A298408 a(n) = 2*a(n-1) - a(n-3) + 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, 6, 20, 53, 130, 277, 574, 1115, 2126, 3862, 7021, 12341, 21553, 36957, 63111, 106224, 178407, 296638, 492231, 811731, 1335994, 2188950, 3583027, 5847108, 9532980, 15512342, 25226123, 40967842, 66506422, 107869832, 174908573, 283452771, 459264017

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, A298409.

Programs

Mathematica

a[0] = 1; a[1] = 1; a[2] = 1;
a[n_] := a[n] = 2*a[n - 1] - a[n - 3] + Sum[k*a[Floor[n/k]], {k, 2, n}];
Table[a[n], {n, 0, 90}]  (* A298408  *)

Python

from functools import lru_cache
@lru_cache(maxsize=None)
def A298408(n):
    if n <= 2:
        return 1
    c, j = 2*A298408(n-1)-A298408(n-3), 2
    k1 = n//j
    while k1 > 1:
        j2 = n//k1 + 1
        c += (j2*(j2-1)-j*(j-1))*A298408(k1)//2
        j, k1 = j2, n//j2
    return c+(n*(n+1)-j*(j-1))//2 # Chai Wah Wu, Mar 31 2021