%I #20 Feb 22 2021 11:43:54
%S 0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,2,0,0,0,0,0,1,0,0,0,0,
%T 0,3,0,0,0,2,0,1,0,0,0,0,0,3,0,0,0,0,0,2,0,1,0,0,0,4,0,0,0,0,0,1,0,0,
%U 0,1,0,5,0,0,0,0,0,1,0,3,0,0,0,3,0,0,0,1,0,3,0,0,0,0,0,4,0,0,0,2,0,1,0,1,0
%N Number of abundant divisors of n.
%C Number of divisors d of n with sigma(d)>2*d (sigma = A000203)
%C a(n)>0 iff n is abundant: a(A005101(n))>0, a(A000396(n))=0 and a(A005100(n))=0; a(A091191(n))=1; a(A091192(n))>1; a(A091193(n))=n and a(m)<>n for m < A091193(n). - _Reinhard Zumkeller_, Dec 27 2003
%H R. Zumkeller, <a href="/A080224/b080224.txt">Table of n, a(n) for n = 1..10000</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/AbundantNumber.html">Abundant Number.</a>
%F a(n) + A080225(n) + A080226(n) = A000005(n).
%F From _Antti Karttunen_, Nov 14 2017: (Start)
%F a(n) = Sum_{d|n} A294937(d).
%F a(n) = A294929(n) + A294937(n).
%F a(n) = 1 iff A294930(n) = 1.
%F (End)
%e Divisors of n=24: {1,2,3,4,6,8,12,24}, two of them are abundant: 12=A005101(1) and 24=A005101(4), therefore a(24)=2.
%p A080224 := proc(n)
%p a := 0 ;
%p for d in numtheory[divisors](n) do
%p if numtheory[sigma](d) > 2*d then
%p a := a+1 ;
%p end if;
%p end do:
%p a;
%p end proc:
%p seq(A080224(n),n=1..80) ; # _R. J. Mathar_, Feb 22 2021
%t Table[Count[Divisors[n],_?(DivisorSigma[1,#]>2#&)],{n,110}] (* _Harvey P. Dale_, Jun 14 2013 *)
%o (PARI) a(n) = sumdiv(n, d, sigma(d)>2*d) \\ _Michel Marcus_, Mar 09 2013
%Y Cf. A000005, A000203, A005101, A080225, A080226, A187795, A294890, A294929, A294930, A294937.
%Y Cf. also A294904.
%K nonn
%O 1,24
%A _Reinhard Zumkeller_, Feb 07 2003