%I #63 Oct 20 2024 02:09:21
%S 4,11,15,22,29,33,40,44,51,58,62,69,76,80,87,91,98,105,109,116,120,
%T 127,134,138,145,152,156,163,167,174,181,185,192,199,203,210,214,221,
%U 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
%N 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.
%C See A190508.
%C From _Clark Kimberling_, Nov 13 2022: (Start)
%C 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:
%C (1) v o u, defined by (v o u)(n) = v(u(n));
%C (2) u o v';
%C (3) v o u';
%C (4) v' o u'.
%C 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.
%C 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
%C 1/(r*r') + 1/(r*s') + 1/(s*s') + 1/(s*r') = 1.
%C 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.
%C (1) v o u = (2, 6, 8, 13, 17, 20, 24, 26, 31, 35, 38, 42, ...) = A356217
%C (2) v' o u = (1, 5, 7, 10, 14, 16, 19, 21, 25, 28, 30, 34, ...) = A356218
%C (3) v o u' = (4, 11, 15, 22, 29, 33, 40, 44, 51, 58, 62, 76, ...) = this sequence
%C (4) v' o u' = (3, 9, 12, 18, 23, 27, 32, 36, 41, 47, 50, 56, ...) = A356220
%C (End)
%H Vincenzo Librandi, <a href="/A190509/b190509.txt">Table of n, a(n) for n = 1..10000</a>
%H Weiru Chen and Jared Krandel, <a href="https://arxiv.org/abs/1810.11938">Interpolating Classical Partitions of the Set of Positive Integers</a>, arXiv:1810.11938 [math.NT], 2018. See sequence D1 p. 4. Also in <a href="https://doi.org/10.1007/s11139-019-00196-3">The Ramanujan Journal</a>, (2020).
%F A190508: a(n) = n + [nr] + [nr^2] + [nr^3];
%F A190509: b(n) = [n/r] + n + [nr] + [nr^2];
%F A054770: c(n) = [n/r^2] + [n/r] + n + [nr];
%F A190511: d(n) = [n/r^3] + [n/r^2] + [n/r] + n.
%F 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
%F a(n) = A000201(n) + A003623(n). - _Primoz Pirnat_, Jan 08 2021
%p 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
%t (See A190508.)
%t Table[3 Floor[n (Sqrt[5] + 1) / 2] + n, {n, 1, 100}] (* _Vincenzo Librandi_, Nov 01 2018 *)
%o (PARI) a(n) = 3*floor(n*(sqrt(5)+1)/2) + n; \\ _Michel Marcus_, Sep 10 2017; after _Michel Dekking_'s formula
%o (Magma) [3*Floor(n*(Sqrt(5)+1)/2) + n: n in [1..80]]; // _Vincenzo Librandi_, Nov 01 2018
%o (Python)
%o from math import isqrt
%o def A190509(n): return n+((m:=n+isqrt(5*n**2))&-2)+(m>>1) # _Chai Wah Wu_, Aug 10 2022
%Y Cf. A054770, A190508, A190511.
%Y Cf. A000201, A001622, A003623, A340429.
%K nonn
%O 1,1
%A _Clark Kimberling_, May 11 2011