OFFSET
1,1
COMMENTS
Dickson proves that there are only a finite number of odd primitive abundant numbers having n distinct prime factors. SequenceA188342 lists the smallest such numbers. -T. D. Noe, Mar 28 2011
SequenceA188439 sorts the numbers in this sequence by the number of distinct prime factors. Eight numbers have exactly three prime factors; 576 have exactly four prime factors. -T. D. Noe, Apr 04 2011
Any multiple of an abundant number (A005101) is again an abundant number. Primitive abundant numbers (A091191) are those not of this form, i.e., without an abundant proper divisor. We don't know any odd perfect number (A000396), so the (odd) terms here have only deficient proper divisors (A071395), and their prime factors p are less than sigma(n/p)/deficiency(n/p). SeeA005231 (odd abundant numbers) for an explanation why all terms have at least 3 distinct prime factors, and 5 prime factors when counted with multiplicity (A001222), whence a(1) = 3^3*5*7. All known terms are semiperfect (A005835, and thus inA006036): no odd weird number (A006037) is known, but if it exists, the smallest one is in this sequence. -M. F. Hasler, Jul 28 2016
So far, a(173) = 351351 is the only known term ofA122036, i.e., which can't be written as sum of its proper divisors > 1. -M. F. Hasler, Jan 26 2020
REFERENCES
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
T. D. Noe,Table of n, a(n) for n = 1..10000
L. E. Dickson,Finiteness of the odd perfect and primitive abundant numbers with n distinct prime factors, American Journal of Mathematics 35 (1913), pp. 413-422.
Jacob Liddy,An algorithm to determine all odd primitive abundant numbers with d prime divisors, Honors Research Projects (2018), 728.
Eric Weisstein's World of Mathematics,Primitive Abundant Number
MAPLE
isA005101 := proc(n) is(numtheory[sigma](n) > 2*n ); end proc:
isA005100 := proc(n) is(numtheory[sigma](n) < 2*n ); end proc:
isA006038 := proc(n) local d; if type(n, 'odd') and isA005101(n) then for d in numtheory[divisors](n) minus {1, n} do if not isA005100(d) then return false; end if; end do: return true; else false; end if; end proc:
n := 1 ; for a from 1 by 2 do if isA006038(a) then printf("%d %d\n", n, a) ; n := n+1 ; end if; end do: #R. J. Mathar, Mar 28 2011
MATHEMATICA
t = {}; n = 1; While[Length[t] < 50, n = n + 2; If[DivisorSigma[1, n] > 2 n && Intersection[t, Divisors[n]] == {}, AppendTo[t, n]]]; t (*T. D. Noe, Mar 28 2011 *)
PROG
(PARI) is(n)=n%2 && sumdiv(n, d, sigma(d, -1)>2)==1 \\Charles R Greathouse IV, Jun 10 2013
(PARI) is_A006038(n)=bittest(n, 0) && sigma(n)>2*n && !for(i=1, #f=factor(n)[, 1], sigma(n\f[i], -1)>2&&return) \\ More than 5 times faster. -M. F. Hasler, Jul 28 2016
(Haskell)
a006038 n = a006038_list !! (n-1)
a006038_list = filter f [1, 3 ..] where
f x = sum pdivs > x && all (<= 0) (map (\d -> a000203 d - 2 * d) pdivs)
where pdivs = a027751_row x
--Reinhard Zumkeller, Jan 31 2014
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved