A298356 a(n) = a(n-1) + a(n-2) + a([n/2]) + a([n/3]) + ... + a([n/n]), where a(0) = 1, a(1) = 1, a(2) = 1.

1, 1, 1, 4, 8, 16, 32, 57, 103, 178, 308, 514, 874, 1441, 2394, 3926, 6462, 10531, 17231, 28001, 45614, 74026, 120258, 194903, 316210, 512171, 830007, 1343883, 2176578, 3523150, 5704107, 9231637, 14942711, 24181525, 39135483, 63328289, 102482212, 165828942

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] = 1; a[2] = 1;
a[n_] := a[n] = a[n - 1] + a[n - 2] + Sum[a[Floor[n/k]], {k, 2, n}];
Table[a[n], {n, 0, 30}]  (* A298356 *)

Python

from functools import lru_cache
@lru_cache(maxsize=None)
def A298356(n):
    if n <= 2:
        return 1
    c, j = A298356(n-1)+A298356(n-2), 2
    k1 = n//j
    while k1 > 1:
        j2 = n//k1 + 1
        c += (j2-j)*A298356(k1)
        j, k1 = j2, n//j2
    return c+n-j+1 # Chai Wah Wu, Mar 31 2021