden ratio, tau (A001622); see Comments.

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A265797 Denominator of lower primes-only best approximates (POBAs) to the golden ratio, tau (A001622); see Comments.

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A265798 Numerators of upper primes-only best approximates (POBAs) to the golden ratio, tau (A001622); see Comments.

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A019587 The left budding sequence: number of i such that 0 < i <= n and 0 < {phii} <= {phin}, where {} denotes the fractional part and phi = A001622.