a new sequence

[Max]

On 5/16/06, Emeric Deutsch <deutsch at duke.poly.edu> wrote:

> Values of n such that the number of distinct sums of
> distinct divisors of n is less than 2^tau(n) - 1 (clearly,
> the latter is the largest number of possible distinct sums).

I would include also the sum of empty subset of divisor (obviously,
this sum equals 0) so that the maximum number of possible distinct
sums become 2^tau(n). But that does not affect the definition of your
sequence though.

> Equivalently: values of n for which there exist two subsets
> of the set of divisors of n, having the same sum.

[...]

> Interestingly, the displayed terms agree with the displayed
> terms of A051774 (shown below), except that the new
> sequence does not contain the term 175.

[...]

> ---------------
> A051774  Contracted numbers.
> 6, 12, 18, 20, 24, 28, 30, 36, 40, 42, 45, 48, 54, 56, 60, 63, 66, 70,
> 72, 78, 80, 84, 88, 90, 96, 99, 100, 102, 104, 105, 108, 110, 112, 114,
> 117, 120, 126, 130, 132, 135, 138, 140, 144, 150, 154, 156, 160, 162, 165,
> 168, 170, 174, 175, 176, 180 (list)
> OFFSET  1,1
> COMMENT  n is said to be contracted if and only if there exist
> distinct divisors d_1<d_2<...<d_k such that
> d_1+d_2+...+d_(k-1) >= d_k.

Your sequence is a subsequence of A051774.
Indeed, suppose that n belongs to your sequence. Then there are two
distinct subsets of divisors of n with equal sums. After removing
elements that appear in both subsets we still have two non-empty
subsets with equal sums, say,
d_1 + ... + d_m = e_1 + ... + e_n
where all divisors d_1, ..., d_m, e_1, ..., e_n are distinct.
Without loss of generality we can assume that d_1 is the maximum
divisor among all divisors in both subsets. In particular, we have e_1
< d_1, ..., e_n < d_m and
d_m <= e_1 + ... + e_n
meaning that n belongs to A051774.

Max