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

%I A218621 #35 Feb 21 2013 12:37:38
%S A218621 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,
%T A218621 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,
%U A218621 1,4,3,10,1,8,1,2,5,4,7,6,1,16,9,2,1,4,5
%N A218621 a(n) = unique divisor d of n such that d + (n/d - 1)/2 is minimal and integral.
%C A218621 Differs from A079891 starting at a(18).
%C A218621 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.
%C A218621 Conjecture. For n, p, v, j natural numbers, the conditions on a(n) seem to be the following:
%C A218621 1. If n is an odd prime, then a(n) = 1.
%C A218621 2. If n is odd and composite, then
%C A218621      a(n) = max(p : p | n, p <= sqrt(n), p is a prime).
%C A218621 3. If n is equal to a power of 2, then a(n) = n.
%C A218621 4. If n = 2^j*v, with v odd, v>1 and j>1, then a(n) = 2^j.
%C A218621 5. If n = 2*v, with v odd and composite, then
%C A218621      a(n) = 2*p, where p is the least prime such that p | v.
%C A218621 6. If n = 2*p, for p an odd prime, then a(n) = 2.
%H A218621 T. D. Noe, <a href="/A218621/b218621.txt">Table of n, a(n) for n = 1..2048</a>
%t A218621 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 *)
%Y A218621 Cf. A000217, A109814, A118235, A141419, A209260, A212652.
%K A218621 nonn
%O A218621 1,2
%A A218621 _L. Edson Jeffery_, Feb 18 2013