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%I #41 Jun 04 2017 10:09:34
%S 1,2,3,4,6,8,10,12,15,18,21,24,26,32,39,42,60,65,72,84,96,160,182,336,
%T 455,512,672,896,960,992,1023,1280,1536,1848,2040,2688,4092,5472,5920,
%U 7808,7936,10416,11934,16352,16380,18720,20384,21824,23424,24564,29127,30240,33792,36720,41440
%N Let sigma_m (n) be result of applying sum-of-divisors function m times to n; call n (m,k)-perfect if sigma_m (n) = k*n; sequence gives the (4,k)-perfect numbers.
%C Similarly to A019278, 2 and sigma(2) are both terms, and this can happen also for further iterations of sigma on known terms, thus providing new terms outside the current known range. - _Michel Marcus_, May 01 2017
%C Also it occurs that many divisors of A019278 are terms of this sequence. - _Michel Marcus_, May 28 2017
%H Michel Marcus, <a href="/A019293/b019293.txt">Table of n, a(n) for n = 1..320</a>
%H Graeme L. Cohen and Herman J. J. te Riele, <a href="http://projecteuclid.org/euclid.em/1047565640">Iterating the sum-of-divisors function</a>, Experimental Mathematics, 5 (1996), pp. 93-100.
%H Michel Marcus, <a href="/A019293/a019293_1.log.txt">Unexhaustive list of terms</a>
%e 10 is a term because applying sigma four times we see that 10 -> 18 -> 39 -> 168 -> 120, and 120 = 12*10. So 10 is a (4,12)-perfect number.
%o (PARI) isok(n) = denominator(sigma(sigma(sigma(sigma(n))))/n) == 1; \\ _Michel Marcus_, Apr 29 2017
%Y Cf. A019276, A019278, A019292, A129076.
%K nonn
%O 1,2
%A _N. J. A. Sloane_
%E Corrected by _Michel Marcus_, Apr 29 2017
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