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

1,2

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

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 *)

Crossrefs

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

Sequence in context: A357689 A324873 A325566 * A079891 A108738 A064405

Adjacent sequences: A218618 A218619 A218620 * A218622 A218623 A218624

Keyword

nonn

Author

L. Edson Jeffery, Feb 18 2013

Status

approved