A190508 -id:A190508 - cp's OEIS frontend

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.

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

Clark Kimberling, May 11 2011

nonn

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)

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
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).

Cf. A054770, A190508, A190511.

Cf. A000201, A001622, A003623, A340429.

[3*Floor(n*(Sqrt(5)+1)/2) + n: n in [1..80]]; // Vincenzo Librandi, Nov 01 2018

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

(See A190508.)
Table[3 Floor[n (Sqrt[5] + 1) / 2] + n, {n, 1, 100}] (* Vincenzo Librandi, Nov 01 2018 *)

a(n) = 3*floor(n*(sqrt(5)+1)/2) + n; \\ Michel Marcus, Sep 10 2017; after Michel Dekking's formula

from math import isqrt
def A190509(n): return n+((m:=n+isqrt(5*n**2))&-2)+(m>>1) # Chai Wah Wu, Aug 10 2022

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

A347068 Rectangular array (T(n,k)), by downward antidiagonals: T(n,k) = position of k in the ordering of {h*r^m, r = 1/(golden ratio), h >= 1, 0 <= m <= n}.

Original entry on oeis.org

2, 5, 4, 7, 10, 8, 10, 14, 18, 14, 13, 20, 26, 31, 25, 15, 26, 36, 46, 53, 42, 18, 30, 47, 63, 79, 88, 71, 20, 36, 55, 81, 107, 132, 146, 117, 23, 40, 65, 96, 136, 178, 219, 239, 193, 26, 46, 73, 112, 162, 225, 294, 359, 391, 315, 28, 52, 84, 127, 189, 269

Offset: 1

Clark Kimberling, Sep 02 2021

Row 1: A001950 (upper Wythoff sequence);
row 2: A283234;
row 3: A190508;
col 1: A020956.

			Corner:
    2,   5,   7,  10,  13,  15,  18,  20,  23, ...
    4,  10,  14,  20,  26,  30,  36,  40,  46, ...
    8,  18,  26,  36,  47,  55,  65,  73,  84, ...
   14,  31,  46,  63,  81,  96, 112, 127, 145, ...
   25,  53,  79, 107, 136, 162, 189, 215, 244, ...
   42,  88, 132, 178, 225, 269, 314, 358, 405, ...
   71, 146, 219, 294, 370, 443, 517, 590, 666, ...
   ...

Cf. A001950, A020956, A283234, A190508, A347065, A347066, A347067, A347069.

z = 1000; r = N[(-1+Sqrt[5])/2];
s[m_] := Range[z] r^m; t[0] = s[0];
t[n_] := Sort[Union[s[n], t[n - 1]]]
row[n_] := Flatten[Table[Position[t[n], N[k]], {k, 1, z}]]
TableForm[Table[row[n], {n, 1, 10}]] (* A347068, array *)
w[n_, k_] := row[n][[k]];
Table[w[n - k + 1, k], {n, 12}, {k, n, 1, -1}] // Flatten (* A347068, sequence *)

A190511 n+[nr/u]+[ns/u]+[nt/u]; r=golden ratio, s=r^2, t=r^3, u=r^4.

Original entry on oeis.org

1, 3, 5, 7, 10, 12, 14, 16, 19, 21, 23, 25, 28, 30, 32, 34, 37, 39, 41, 43, 45, 48, 50, 52, 54, 57, 59, 61, 63, 66, 68, 70, 72, 75, 77, 79, 81, 83, 86, 88, 90, 92, 95, 97, 99, 101, 104, 106, 108, 110, 113, 115, 117, 119, 121, 124, 126, 128, 130, 133, 135, 137, 139, 142, 144, 146, 148, 151, 153, 155, 157, 159, 162, 164, 166, 168, 171, 173

Offset: 1

Clark Kimberling, May 11 2011

nonn

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

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A190508, A190509, A054770.

(See A190508.)
Module[{g=GoldenRatio,g4},g4=g^4;Table[n+Floor[(n*g)/g4]+Floor[(n*g^2)/g4]+Floor[(n*g^3)/g4],{n,80}]] (* Harvey P. Dale, Apr 03 2025 *)

a(n) = A022839(n)-1. - Michel Dekking, May 04 2019

cp's OEIS Frontend

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.

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A347068 Rectangular array (T(n,k)), by downward antidiagonals: T(n,k) = position of k in the ordering of {h*r^m, r = 1/(golden ratio), h >= 1, 0 <= m <= n}.

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A190511 n+[nr/u]+[ns/u]+[nt/u]; r=golden ratio, s=r^2, t=r^3, u=r^4.

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