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Let x be the sum of the divisors d_i of k such that d_i | sigma(k). Sequence lists the numbers k for which x^3 = sigma(k).
2

%I #11 Mar 28 2017 08:17:25

%S 1,690,714,75432,81172,81192,81624,82248,84196,305320,312040,315880,

%T 619542,639198,646758,665874,684342,737694,743958,750114,751626,

%U 761454,762966,763614,4349280,4651680,4789920,4939680,4981920,5259936,5325216,5428896,5474976

%N Let x be the sum of the divisors d_i of k such that d_i | sigma(k). Sequence lists the numbers k for which x^3 = sigma(k).

%C Subset of A020477.

%H Giovanni Resta, <a href="/A284284/b284284.txt">Table of n, a(n) for n = 1..200</a>

%e Divisors of 690 are 1, 2, 3, 5, 6, 10, 15, 23, 30, 46, 69, 115, 138, 230, 345, 690 and sigma(690) = 1728. Then:

%e 1728 / 1 = 1728, 1728 / 2 = 864, 1728 / 3 = 576, 1728 / 6 = 288 and (1 + 2 + 3 + 6)^2 = 12^3 = 1728.

%p with(numtheory): P:=proc(q) local a,k,n,x;

%p for n from 1 to q do a:=sort([op(divisors(n))]); x:=0;

%p for k from 1 to nops(a)-1 do if type(sigma(n)/a[k],integer) then x:=x+a[k]; fi; od;

%p if x^3=sigma(n) then print(n); fi; od; end: P(10^6);

%t Select[Range[10^5], (d = DivisorSigma[1, #]; IntegerQ[ d^(1/3)] && d == DivisorSigma[1, GCD[d, #]]^3) &] (* _Giovanni Resta_, Mar 28 2017 *)

%Y Cf. A000203, A020477, A284283.

%K nonn

%O 1,2

%A _Paolo P. Lava_, Mar 24 2017

%E a(1), a(25)-a(33) from _Giovanni Resta_, Mar 28 2017