A341240 -id:A341240 - cp's OEIS frontend

A341239 a(n) = floor(rfloor(sn)), where r = 1 + sqrt(2) and s = sqrt(2).

2, 4, 9, 12, 16, 19, 21, 26, 28, 33, 36, 38, 43, 45, 50, 53, 57, 60, 62, 67, 70, 74, 77, 79, 84, 86, 91, 94, 98, 101, 103, 108, 111, 115, 118, 120, 125, 127, 132, 135, 137, 142, 144, 149, 152, 156, 159, 161, 166, 168, 173, 176, 178, 183, 185, 190, 193, 197

Offset: 1

Clark Kimberling, Feb 07 2021

nonn

Conjecture: 1 < r*s*n - a(n) < 3 for n >= 1.
From Clark Kimberling, Dec 27 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 (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 A184922.
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 A341239, u, v, u', v', are the Beatty sequences given by u(n) = floor(n*sqrt(2)) and v(n) = floor((1+sqrt(2))/2)*n), so that r = sqrt(2), s = (1+sqrt(2))/2, r' = (2+sqrt(2))/2, s' = 1 + 1/sqrt(2).
(1) v o u = (2, 4, 9, 12, 16, 19, 21, 26, 28, 33, 36, 38, ...) = A341239
(2) v' o u = (1, 3, 6, 8, 11, 13, 15, 18, 20, 23, 25, 27, ...) = A286666
(3) v o u' = (7, 14, 24, 31, 41, 48, 55, 65, 72, 82, 89, ...) = A188383
(4) v' o u' = (5, 10, 17, 22, 29, 34, 39, 46, 51, 58, 63, ...) = A098021
(End)

Cf. A184922, A341240; A098021, A188383, A341239, A286666.

z = 140; r = 1 + Sqrt[2]; s = Sqrt[2]; f[x_] := Floor[r*Floor[s*x]];
Table[f[n], {n, 1, z}]

a(n) = floor(r*floor(s*n)), where r = 1 + sqrt(2) and s = sqrt(2).

A339828 a(n) = 4a(n-1) - 2a(n-2) + a(n-3) - 4*a(n-4) + a(n-5) for n >= 6, where a(1) = 1, a(2) = 3, a(3) = 5, a(4) = 16, a(5) = 53.

Original entry on oeis.org

1, 2, 5, 16, 53, 179, 610, 2081, 7103, 24250, 82793, 282671, 965098, 3295049, 11249999, 38409898, 131139593, 447738575, 1528675114, 5219223305, 17819542991, 60839725354, 207719815433, 709199811023, 2421359613226, 8267038830857, 28225436096975, 96367666726186

Offset: 1

Clark Kimberling, Feb 07 2021

nonn

Index entries for linear recurrences with constant coefficients, signature (4,-2,1,-4,2).

Cf. A184922, A341239, A341240.

z = 40; r = Sqrt[2]; s = 1 + Sqrt[2]; f[x_] := Floor[r*Floor[s*x]];
Table[f[n], {n, 1, z}] m (* A184922 *)
a[1] = 1; a[n_] := f[a[n - 1]]; Table[a[n], {n, 1, z}] (* A339828 *)

Let f(n) = floor(r*floor(s*n)) = A184922(n), where r = sqrt(2) and s = r + 1. Let a(1) = 1. Then a(n) = f(a(n-1)) for n >= 2.
Also, a(n) = 4*a(n-1) - 2*a(n-2) + a(n-3) - 4*a(n-4) + a(n-5) for n >= 6, where a(1) = 1, a(2) = 3, a(3) = 5, a(4) = 16, a(5) = 53.
G.f.: x*(-x^4 + x^3 + x^2 + 2*x - 1)/((x - 1)*(x^2 + x + 1)*(2*x^2 - 4*x + 1)). - Chai Wah Wu, Feb 15 2021

A341250 a(n) = 5a(n-1) - 4a(n-3) for n >= 4, where a(1) = 1, a(2) = 3, a(3) = 13.

Original entry on oeis.org

1, 3, 13, 61, 293, 1413, 6821, 32933, 159013, 767781, 3707173, 17899813, 86427941, 417311013, 2014955813, 9729067301, 46976092453, 226820639013, 1095186925861, 5288030259493, 25532868741413, 123283596003621, 595265858980133, 2874197819935013

Offset: 1

Clark Kimberling, Feb 13 2021

Index entries for linear recurrences with constant coefficients, signature (5,0,-4).

Cf. A218989, A339828, A341240, A341249.

z = 40; r = 2 + Sqrt[2]; s = Sqrt[2]; f[x_] := Floor[r*Floor[s*x]];
Table[f[n], {n, 1, z}] (* A341249 *)
a[1] = 1; a[n_] := f[a[n - 1]];
Table[a[n], {n, 1, z}] (* A341250 *)

Let f(n) = floor(r*floor(s*n)) = A341249(n), where r = 2 + sqrt(2) and s = sqrt(2).  Let a(1) = 1. Then a(n) = f(a(n-1)) for n >= 2.
a(n) = (A218989(n-2) + 1)/2. - Hugo Pfoertner, Feb 13 2021
G.f.: x*(-2*x^2 - 2*x + 1)/(4*x^3 - 5*x + 1). - Chai Wah Wu, Feb 15 2021

A341255 Let f(n) = floor(rfloor(rn)) = A341254(n), where r = (2 + sqrt(5))/2. Let a(1) = 1. Then a(n) = f(a(n-1)) for n >= 2.

Original entry on oeis.org

1, 4, 16, 69, 309, 1385, 6212, 27866, 125008, 560793, 2515754, 11285842, 50629052, 227125366, 1018899829, 4570853893, 20505161277, 91987547377, 412662390616, 1851231536059, 8304750512850, 37255675336820, 167131492108634, 749763234780300, 3363488838254558

Offset: 1

Clark Kimberling, Feb 13 2021

Cf. A339828, A341240, A341249.

z = 40; u = GoldenRatio;
r = u + 1/2; f[x_] := Floor[r*Floor[r*x]];
Table[f[n], {n, 1, z}]  (* A341254 *)
a[1] = 1; a[n_] := f[a[n - 1]];
Table[a[n], {n, 1, z}]  (* A341255 *)

A341248 a(n) = 5a(n-1) - 4a(n-3) for n >= 4, where a(1) = 1, a(2) = 4, a(3) = 18.

Original entry on oeis.org

1, 4, 18, 86, 414, 1998, 9646, 46574, 224878, 1085806, 5242734, 25314158, 122227566, 590166894, 2849577838, 13758978926, 66434227054, 320772823918, 1548828203886, 7478404111214, 36108929260398, 174349333486446, 841833050987374, 4064729537895278

Offset: 1

Clark Kimberling, Feb 13 2021

Index entries for linear recurrences with constant coefficients, signature (5,0,-4).

Cf. A339828, A341240, A341250.

z = 40; r = Sqrt[2]; s = 2 + Sqrt[2]; f[x_] := Floor[r*Floor[s*x]];
Table[f[n], {n, 1, z}] (* A022804 *)
a[1] = 1; a[n_] := f[a[n - 1]];
t = Table[a[n], {n, 1, z}] (* A341248 *)

Let f(n) = floor(r*floor(s*n)) = A022804(n), where r = sqrt(2) and s = r + 2. Let a(1) = 1. Then a(n) = f(a(n-1)) for n >= 2.

G.f.: x*(1 - x - 2*x^2)/(1 - 5*x + 4*x^3). - Stefano Spezia, Feb 13 2021

cp's OEIS Frontend

A341239 a(n) = floor(r*floor(s*n)), where r = 1 + sqrt(2) and s = sqrt(2).

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A339828 a(n) = 4*a(n-1) - 2*a(n-2) + a(n-3) - 4*a(n-4) + a(n-5) for n >= 6, where a(1) = 1, a(2) = 3, a(3) = 5, a(4) = 16, a(5) = 53.

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A341250 a(n) = 5*a(n-1) - 4*a(n-3) for n >= 4, where a(1) = 1, a(2) = 3, a(3) = 13.

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A341255 Let f(n) = floor(r*floor(r*n)) = A341254(n), where r = (2 + sqrt(5))/2. Let a(1) = 1. Then a(n) = f(a(n-1)) for n >= 2.

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Mathematica

A341248 a(n) = 5*a(n-1) - 4*a(n-3) for n >= 4, where a(1) = 1, a(2) = 4, a(3) = 18.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

Formula

A341239 a(n) = floor(rfloor(sn)), where r = 1 + sqrt(2) and s = sqrt(2).

A339828 a(n) = 4a(n-1) - 2a(n-2) + a(n-3) - 4*a(n-4) + a(n-5) for n >= 6, where a(1) = 1, a(2) = 3, a(3) = 5, a(4) = 16, a(5) = 53.

A341250 a(n) = 5a(n-1) - 4a(n-3) for n >= 4, where a(1) = 1, a(2) = 3, a(3) = 13.

A341255 Let f(n) = floor(rfloor(rn)) = A341254(n), where r = (2 + sqrt(5))/2. Let a(1) = 1. Then a(n) = f(a(n-1)) for n >= 2.

A341248 a(n) = 5a(n-1) - 4a(n-3) for n >= 4, where a(1) = 1, a(2) = 4, a(3) = 18.