A298407 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) = 2, a(2) = 3.

1, 2, 3, 9, 23, 52, 113, 223, 431, 794, 1442, 2532, 4433, 7589, 12924, 21730, 36411, 60440, 100125, 164816, 270863, 443390, 724846, 1181713, 1925113, 3130488, 5087530, 8258585, 13400782, 21728136, 35221342, 57065559, 92441545, 149701409, 242400952, 392424193

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, A298406.

Programs

Mathematica

a[0] = 1; a[1] = 2; a[2] = 3;
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}]  (* A298407  *)

Python

from functools import lru_cache
@lru_cache(maxsize=None)
def A298407(n):
    if n <= 2:
        return n+1
    c, j = 2*A298407(n-1)-A298407(n-3), 2
    k1 = n//j
    while k1 > 1:
        j2 = n//k1 + 1
        c += (j2-j)*A298407(k1)
        j, k1 = j2, n//j2
    return c+2*(n-j+1) # Chai Wah Wu, Mar 31 2021