%I
%S 1,2,3,4,5,7,8,9,10,11,13,14,15,16,17,19,21,22,23,25,26,27,29,31,32,
%T 33,34,35,37,38,39,41,43,44,46,47,49,50,51,52,53,55,57,58,59,61,62,64,
%U 65,67,68,69,71,73,74,75,76,77,79,81,82,83,85,86,87,89,91,92,93
%N Distended numbers.
%C Let 1 = d_1 < d_2 < ... < d_k = n be the k distinct divisors of n. Then n is said to be distended if and only if d_1+d_2+...+d_m < d_(m+1) for all 0 < m < k.
%C By definition, all distended numbers are deficient (A005100). For this, it suffices to consider the case for m = k1, then the sum of divisors = d_1+d_2+...+d_(k1) < d_(m+1) = d_k = n.  _Jaycob Coleman_, _Michel Marcus_, Oct 24 2013
%C If n is distended and defined as above, then 2^r(d_(mr)+d_(mr1)+...+d_1) < d_(m+1) for all 0 < m < k and 0 <= r < m, which for r = 0 is the definition above. When r = m1 the inequality reduces to 2^(m1) < d_(m+1) for all 0 < m < k. In particular, 2^k < 4n.  _Jaycob Coleman_, Oct 29 2013
%C Contains A000961.  _Robert Israel_, Dec 20 2015
%H Ivan Neretin, <a href="/A051772/b051772.txt">Table of n, a(n) for n = 1..10000</a>
%p filter:= proc(n) local F,L;
%p F:= sort(convert(numtheory:divisors(n),list));
%p L:= ListTools:PartialSums(F);
%p max(L[1..2]  F[2..1]) < 0;
%p end proc:
%p select(filter, [$1..1000]); # _Robert Israel_, Dec 20 2015
%t Select[Range[93], Min[Rest[d = Divisors[#]]  Most[Accumulate[d]]] > 0 &] (* _Ivan Neretin_, Dec 19 2015 *)
%o (PARI) isok(n) = {d = divisors(n); k = #d; for (m = 1, k1, if (sum(j = 1, m, d[j]) >= d[m+1], return (0));); return (1);} \\ _Michel Marcus_, Sep 04 2013
%Y Cf. A000961, A005100, A051773, A051774.
%K nonn
%O 1,2
%A Alexander Benjamin Schwartz (QBOB(AT)aol.com), Dec 08 1999
