A218621 a(n) = unique divisor d of n such that d + (n/d - 1)/2 is minimal and integral.

1, 2, 1, 4, 1, 2, 1, 8, 3, 2, 1, 4, 1, 2, 3, 16, 1, 2, 1, 4, 3, 2, 1, 8, 5, 2, 3, 4, 1, 6, 1, 32, 3, 2, 5, 4, 1, 2, 3, 8, 1, 6, 1, 4, 5, 2, 1, 16, 7, 10, 3, 4, 1, 6, 5, 8, 3, 2, 1, 4, 1, 2, 7, 64, 5, 6, 1, 4, 3, 10, 1, 8, 1, 2, 5, 4, 7, 6, 1, 16, 9, 2, 1, 4, 5

Offset: 1

Author

L. Edson Jeffery, Feb 18 2013

Keywords

• nonn

Comments

Differs from A079891 starting at a(18).

For integers M, k, with 0<=k<=M, consider a representation of n as n = T(M) - T(M-k) = M + (M-1) + ... + (M-k+1), in which k is maximal, where T(r) = r*(r+1)/2 is the r-th triangular number. Then k = A109814(n), and M = A212652(n) = a(n) + (n/a(n) - 1)/2 is minimal.

Conjecture. For n, p, v, j natural numbers, the conditions on a(n) seem to be the following:

1. If n is an odd prime, then a(n) = 1.

2. If n is odd and composite, then

a(n) = max(p : p | n, p <= sqrt(n), p is a prime).

3. If n is equal to a power of 2, then a(n) = n.

4. If n = 2^j*v, with v odd, v>1 and j>1, then a(n) = 2^j.

5. If n = 2*v, with v odd and composite, then

a(n) = 2*p, where p is the least prime such that p | v.

6. If n = 2*p, for p an odd prime, then a(n) = 2.

Links

• T. D. Noe, Table of n, a(n) for n = 1..2048

Crossrefs

Cf. A000217, A109814, A118235, A141419, A209260, A212652.

Programs

Mathematica

Table[d = Divisors[n]; mn = Infinity; best = 0; Do[q = i + (n/i - 1)/2; If[IntegerQ[q] && q < mn, mn = q; best = i], {i, d}]; best, {n, 100}] (* T. D. Noe, Feb 21 2013 *)