A300826 a(n) = n/A125746(n), where A125746(n) gives the smallest divisor d of n such that the sum which includes d and all smaller divisors is >= n.

1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 3, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 3, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 3, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 3, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 3, 1, 1, 1, 2, 1, 3, 1, 1, 1, 1, 1, 3, 1, 1, 1, 2, 1, 2, 1, 2, 1

Offset: 1

Author

Antti Karttunen, Mar 21 2018

Keywords

• nonn

Comments

Records occur at 1, 6, 24, 120, 240, 504, 1260, 2520, 5040, 15120, 50400, 55440, ... and they are 1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 14, 15, ...

Links

• Antti Karttunen, Table of n, a(n) for n = 1..65537

Crossrefs

Cf. A125746.

Cf. A005100 (positions of ones), A023196 (positions of terms > 1).

Programs

PARI

A300826(n) = { my(k=0,s=0); fordiv(n,d, k++; s += d; if(s>=n,return(n/d))); };

Formula

a(n) = n/A125746(n).