cp's OEIS Frontend

Every positive integer occurs exactly twice. Taking a Lucas number (A000032) of terms L(n) starting at a(0), the last two terms are a pair of Fibonacci numbers (A000045). If n is even, then the last two terms are F(n+1) followed by F(n-1), if n is odd they are F(n-1) followed by F(n+1), where F is the Fibonacci sequence. For example, the first L(4) = 7 terms of this sequence are (1,2,1,3,4,5,2) and the last members are 5 and 2 which are equal to F(5) and F(3). Note also that L(n) = F(n-1) + F(n+1).

A000211(n) = Lucas(n) + 2 is a term for all n > 2, since the representation of Lucas(n) + 2 is 10...01 with n-1 0's between the two 1's.

See A190508.

From Clark Kimberling, Nov 13 2022: (Start)

This is the third of four sequences that partition the positive integers. Suppose that u = (u(n)) and v = (v(n)) are increasing sequences of positive integers. Let u' and v' be their (increasing) complements, and consider these four sequences:

(1) v o u, defined by (v o u)(n) = v(u(n));

(2) u o v';

(3) v o u';

(4) v' o u'.

Every positive integer is in exactly one of the four sequences. For the reverse composites, u o v, u o v', u' o v, u' o v', see A356104 to A356107.

Assume that if w is any of the sequences u, v, u', v', then lim_{n->oo} w(n)/n exists and defines the (limiting) density of w. For w = u,v,u',v', denote the densities by r,s,r',s'. Then the densities of sequences (1)-(4) exist, and

1/(r*r') + 1/(r*s') + 1/(s*s') + 1/(s*r') = 1.

For this sequence, u, v, u', v', are the Beatty sequences given by u(n) = floor(n*(1+sqrt(5))/2) and v(n) = floor(n*sqrt(5)), so that r = (1+sqrt(5))/2, s = sqrt(5), r' = (3+sqrt(5))/2, s' = (5 + sqrt(5))/4.

(1) v o u = (2, 6, 8, 13, 17, 20, 24, 26, 31, 35, 38, 42, ...) = A356217

(2) v' o u = (1, 5, 7, 10, 14, 16, 19, 21, 25, 28, 30, 34, ...) = A356218

(3) v o u' = (4, 11, 15, 22, 29, 33, 40, 44, 51, 58, 62, 76, ...) = this sequence

(4) v' o u' = (3, 9, 12, 18, 23, 27, 32, 36, 41, 47, 50, 56, ...) = A356220

(End)

This is one of four sequences that partition the positive integers. In general, suppose that r, s, t, u are positive real numbers for which the sets {i/r: i>=1}, {j/s: j>=1}, {k/t: k>=1}, {h/u: h>=1} are pairwise disjoint. Let a(n) be the rank of n/r when all the numbers in the four sets are jointly ranked. Define b(n), c(n), d(n) as the ranks of n/s, n/t, n/u, respectively.

It is easy to prove that

a(n)=n+[n*s/r]+[n*t/r]+[n*u/r],

b(n)=n+[n*r/s]+[n*t/s]+[n*u/s],

c(n)=n+[n*r/t]+[n*s/t]+[n*u/t],

d(n)=n+[n*r/u]+[n*s/u]+[n*t/u], where []=floor.

Taking r=golden ratio, s=r^2, t=r^3, u=r^4 gives a=A190508, b=A190509, c=A054770, d=A190511.

A190508: a(n)=n+[nr]+[nr^2]+[nr^3]

A190509: b(n)=[n/r]+n+[nr]+[nr^2]

A054770: c(n)=[n/r^2]+[n/r]+n+[nr]

A190511: d(n)=[n/r^3]+[n/r^2]+[n/r]+n

Are the differences between successive terms always 2 or 3? - Harvey P. Dale, Apr 03 2025

Convolution of the sequences A003263 and A033999.

Positions of 0: 1, 2, 5, 6, 8, 9, 12, 13, ... = A287775(n) - 1 (conjecture).

From Michel Dekking, Dec 30 2017: (Start)

Proof of the 'positions of 0' conjecture: let (z(n))=1,2,5,6,8,9,12,... be the positions of 0. The crucial observation is that if a number n is the sum of distinct Lucas parts greater than 1, then n+1 is a sum of Lucas parts. This implies that (z(2n))=2,6,9,13,... is the sequence of numbers A054770 that are not a sum of Lucas numbers. We see there that Ian Agol proved that b(n):=A054770(n)=floor(phi*n)+2n-1. But then the sequence of first differences (b(n+1)-b(n)) equals the Fibonacci word on the alphabet {4,3}, yielding that (z(2n)-z(2n-1)) equals the Fibonacci word on {3,2}, and we already know that z(2n+1)-z(2n)=1 for all n. On the other hand, A287775 has the same first difference sequence given by A108103. Since A287775(1)=2, the conjecture follows. (End)

Positions of 1: 0, 3, 4, 10, 15, 28, 44, 75, ... = A001350(n+1) - 1 (conjecture).

From Michel Dekking, Aug 26 2019: (Start)

This sequence is a generalized Beatty sequence. We know that A054770, the sequence of numbers whose Lucas representation includes L_0=2, is equal to A054770(n) = A000201(n) + 2*n - 1 = floor((phi+2)*n) - 1.

One also easily checks that the numbers 3-phi and phi+2 form a Beatty pair. This implies that the sequence with terms floor((3-phi)*n)-1 is the complement of A054770 in the natural numbers 0,1,2,...

It follows that a(n) = 3*n - floor(n*phi) - 2.

(End)