A330262 Start with an empty stack S; for n = 1, 2, 3, ..., interpret the binary representation of n from left to right as follows: in case of bit 1, push the number 1 on top of S, in case of bit 0, replace the two numbers on top of S, say u on top of v, with v-u; a(n) gives the number on top of S after processing n.

1, 0, 1, 1, 1, 0, 1, 0, 1, -1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 2, 1, 0, 1, 0, 1, -1, 1, 1, 1, 0, 1, 0, 1, -1, 1, -1, 1, 0, 1, -1, 1, -2, 1, 0, 1, 0, 1, 1, 1, 0, 1, 2, 1, 0, 1, 0, 1, -1, 1, 1, 1, 0, 1, 1, 1, -1, 1, 0, 1, 0, 1, -3, 1, -3, 1, 1, 1, 0, 1, 2, 1, -2, 1

Offset

1,20

Comments

This sequence is a variant of A330261.

After processing n, S has A268289(n) elements.

Every integer appears infinitely many times in the sequence:

- the proof is similar to that found in A330261,

- see A330265 for the values in order of appearance.

Links

Rémy Sigrist, Table of n, a(n) for n = 1..8192

Rémy Sigrist, Scatterplot of the first 2^20 terms

Rémy Sigrist, PARI program for A330262

Example

The first terms, alongside the binary representation of n and the evolution of stack S, are:
  n  a(n)  bin(n)  S
  -  ----  ------  ------------------------------------------------------------
  1     1       1  () -> (1)
  2     0      10  (1) -> (1,1) -> (0)
  3     1      11  (0) -> (0,1) -> (0,1,1)
  4     1     100  (0,1,1) -> (0,1,1,1) -> (0,1,0) -> (0,1)
  5     1     101  (0,1) -> (0,1,1) -> (0,0) -> (0,0,1)
  6     0     110  (0,0,1) -> (0,0,1,1) -> (0,0,1,1,1) -> (0,0,1,0)
  7     1     111  (0,0,1,0) -> (0,0,1,0,1) -> (0,0,1,0,1,1) -> (0,0,1,0,1,1,1)

Prog

(PARI) See Links section.

Crossrefs

Cf. A268289, A308551, A330261, A330265.

Sequence in context: A355824 A355826 A355819 * A098055 A344739 A092111

Adjacent sequences: A330259 A330260 A330261 * A330263 A330264 A330265

Keyword

sign,base

Author

Rémy Sigrist, Dec 07 2019

Status

approved