cp's OEIS Frontend

From Jon E. Schoenfield, Jun 13 2010: (Start)

Given a value of n, one way to obtain a set of residue classes having no zero subset sum is simply to select the first k positive integers, where k is the largest integer such that 1+2+3+...+k < n; for any n > 0, this yields a set of k residue classes where k = round(sqrt(2*n)) - 1, which serves as a first lower bound for a(n).

For n > 5, another simple method is to select the residue classes 1, n-2, and 3..k, where k is the largest integer such that (1+2+3+...+k) - 2 < n; this yields a set of k residue classes where k = round(sqrt(2*n+4)) - 1. This second lower bound for a(n) is an improvement over the first bound at n = 6, 9, 10, 14, 15, 20, 21, ... (i.e., values of n that are triangular numbers or one less than triangular numbers; these are the terms of A117142 that exceed 5).

For even values of n, a third method is to select residue classes m = n/2, 1, m+1, 2, m+2, 3, m+3, etc., until k = floor(sqrt(2n-3)) residue classes have been selected. This third lower bound for a(n) is an improvement over the other two for n = 26, 34, 42, 52, 62, 64, 74, 76, 86, 88, 100, 102, 114, 116, 118, ... (i.e., even numbers whose difference d from the next higher triangular number T(k) = k*(k+1)/2 satisfies 1 < d < k/2-1).

Let S = {25, 75, 375, 525, 1125, 1375, ...} be the set of numbers that are 25 times an odd triangular number, i.e., numbers of the form 25*T(j) = 25*j*(j+1)/2 where (j+3) mod 4 < 2. For values of n in S, a fourth method is to let m = n/5 and select residue classes 1..j, m+1..m+j, 2*m+1..2*m+j, 3*m+1..3*m+j, 4*m+1..4*m+j, m, and 2*m; this yields a set of 5*j+2 residue classes, which gives sqrt(2*n + 25/4) - 1/2 as an improved lower bound for a(n) for n in S: a(25) >= 7, a(75) >= 12, a(375) >= 27, a(525) >= 32, a(1125) >= 47, a(1375) >= 52, etc.

For each of the first 87 terms in the sequence, an exhaustive search determined that a(n) either equals the first lower bound (round(sqrt(2*n))-1) or exceeds it by only 1, and that at least one of the four methods above yields a maximal solution.

The sequence is not nondecreasing; a(n) < a(n-1) at n = 43, 53, 63, and 87 (and, if it continues to be the case that at least one of the four methods above yields a maximal solution, the next decreases occur at n = 89, 101, 103, 115, 117, 131, 133, 147, 149, 151, ...).

Does there exist any value of n for which none of the four methods above yields a maximal solution? (End)

Curtz (1965), page 15, from right to left, gives (F1):

1/2;

1/4, 3/4;

1/8, 4/8, 7/8;

1/16, 5/16, 11/16, 15/16;

... .

Numerators + Denominators = (C) =

3;

5, 7;

9, 12, 15;

17, 21, 27, 31;

... .

This is the current sequence without powers of 2.

The triangle (P) for a(n) is

1;

1, 2;

2, 3, 4;

4, 5, 7, 8;

8, 9, 12, 15, 16;

... .

(C) is the core of (P).

Extension of (F1). (F2) =

0/1;

0/1, 1/1;

0/2, 1/2, 2/2;

0/4, 1/4, 3/4, 4/4;

0/8, 1/8, 4/8, 7/8, 8/8;

... .

(Mentioned, without 0's, op. cit., page 16.)

a(n) = Numerators + Denominators.

Row sums of triangle (P): A084858(n).

From right to left, with alternating signs: 1, 1, 3, 2, 12, 8, 48, 32, ..., see A098646.

For triangle (C), row sums give A167667(n+1).

From right to left, with alternating signs: A098646(n).

Rank of A016116(n): 0 together with A117142.