a property of multisets: unit fractions with a twist

[Maximilian Hasler]

I offer the comment:

%C A000001 Also, number of multisets A={ a_1, ..., a_n } such that
P[n](A) = 1+sum( P[i](A), i=1..n-1), where P[i] is the i-th symmetric
polynomial (P[1](A)=sum(a_i), P[n](A)=product(a_i)).

Maximilian

On Dec 10, 2007 10:05 PM,  <hv at crypt.org> wrote:
> %I A000001
> %S A000001 1,1,5,43,875,49506
> %N A000001 The number of multisets A={a_1,a_2,...a_n} such that prod{1+1/a_i}=2
> %e A000001 The multiset A={3,3,8} has prod{1+1/a_i}=4/3*4/3*9/8=2; a(3)=5 because there are 5 such sets with 3 elements (the others being {2,4,15}, {2,5,9}, {2,6,7}, {3,4,5})
> %C A000001 For given n, the largest element appears in the set {2, 4, 16, 256, ... 2^2^(n-2), 2^2^(n-1)-1}
> %C A000001 If k is in A, then there are tau(k(k+1))/2 possible n+1-element sets {A-k \union {k+x, k+y}} that also have the property, where xy=k(k+1)
> %K A000001 nonn,more,new
> %O A000001 1,3
> %Y A000001 Cf. A002966, the parallel sequence for sum{1/a_i}=1
> %A A000001 Hugo van der Sanden (hv at crypt.org)
>
> I'd welcome confirmation of the last term, and more terms: I don't have
> time to do more with this right now.
>
> Inspired by an aside in A066218, this actually originated as: find n such
> that 2 tau(n) = sum_{d | n} tau(d); these multisets are the powers+1 in
> the prime factorisation of any such n.
>
> This seems like quite an interesting set to investigate further: it is not
> surprising that it has parallels with sum{1/a_i}=1, such as that hinted
> at in the second comment (parallel to (tau(k^2)+1)/2 in the sum case).
>
> The sets contributing to a(1)..a(4) are:
>
> <1> ; <2 3> ; <2 4 15> <2 5 9> <2 6 7> <3 3 8> <3 4 5> ;
> <2 4 16 255> <2 4 17 135> <2 4 18 95> <2 4 19 75> <2 4 20 63> <2 4 21 55>
> <2 4 23 45> <2 4 25 39> <2 4 27 35> <2 4 30 31> <2 5 10 99> <2 5 11 54>
> <2 5 12 39> <2 5 14 27> <2 5 15 24> <2 5 18 19> <2 6 8 63> <2 6 9 35>
> <2 6 11 21> <2 6 14 15> <2 7 7 48> <2 7 8 27> <2 7 9 20> <2 7 12 13>
> <2 8 9 15> <2 9 10 11> <3 3 9 80> <3 3 10 44> <3 3 11 32> <3 3 12 26>
> <3 3 14 20> <3 3 16 17> <3 4 6 35> <3 4 7 20> <3 4 8 15> <3 4 10 11>
> <3 5 5 24> <3 5 6 14> <3 5 8 9> <3 6 7 8> <4 4 5 15> <4 5 5 9> <4 5 6 7>
>
> .. and I'd include them in a sequence of their own if I could think of
> a useful way to arrange it. (The original 2tau(n) interpretation may
> also be worth submitting.)
>
> Hugo
>