Review: A068067

[David Wilson]

I have simplified the description of A068087, provided a formula,
and tightened up some of the observations.


I recommend replacing these lines:

%N A068067 The number of subsets of {1,2,3,...,m} such that the sum is 0 mod
n, 0 < m <= n.
%C A068067 The empty set is not included. It seems that the only values that
f(n) can assume are powers of two and one less than a power of two. The
first time the value k shows up is when n equals 1, 3, 30, 15, 0, 0, 210,
105, ..., . f(k) = 0 when k = 2^m.
%K A068067 nonn,uned


with these lines:

%N A068067 Number of m with 1<=m<=n and n | 1+2+...+m.
%F A068067 a(n) = 2^(f(n)-1)-1 if n even, 2^f(n) if n odd; where f(n) =
A001221(n) = number of prime divisors of n.
%C A068067 a(n) = 0 iff n = 2^k with k >= 1.
%C A068087 Least n with a(n) = 2^k is p(k+1)#/2 = A002110(A000040(k+1))/2.
Least n with a(n) = 2^k-1 != 1 is p(k+1)#.
%K A068067 nonn