A367315 a(1) = 1, and for any n > 0, a(2*n) is the number of k's among 1..n such that a(k) <= a(n), a(2*n+1) is the number of k's among 1..n such that a(k) >= a(n).

1, 1, 1, 2, 2, 3, 3, 4, 1, 5, 2, 6, 1, 7, 2, 8, 1, 4, 9, 10, 1, 7, 7, 12, 1, 5, 13, 14, 1, 9, 10, 16, 1, 6, 17, 14, 6, 19, 1, 20, 1, 7, 21, 19, 5, 20, 6, 24, 1, 8, 25, 18, 10, 27, 1, 28, 1, 9, 29, 26, 6, 28, 5, 32, 1, 10, 33, 22, 14, 35, 1, 34, 4, 23, 17, 38

Offset: 1

Author

Rémy Sigrist, Nov 14 2023

Keywords

• nonn

Comments

This sequence is unbounded: for any m > 0, a(4*m) + a(4*m+1) >= 2*m + 1, as both terms are positive, at least one of them must be >= m.

This sequence contains infinitely many 1's: if a(n) > a(k) for all k < n, then a(2*n + 1) = 1, and as the sequence is unbounded, we have infinitely many such n.

This sequence contains all positive integers: for any n > 0, if k is the index of the n-th 1, then a(2*k) = n.

Examples

			a(1) = 1 by definition.
a(1) <= a(1) hence a(2) = 1.
a(1) >= a(1) hence a(3) = 1.
a(1) and a(2) <= a(2) hence a(4) = 2.
a(1), a(2), a(3) and a(4) <= a(4) hence a(8) = 4.
a(1), a(2) and a(3) < a(4), a(4) >= a(4) hence a(9) = 1.
		

Crossrefs

Cf. A272727.

Programs

PARI

{ for (n = 1, #a = vector(76), print1 (a[n] = if (n==1, 1, sum (k=1, n\2, if (n%2==0, a[k] <= a[n\2], a[k] >= a[n\2])))", ")) }

Formula

a(2*n) + a(2*n+1) >= n + 1.