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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.
2
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
1,24
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
FORMULA
a(n) = A000005(n) - A125747(n).
a(n) > 0 if and only if n is in A023196.
PROG
(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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
David A. Corneth, Apr 01 2018
STATUS
approved