OFFSET
1,4
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.
FORMULA
a(2*n) + a(2*n+1) >= n + 1.
EXAMPLE
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.
PROG
(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])))", ")) }
CROSSREFS
KEYWORD
nonn
AUTHOR
Rémy Sigrist, Nov 14 2023
STATUS
approved