cp's OEIS Frontend

Theorem. For every even integer m there exists a representation of the form m=a(r)+a(s). If A(x) is the counting function of a(n)<=x, then A(x)=O(sqrt(x))and Omega(sqrt(x)). Conjecture. The sequence is minimal in the following sense: if any sequence has the counting function B(x)<=A(x) for all x>=1 and B(x) < A(x) for x>=x_0, then there exists an even integer N which is not expressible as a sum of two terms of such sequence.

Every positive odd integer m == 3 (mod 16) is a unique sum of the form a(s) + 2a(t), while other odd integers are not expressible in such form.

Note that a(4n)=a(4n-2), a(4n-1)=a(4n-3).

Theorem. An odd number is in the sequence iff in its binary expansion all digits on k-th positions from the end, k=3, 5, 7, ..., are zeros. For example, 169, 171 have binary expansions 10101001, 10101011. Thus both of them are in the sequence. If A(x) is the counting function of a(n) <= x, then A(x) = O(sqrt(x)) and Omega(sqrt(x)).

Every positive odd integer m==3 (mod 2^(2r-1)) is a unique sum of the form a(2^(r-1)*(s-1)+1) + 2a(2^(r-1)*(t-1)+1), r=1,2,..., while the other odd integers are not expressible in such a form (see also the comment to A145818). Problem: Let B be the set of analytical functions f(x), f(0)=0, abs(x) > 1, with (0,1)-Taylor coefficients. Let F(x) be in B. It could be proved that the equation f(x)*f(x^2) = F(x) has no more than one solution in B. In case of F(x) = x^3/(1-x^(2^r)), r=1,2,..., it has one solution, which leads to A145812, A145818, A145849, A145850, etc. Find the conditions of the solvability of this equation in B and give, if it is possible, other examples. [Vladimir Shevelev, Oct 21 2008]

To get the decomposition of odd m as sum a(l) + 2a(s), write m-2 as Sum b_j 2^j, then a(l) = 1 + Sum_{j odd} b_j 2^j. This algorithm follows from our comment to A088442 and the algorithm of calculation of A088442(n). For example, if m=81, then we have 79 = 1*2^0 + 1*2^1 + 1*2^2 + 1*2^3 + 1*2^6. Thus a(l) = 1 + (1*2^1 + 1*2^3) = 11 and the required decomposition is 81 = 11 + 2*35, such that a(s)=35. We see that L=4, s=6, i.e., "index coordinates" of 81 are (4,6). Thus we have a one-to-one map of odd integers > 1 to the positive lattice points in the plane. [Vladimir Shevelev, Oct 24 2008]

Theorem. An integer is in the sequence iff in its expansion on base 3 all digits at the k-th position from the end, k=3, 5, 7, ..., are zeros while the first digit from the end is 1. To get the decomposition of m==1(mod 3) as sum a(l)+3a(s), write m-3 as Sum b_j 3^j, then a(l) = 1 + Sum_{j odd} b_j 3^j.

Positive integers such that for every integer m==8 (mod 9) there exists a unique representation of m as a sum of the form a(l)+3a(s).

Positive integers such that for every integer m==4 (mod 9) there exists a unique representation of m as a sum of the form a(l)+3a(s).

Since, e.g., 27=17+2*3+4*1 and 17=a(3),3=a(2),1=a(1), then 27 has "coordinates" (3,2,1). Thus we have a one-to-one map of odd integers >=7 to the positive lattice points in the three-dimensional space.