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A218621
a(n) = unique divisor d of n such that d + (n/d - 1)/2 is minimal and integral.
2
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.
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
KEYWORD
nonn
AUTHOR
L. Edson Jeffery, Feb 18 2013
STATUS
approved