A302110 Let d be the list of A000005(n) = tau(n) divisors of n. Then a(n) is the largest k such that Sum_{i=1..#d-k} d_i > n.

0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 2, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 2, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 2, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 2, 0, 0, 0, 0, 0, 2, 0, 0, 0, 1

Offset: 1

Author

David A. Corneth, Apr 01 2018

Keywords

• nonn
• easy

Comments

Records (0, 1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 13, ...) occur at 1, 6, 24, 120, 240, 720, 1260, 2520, 5040, 15120, 27720, 55440, ... - Antti Karttunen, Apr 02 2018

Links

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

Crossrefs

Cf. A000005, A023196, A125746, A125747, A300826.

Programs

PARI

A125747(n) = { my(k=0,s=0); fordiv(n,d, k++; s += d; if(s>=n,return(k))); };
A302110(n) = (numdiv(n) - A125747(n)); \\ Antti Karttunen, Apr 02 2018

Formula

a(n) = A000005(n) - A125747(n).

a(n) > 0 if and only if n is in A023196.