A374318 For any n > 0, let b_n(n+1) = 0, and for k = 1..n, if b_n(k+1) >= k then b_n(k) = b_n(k+1) - k otherwise b_n(k) = b_n(k+1) + k; a(n) = b_n(1).

0, 1, 1, 0, 2, 1, 1, 2, 0, 1, 1, 2, 0, 1, 1, 2, 2, 1, 1, 0, 2, 1, 1, 0, 2, 1, 1, 0, 2, 1, 1, 0, 2, 1, 1, 0, 2, 1, 1, 0, 2, 1, 1, 2, 2, 1, 1, 2, 2, 1, 1, 2, 0, 1, 1, 2, 0, 1, 1, 2, 0, 1, 1, 2, 0, 1, 1, 2, 0, 1, 1, 2, 0, 1, 1, 2, 0, 1, 1, 2, 0, 1, 1, 2, 0, 1, 1

Offset

0,5

Comments

This sequence is a variant of A008344; here we add or subtract by numbers from n down to 1, there by numbers from 1 up to n.

Apparently, the sequence only contains 0's, 1's and 2's.

Links

Table of n, a(n) for n=0..86.

Rémy Sigrist, Colored representation of b_n(k) for n <= 1000 (where the color at (x, y) is function of b_x(y))

Rémy Sigrist, Log-log scatterplot of the ordinal transform of the first 10000 terms

Formula

Empirically, a(n) = 1 iff n belongs to A042963.

Example

The first terms, alongside the corresponding sequences b_n, are:
  n   a(n)  b_n
  --  ----  ----------------------------------
   0     0  [0]
   1     1  [1, 0]
   2     1  [1, 2, 0]
   3     0  [0, 1, 3, 0]
   4     2  [2, 3, 1, 4, 0]
   5     1  [1, 2, 4, 1, 5, 0]
   6     1  [1, 0, 2, 5, 1, 6, 0]
   7     2  [2, 3, 5, 2, 6, 1, 7, 0]
   8     0  [0, 1, 3, 6, 2, 7, 1, 8, 0]
   9     1  [1, 2, 0, 3, 7, 2, 8, 1, 9, 0]
  10     1  [1, 2, 4, 7, 3, 8, 2, 9, 1, 10, 0]

Prog

(PARI) a(n) = { my (b = 0); forstep (k = n, 1, -1, if (b >= k, b -= k, b += k); ); return (b); }

Crossrefs

Cf. A008344, A042963, A374317.

Sequence in context: A025925 A372510 A240857 * A109066 A079066 A157188

Adjacent sequences: A374315 A374316 A374317 * A374319 A374320 A374321

Keyword

nonn

Author

Rémy Sigrist, Jul 04 2024

Status

approved