A129575 Exponential abundant numbers: integers k for which A126164(k) > k, or equivalently for which A051377(k) > 2k.

900, 1764, 3600, 4356, 4500, 4900, 6084, 6300, 7056, 8100, 8820, 9900, 10404, 11700, 12348, 12996, 14700, 15300, 17100, 19044, 19404, 20700, 21780, 22500, 22932, 25200, 26100, 27900, 29988, 30276, 30420, 30492, 31500, 33300, 33516, 34596, 35280, 36900, 38700, 39600

Offset: 1

Author

Ant King, Apr 28 2007

Keywords

• easy
• nonn

Comments

There are only 52189 exponential abundant numbers less than 50 million, which suggests that these account for approximately 0.1% of all integers.

Includes 36*m for all m coprime to 6 that are not squarefree. - Robert Israel, Feb 19 2019

The asymptotic density of this sequence is Sum_{n>=1} f(A328136(n)) = 0.001043673..., where f(n) = (6/(Pi^2*n))*Product_{prime p|n}(p/(p+1)). - Amiram Eldar, Sep 02 2022

Examples

			The third integer that is exceeded by its proper exponential divisor sum is 3600. Hence a(3) = 3600.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000
• Peter Hagis, Jr., Some Results Concerning Exponential Divisors, Internat. J. Math. & Math. Sci., Vol. 11, No. 2, (1988), pp. 343-350.
• Eric Weisstein's World of Mathematics, e-Divisor.

Crossrefs

Cf. A126164, A051377, A049419, A054979, A054980, A328136.

Programs

Maple

filter:= proc(n)  local L,m,i,j;
  L:= ifactors(n)[2];
  m:= nops(L);
  mul(add(L[i][1]^j, j=numtheory:-divisors(L[i][2])),i=1..m)>2*n
end proc:
select(filter, [$1..10^5]); # Robert Israel, Feb 19 2019

Mathematica

ExponentialDivisors[1]={1};ExponentialDivisors[n_]:=Module[{}, {pr,pows}=Transpose@FactorInteger[n];divpowers=Distribute[Divisors[pows],List];Sort[Times@@(pr^Transpose[divpowers])]];properexponentialdivisorsum[k_]:=Plus@@ExponentialDivisors[k]-k;Select[Range[5 10^4],properexponentialdivisorsum[ # ]># &]
f[p_, e_] := DivisorSum[e, p^# &]; esigma[1] = 1; esigma[n_] := Times @@ f @@@ FactorInteger[n]; Select[Range[40000], esigma[#] > 2*# &] (* Amiram Eldar, May 06 2025 *)

PARI

isok(k) = {my(f = factor(k)); prod(i = 1, #f~, sumdiv(f[i, 2], d, f[i, 1]^d)) > 2*k;} \\ Amiram Eldar, May 06 2025