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A051772 Distended numbers. 5

%I #25 Dec 23 2015 16:20:53

%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 = k-1, then the sum of divisors = d_1+d_2+...+d_(k-1) < 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_(m-r)+d_(m-r-1)+...+d_1) < d_(m+1) for all 0 < m < k and 0 <= r < m, which for r = 0 is the definition above. When r = m-1 the inequality reduces to 2^(m-1) < 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, k-1, 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

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Last modified April 19 18:00 EDT 2024. Contains 371797 sequences. (Running on oeis4.)