A274213 Meta recurrence: a(0) = 1, a(1) = 2, a(2) = 3, a(n) = a(n - a(n-3)) + 3 for n > 2.

1, 2, 3, 6, 6, 6, 4, 5, 6, 9, 9, 9, 9, 9, 9, 7, 8, 9, 12, 12, 12, 12, 12, 12, 12, 12, 12, 10, 11, 12, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 13, 14, 15, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 16, 17, 18, 21, 21, 21, 21, 21, 21, 21

Offset

0,2

Comments

The sequence is constructed by starting with 3*m copies of 3*(m+1), followed by 3*m+1, 3*m+2, 3*m+3, as m varies from 0, 1, 2, ... It is straightforward to check that this construction satisfies the recurrence relation.

The construction shows that the sequence is well defined, every positive integer is in the sequence, and every integer not a proper multiple of 3 appears only once. If t is a multiple of 3, then t appears t-2 times.

In general, the meta recurrence a(n) = a(n-a(n-k))+k with initial conditions a(i) = i+1 for i = 0,...,k-1 has a simple solution and can be constructed starting with k*m copies of k*(m+1), followed by k*m+1, k*m+2, ..., k*(m+1), as m varies from 0, 1, 2, ... This sequence is well defined, every positive integer is in the sequence, and every integer not a proper multiple of k appears once. If t is a multiple of k, then t appears t-k+1 times.

Links

Chai Wah Wu, Table of n, a(n) for n = 0..10000

Mathematica

A274213[n_] := A274213[n] = If[n < 3, n + 1, A274213[n - A274213[n - 3]] + 3]; Array[A274213, 100, 0] (* Paolo Xausa, May 24 2024 *)

Prog

(Python)
A274213_list = [1, 2, 3]
for n in range(3, 10001):
    A274213_list.append(A274213_list[-A274213_list[-3]]+3)
(Magma) I:=[1, 2, 3]; [n le 3 select I[n] else Self(n-Self(n-3))+3 : n in [1..80]]; // Vincenzo Librandi, Jun 18 2016

Crossrefs

Cf. A193358

Sequence in context: A085273 A151850 A290223 * A350315 A078706 A077082

Adjacent sequences: A274210 A274211 A274212 * A274214 A274215 A274216

Keyword

nonn

Author

Chai Wah Wu, Jun 13 2016

Status

approved