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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: A325194 A066087 A294927 * A261488 A010105 A083916

Adjacent sequences:  A080221 A080222 A080223 * A080225 A080226 A080227

KEYWORD

nonn

AUTHOR

Reinhard Zumkeller, Feb 07 2003

STATUS

approved

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Last modified April 20 04:42 EDT 2019. Contains 322294 sequences. (Running on oeis4.)