A194960 a(n) = floor((n+2)/3) + ((n-1) mod 3).

1, 2, 3, 2, 3, 4, 3, 4, 5, 4, 5, 6, 5, 6, 7, 6, 7, 8, 7, 8, 9, 8, 9, 10, 9, 10, 11, 10, 11, 12, 11, 12, 13, 12, 13, 14, 13, 14, 15, 14, 15, 16, 15, 16, 17, 16, 17, 18, 17, 18, 19, 18, 19, 20, 19, 20, 21, 20, 21, 22, 21, 22, 23, 22, 23, 24, 23, 24, 25, 24, 25, 26, 25, 26

Offset: 1

Author

Clark Kimberling, Sep 06 2011

Keywords

• nonn
• easy

Comments

The sequence is formed by concatenating triples of the form (n, n+1, n+2) for n>=1. See A194961 and A194962 for the associated fractalization and interspersion. The sequence can be obtained from A008611 by deleting its first four terms.

The sequence contains every positive integer n exactly min(n,3) times. - Wesley Ivan Hurt, Dec 17 2013

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).

Crossrefs

Cf. A006446, A008611, A010892, A047878, A049347, A130772, A194961, A194962, A194963, A330396.

Programs

Magma

I:=[1,2,3,2]; [n le 4 select I[n] else Self(n-1)+Self(n-3)-Self(n-4): n in [1..100]]; // Vincenzo Librandi, Dec 17 2013

Maple

A194960:=n->floor((n+2)/3)+((n-1) mod 3); seq(A194960(n), n=1..100); # Wesley Ivan Hurt, Dec 17 2013

Mathematica

(* First program *)
p[n_]:= Floor[(n+2)/3] + Mod[n-1, 3]
Table[p[n], {n, 1, 90}]  (* A194960 *)
g[1] = {1}; g[n_]:= Insert[g[n-1], n, p[n]]
f[1] = g[1]; f[n_]:= Join[f[n-1], g[n]]
f[20]  (* A194961 *)
row[n_]:= Position[f[30], n];
u = TableForm[Table[row[n], {n, 1, 5}]]
v[n_, k_]:= Part[row[n], k];
w = Flatten[Table[v[k, n-k+1], {n, 1, 13}, {k, 1, n}]]  (* A194962 *)
q[n_]:= Position[w, n];
Flatten[Table[q[n], {n, 1, 80}]]  (* A194963 *)
(* Other programs *)
CoefficientList[Series[(1 +x +x^2 -2 x^3)/((1+x+x^2) (1-x)^2), {x, 0, 100}], x] (* Vincenzo Librandi, Dec 17 2013 *)
Table[(n+4 -2*ChebyshevU[2*n+4, 1/2])/3, {n,80}] (* G. C. Greubel, Oct 23 2022 *)

PARI

a(n)=(n+2)\3 + (n-1)%3 \\ Charles R Greathouse IV, Sep 02 2015

SageMath

[(n+4 - 2*chebyshev_U(2*n+4, 1/2))/3 for n in (1..80)] # G. C. Greubel, Oct 23 2022

Formula

From R. J. Mathar, Sep 07 2011: (Start)

a(n) = ((-1)^n*A130772(n) + n + 4)/3.

G.f.: x*(1 + x + x^2 - 2*x^3)/((1+x+x^2)*(1-x)^2). (End)

a(n) = A006446(n)/floor(sqrt(A006446(n))). - Benoit Cloitre, Jan 15 2012

a(n) = a(n-1) + a(n-3) - a(n-4). - Vincenzo Librandi, Dec 17 2013

a(n) = a(n-3) + 1, n >= 1, with input a(-2) = 0, a(-1) = 1 and a(0) = 2. Proof trivial. a(n) = A008611(n+3), n >= -2. See the first comment above. - Wolfdieter Lang, May 06 2017

From Guenther Schrack, Nov 09 2020: (Start)

a(n) = n - 2*floor((n-1)/3).

a(n) = (n + 2 + 2*((n-1) mod 3))/3.

a(n) = (3*n + 12 + 2*(w^(2*n)*(1 - w) + w^n*(2 + w)))/9, where w = (-1 + sqrt(-3))/2.

a(n) = (n + 4 + 2*A049347(n))/3.

a(n) = (2*n + 3 - A330396(n-1))/3. (End)

a(n) = (n + 4 - 2*A010892(2*n+4))/3. - G. C. Greubel, Oct 23 2022