A282668 -id:A282668 - cp's OEIS frontend

1, 1, 1, 2, 1, 2, 1, 2, 3, 2, 1, 3, 1, 2, 3, 4, 1, 3, 1, 4, 3, 2, 1, 4, 5, 2, 3, 4, 1, 5, 1, 4, 3, 2, 5, 6, 1, 2, 3, 5, 1, 6, 1, 4, 5, 2, 1, 6, 7, 5, 3, 4, 1, 6, 5, 7, 3, 2, 1, 6, 1, 2, 7, 8, 5, 6, 1, 4, 3, 7, 1, 8, 1, 2, 5, 4, 7, 6, 1, 8, 9, 2, 1, 7, 5, 2, 3

Offset: 1

N. J. A. Sloane

a(n) = sqrt(n) is a new record if and only if n is a square. - Zak Seidov, Jul 17 2009
a(n) = A060775(n) unless n is a square, when a(n) = A033677(n) = sqrt(n) is strictly larger than A060775(n). It would be nice to have an efficient algorithm to calculate these terms when n has a large number of divisors, as for example in A060776, A060777 and related problems such as A182987. - M. F. Hasler, Sep 20 2011
a(n) = 1 when n = 1 or n is prime. - Alonso del Arte, Nov 25 2012
a(n) is the smallest central divisor of n. Column 1 of A207375. - Omar E. Pol, Feb 26 2019
a(n^4+n^2+1) = n^2-n+1: suppose that n^2-n+k divides n^4+n^2+1 = (n^2-n+k)*(n^2+n-k+2) - (k-1)*(2*n+1-k) for 2 <= k <= 2*n, then (k-1)*(2*n+1-k) >= n^2-n+k, or n^2 - (2*k-1)*n + (k^2-k+1) = (n-k+1/2)^2 + 3/4 < 0, which is impossible. Hence the next smallest divisor of n^4+n^2+1 than n^2-n+1 is at least n^2-n+(2*n+1) = n^2+n+1 > sqrt(n^4+n^2+1). - Jianing Song, Oct 23 2022

G. Tenenbaum, pp. 268 ff, in: R. L. Graham et al., eds., Mathematics of Paul Erdős I.

T. D. Noe, Table of n, a(n) for n = 1..10000

Cf. A033677 (n/a(n)), A000196 (sqrt), A027750 (list of divisors), A056737 (n/a(n) - a(n)), A219695 (half of this for odd numbers), A207375 (list the central divisor(s)).
The strictly inferior case is A060775. Cf. also A140271.
Indices of given values: A008578 (1 and the prime numbers: a(n) = 1), A161344 (a(n) = 2), A161345 (a(n) = 3), A161424 (4), A161835 (5), A162526 (6), A162527 (7), A162528 (8), A162529 (9), A162530 (10), A162531 (11), A162532 (12), A282668 (indices of primes).

a033676 n = last $ takeWhile (<= a000196 n) $ a027750_row n
-- Reinhard Zumkeller, Jun 04 2012

A033676 := proc(n) local a,d; a := 0 ; for d in numtheory[divisors](n) do if d^2 <= n then a := max(a,d) ; end if; end do: a; end proc: # R. J. Mathar, Aug 09 2009

largestDivisorLEQR[n_Integer] := Module[{dvs = Divisors[n]}, dvs[[Ceiling[Length@dvs/2]]]]; largestDivisorLEQR /@ Range[100] (* Borislav Stanimirov, Mar 28 2010 *)
Table[Last[Select[Divisors[n],#<=Sqrt[n]&]],{n,100}] (* Harvey P. Dale, Mar 17 2017 *)

A033676(n) = {local(d);if(n<2,1,d=divisors(n);d[(length(d)+1)\2])} \\ Michael B. Porter, Jan 30 2010

from sympy import divisors
def A033676(n):
    d = divisors(n)
    return d[(len(d)-1)//2]  # Chai Wah Wu, Apr 05 2021

a(n) = n / A033677(n).
a(n) = A161906(n,A038548(n)). - Reinhard Zumkeller, Mar 08 2013
a(n) = A162348(2n-1). - Daniel Forgues, Sep 29 2014

cp's OEIS Frontend

A033676 Largest divisor of n <= sqrt(n).

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