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A190509
a(n) = n + [nr/s] + [nt/s] + [nu/s] where r=golden ratio, s=r^2, t=r^3, u=r^4, and [] represents the floor function.
12
4, 11, 15, 22, 29, 33, 40, 44, 51, 58, 62, 69, 76, 80, 87, 91, 98, 105, 109, 116, 120, 127, 134, 138, 145, 152, 156, 163, 167, 174, 181, 185, 192, 199, 203, 210, 214, 221, 228, 232, 239, 243, 250, 257, 261, 268, 275, 279, 286, 290, 297, 304, 308, 315, 319, 326, 333, 337, 344, 351, 355, 362, 366, 373, 380, 384, 391, 398, 402, 409
OFFSET
1,1
COMMENTS
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)
LINKS
Weiru Chen and Jared Krandel, Interpolating Classical Partitions of the Set of Positive Integers, arXiv:1810.11938 [math.NT], 2018. See sequence D1 p. 4. Also in The Ramanujan Journal, (2020).
FORMULA
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.
a(n) = 3*A000201(n)+n, since r/s = 1/r = r-1, and u/s = r^2 = r+1. - Michel Dekking, Sep 06 2017
a(n) = A000201(n) + A003623(n). - Primoz Pirnat, Jan 08 2021
MAPLE
r:=(1+sqrt(5))/2: s:=r^2: t:=r^3: u:=r^4: a:=n->n+floor(n*r/s)+floor(n*t/s)+floor(n*u/s): seq(a(n), n=1..70); # Muniru A Asiru, Nov 01 2018
MATHEMATICA
(See A190508.)
Table[3 Floor[n (Sqrt[5] + 1) / 2] + n, {n, 1, 100}] (* Vincenzo Librandi, Nov 01 2018 *)
PROG
(PARI) a(n) = 3*floor(n*(sqrt(5)+1)/2) + n; \\ Michel Marcus, Sep 10 2017; after Michel Dekking's formula
(Magma) [3*Floor(n*(Sqrt(5)+1)/2) + n: n in [1..80]]; // Vincenzo Librandi, Nov 01 2018
(Python)
from math import isqrt
def A190509(n): return n+((m:=n+isqrt(5*n**2))&-2)+(m>>1) # Chai Wah Wu, Aug 10 2022
CROSSREFS
KEYWORD
nonn
AUTHOR
Clark Kimberling, May 11 2011
STATUS
approved