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Conjecture: every term of the difference sequence is in {2,3,4,5,6}, and each occurs infinitely many times.

From Clark Kimberling, Jul 29 2022: (Start)

This is the first of four sequences that partition the positive integers. Starting with a general overview, suppose that u = (u(n)) and v = (v(n)) are increasing sequences of positive integers. Let u' and v' be their complements, and assume that the following four sequences are infinite:

(1) u ^ v = intersection of u and v (in increasing order);

(2) u ^ v';

(3) u' ^ v;

(4) u' ^ v'.

Every positive integer is in exactly one of the four sequences. For A351415, 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) u ^ v = (4, 6, 8, 11, 17, 22, 24, 29, 33, 35, 38, 40, 42, ...) = A351415

(2) u ^ v' = (1, 3, 9, 12, 14, 16, 19, 21, 25, 27, 30, 32, ...) = A356101

(3) u' ^ v = (2, 13, 15, 20, 26, 31, 44, 49, 60, 62, 73, 78, ...) = A356102

(4) u' ^ v' = (5, 7, 10, 18, 23, 28, 34, 36, 39, 41, 47, 52, ...) = A356103

(End)

This is the first 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) v' o u;

(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 A356217 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.

This is the fourth of four sequences that partition the positive integers. See A356217.

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

This is the second of four sequences that partition the positive integers. See A351415.

This is the third of four sequences that partition the positive integers. See A351415.

This is the fourth of four sequences that partition the positive integers. See A351415.

This is the second of four sequences that partition the positive integers. See A356217.