A051772 Distended numbers.

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, 33, 34, 35, 37, 38, 39, 41, 43, 44, 46, 47, 49, 50, 51, 52, 53, 55, 57, 58, 59, 61, 62, 64, 65, 67, 68, 69, 71, 73, 74, 75, 76, 77, 79, 81, 82, 83, 85, 86, 87, 89, 91, 92, 93

Offset: 1

Author

Alexander Benjamin Schwartz (QBOB(AT)aol.com), Dec 08 1999

Keywords

• nonn

Comments

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.

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

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

Contains A000961. - Robert Israel, Dec 20 2015

Links

• Ivan Neretin, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000961, A005100, A051773, A051774.

Programs

Maple

filter:= proc(n) local F,L;
  F:= sort(convert(numtheory:-divisors(n),list));
  L:= ListTools:-PartialSums(F);
  max(L[1..-2] - F[2..-1]) < 0;
end proc:
select(filter, [$1..1000]); # Robert Israel, Dec 20 2015

Mathematica

Select[Range[93], Min[Rest[d = Divisors[#]] - Most[Accumulate[d]]] > 0 &] (* Ivan Neretin, Dec 19 2015 *)

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