|
%I #10 Apr 15 2018 14:21:12
%S 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,
%T 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,
%U 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
%N 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.
%C 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
%H Antti Karttunen, <a href="/A302110/b302110.txt">Table of n, a(n) for n = 1..65537</a>
%F a(n) = A000005(n) - A125747(n).
%F a(n) > 0 if and only if n is in A023196.
%o (PARI)
%o A125747(n) = { my(k=0,s=0); fordiv(n,d, k++; s += d; if(s>=n,return(k))); };
%o A302110(n) = (numdiv(n) - A125747(n)); \\ _Antti Karttunen_, Apr 02 2018
%Y Cf. A000005, A023196, A125746, A125747, A300826.
%K nonn,easy
%O 1,24
%A _David A. Corneth_, Apr 01 2018
|
