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 A080224 Number of abundant divisors of n. 12
 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, 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, 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,24 COMMENTS Number of divisors d of n with sigma(d)>2*d (sigma = A000203) 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 LINKS R. Zumkeller, Table of n, a(n) for n = 1..10000 Eric Weisstein's World of Mathematics, Abundant Number. FORMULA a(n) + A080225(n) + A080226(n) = A000005(n). From Antti Karttunen, Nov 14 2017: (Start) a(n) = Sum_{d|n} A294937(d). a(n) = A294929(n) + A294937(n). a(n) = 1 iff A294930(n) = 1. (End) EXAMPLE 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. MATHEMATICA Table[Count[Divisors[n], _?(DivisorSigma[1, #]>2#&)], {n, 110}] (* Harvey P. Dale, Jun 14 2013 *) PROG (PARI) a(n) = sumdiv(n, d, sigma(d)>2*d)  \\ Michel Marcus, Mar 09 2013 CROSSREFS Cf. A000005, A000203, A005101, A080225, A080226, A187795, A294890, A294929, A294930, A294937. Cf. also A294904. Sequence in context: A060862 A066087 A294927 * A261488 A010105 A083916 Adjacent sequences:  A080221 A080222 A080223 * A080225 A080226 A080227 KEYWORD nonn,changed AUTHOR Reinhard Zumkeller, Feb 07 2003 STATUS approved

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