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%I #23 Feb 14 2019 16:34:01
%S 120,672,4320,4680,26208,523776,20427264,29795040,34369920,96445440,
%T 197064960,459818240,557107200
%N Integers m such that the sum of the first k divisors is equal to 2*m for some k less than the number of divisors of m.
%C 3-perfect numbers (A005820) are terms.
%C All known terms of A055153 (abundancy 7/2) are terms.
%C 1907020800 (with abundancy 23/6) is a term too.
%C A055153 is a subsequence, because no term of that sequence may be odd and so for each k in A055153 we have 2*k = sigma(k) - k - k/2. - _Charlie Neder_, Feb 12 2019
%o (PARI) isok(n) = {if (sigma(n) < 2*n, return (0)); my(d = divisors(n), s = 0); for (k=1, #d-1, s += d[k]; if (s == 2*n, return (1)); if (s > 2*n, break);); return (0);}
%o (PARI) is(n) = my(d = divisors(n), s = vecsum(d) - d[#d]); forstep(i = #d-1, 1, -1, if(s <= 2*n, return(s == 2*n)); s-=d[i]); 0 \\ _David A. Corneth_, Feb 11 2019
%Y Cf. A005820 (3-perfect numbers), A055153 (abundancy 7/2).
%Y Cf. A064510, A194472 (both with equal to m rather than to 2*m).
%K nonn,more
%O 1,1
%A _Michel Marcus_, Feb 11 2019
%E a(11)-a(13) from _Jinyuan Wang_, Feb 11 2019
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