A298357 a(n) = a(n-1) + a(n-2) + a([n/2]) + a([n/3]) + ... + a([n/n]), where a(0) = 1, a(1) = 2, a(2) = 3.

1, 2, 3, 9, 19, 37, 74, 131, 238, 410, 710, 1184, 2014, 3320, 5516, 9044, 14888, 24262, 39698, 64510, 105089, 170545, 277057, 449027, 728502, 1179967, 1912216, 3096110, 5014519, 8116824, 13141430, 21268343, 34425826, 55710704, 90162442, 145899135, 236104060

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.

Programs

Mathematica

a[0] = 1; a[1] = 2; a[2] = 3;
a[n_] := a[n] = a[n - 1] + a[n - 2] + Sum[a[Floor[n/k]], {k, 2, n}];
Table[a[n], {n, 0, 30}]  (* A298357 *)

Python

from functools import lru_cache
@lru_cache(maxsize=None)
def A298357(n):
    if n <= 2:
        return n+1
    c, j = A298357(n-1)+A298357(n-2), 2
    k1 = n//j
    while k1 > 1:
        j2 = n//k1 + 1
        c += (j2-j)*A298357(k1)
        j, k1 = j2, n//j2
    return c+2*(n-j+1) # Chai Wah Wu, Mar 31 2021