A298369 a(n) = a(n-1) + a(n-2) + 2*a(floor(n/2)) + 3*a(floor(n/3)) + ... + n*a(floor(n/n)), where a(0) = 1, a(1) = 1, a(2) = 1.

1, 1, 1, 7, 17, 38, 87, 164, 318, 576, 1040, 1773, 3134, 5241, 8877, 14728, 24579, 40298, 66585, 108610, 178004, 289717, 472312, 766643, 1247081, 2021980, 3281557, 5316888, 8619474, 13957420, 22611507, 36603571, 59270152, 95931095, 155290091, 251310597

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[k*a[Floor[n/k]], {k, 2, n}];
Table[a[n], {n, 0, 30}]  (* A298369 *)

Python

from functools import lru_cache
@lru_cache(maxsize=None)
def A298369(n):
    if n <= 2:
        return 1
    c, j = A298369(n-1)+A298369(n-2), 2
    k1 = n//j
    while k1 > 1:
        j2 = n//k1 + 1
        c += (j2*(j2-1)-j*(j-1))*A298369(k1)//2
        j, k1 = j2, n//j2
    return c+(n*(n+1)-j*(j-1))//2 # Chai Wah Wu, Mar 31 2021