A298406 a(n) = 2*a(n-1) - a(n-3) + a(floor(n/2)) + a(floor(n/3)) + ... + a(floor(n/n)), where a(0) = 1, a(1) = 1, a(2) = 1.

1, 1, 1, 3, 8, 19, 42, 84, 163, 301, 547, 961, 1682, 2879, 4902, 8241, 13807, 22917, 37962, 62487, 102690, 168096, 274798, 448000, 729829, 1186797, 1928729, 3130905, 5080360, 8237339, 13352743, 21634097, 35045477, 56753250, 91896553, 148771833, 240830555

Offset: 0

Author

Clark Kimberling, Feb 10 2018

Keywords

• nonn
• easy

Comments

a(n)/a(n-1) -> (1 + sqrt(5))/2, the golden ratio (A001622), so that (a(n)) has the growth rate of the Fibonacci numbers (A000045). See A298338 for a guide to related sequences.

Links

• Clark Kimberling, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A001622, A000045, A298338, A298407.

Programs

Mathematica

a[0] = 1; a[1] = 1; a[2] = 1;
a[n_] := a[n] = 2*a[n - 1] - a[n - 3] + Sum[a[Floor[n/k]], {k, 2, n}];
Table[a[n], {n, 0, 90}]  (* A298406  *)

Python

from functools import lru_cache
@lru_cache(maxsize=None)
def A298406(n):
    if n <= 2:
        return 1
    c, j = 2*A298406(n-1)-A298406(n-3), 2
    k1 = n//j
    while k1 > 1:
        j2 = n//k1 + 1
        c += (j2-j)*A298406(k1)
        j, k1 = j2, n//j2
    return c+n-j+1 # Chai Wah Wu, Mar 31 2021