A309262 a(0) = 0, a(1) = 1, and for any n > 1, a(n) = Sum_{k > 1} a(floor(n/k^2)).

0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 3, 3, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 1, 1, 1, 1, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 8, 8, 8, 8, 8, 8, 8, 8, 5, 5, 5, 5, 5, 5, 5, 5, 5, 7, 7, 7, 7, 7, 7

Offset

0,17

Comments

For any n > 1 and k > A000196(n), a(floor(n/k^2)) = a(0) = 0, hence the series in the name is well defined.

This sequence is a variant of A022825; here we sum terms of the form a(floor(n/k^2)), there terms of the form a(floor(n/k)).

Links

Rémy Sigrist, Table of n, a(n) for n = 0..10000

Example

a(5) = a(floor(5/2^2)) = a(1) = 1.

Mathematica

Join[{0}, Clear[a]; a[0]=0; a[1]=1; a[n_]:=a[n]=Sum[a[Floor[n/k^2]], {k, 2, n}]; Table[a[n], {n, 1, 100}]] (* Vincenzo Librandi, Jul 22 2019 *)

Prog

(PARI) a(n) = if (n<=1, n, sum (k=2, sqrtint(n), a(n\k^2)))
(Python)
from functools import lru_cache
from math import isqrt
@lru_cache(maxsize=None)
def A309262(n):
    if n<2: return n
    c, j = 0, 2
    while (j2:=j**2) <= n:
        k = n//j2
        m = isqrt(n//k)
        c += A309262(k)*(-j+(j:=m+1))
    return c # Chai Wah Wu, May 17 2026

Crossrefs

Cf. A000196, A022825.

Sequence in context: A328143 A278402 A276415 * A064983 A124933 A133884

Adjacent sequences: A309259 A309260 A309261 * A309263 A309264 A309265

Keyword

nonn

Author

Rémy Sigrist, Jul 19 2019

Status

approved