A377499 a(n) is the median of the divisors of 2n-1.

1, 2, 3, 4, 3, 6, 7, 4, 9, 10, 5, 12, 5, 6, 15, 16, 7, 6, 19, 8, 21, 22, 7, 24, 7, 10, 27, 8, 11, 30, 31, 8, 9, 34, 13, 36, 37, 10, 9, 40, 9, 42, 11, 16, 45, 10, 17, 12, 49, 10, 51, 52, 11, 54, 55, 20, 57, 14, 11, 12, 11, 22, 15, 64, 23, 66, 13, 12, 69, 70, 25, 12, 17, 14, 75, 76

Offset

1,2

Comments

From Rémi Guillaume, Nov 26 2024 and Dec 05 2024: (Start)

2n-1 has only odd divisors; so the sum of any two of them is even.

a(n) and A219695(n) have opposite parity.

a(n) and n have the same parity.

a(n) = sqrt(2n-1) iff 2n-1 = (2j+1)^2 for some j >= 0, iff n is a centered square (A001844(j)); in this case, the two "median" divisors coincide with 2j+1, so their mean a(n) = 2j+1 and A219695(n) = 0.

More generally, with s a nonnegative integer:

If j >= s and n is the centered square A001844(j), then a(n-2s^2) <= 2j+1 and A219695(n-2s^2) <= 2s.

If j > (s^2)/2 and n = A001844(j), then a(n-2s^2) = 2j+1 and A219695(n-2s^2) = 2s. (P)

Basis of the proofs: 2(n-2s^2)-1 = (2j+1)^2-(2s)^2.

If j = s and n = A001844(j), then n-2s^2 = 2j+1 and 2(n-2s^2)-1 = 4j+1.

(End)

Links

Table of n, a(n) for n=1..76.

Formula

a(n) = (A033677(2n-1) + A033676(2n-1))/2.

a(n) = A063655(2n-1)/2.

a(n) = sqrt((2n-1) + A219695(n)^2).

a(n) = n iff 2n-1 is 1 or prime (n is 1 or in A006254); in this case, A219695(n) = n-1.

From Rémi Guillaume, Nov 21 2024: (Start)

a(n) = A361565(2n-1).

sqrt(2n-1) <= a(n) <= n.

a(n) = (A377865(n) + A377864(n))/2.

a(n) = A377864(n) + A219695(n).

a(n) = A377865(n) - A219695(n). (End)

Example

From Michael De Vlieger, Nov 01 2024: (Start)
Let u = 2*n-1, let factor d <= sqrt(u) be the largest such, and let D = u/d.
For n = 2, u = 2*2-1 = 3, d = 1, D = 3, so a(2) = (1+3)/2 = 2.
For n = 5, u = 2*5-1 = 9 is a perfect square and d = D = 3, so a(5) = (3+3)/2 = 3.
For n = 8, u = 2*8-1 = 15, d = 3, D = 5, so a(8) = (3+5)/2 = 4, etc. (End)

Mathematica

{1}~Join~Table[u = 2*n + 1; (# + u/#)/2 &@ #[[Floor[Length[#]/2] ]] &@ Divisors[u], {n, 2, 120}] (* Michael De Vlieger, Nov 01 2024 *)

Prog

(Python)
from sympy import divisors
def A377499(n): return (d:=(f:=divisors(m:=(n<<1)-1))[len(f)-1>>1])+m//d>>1 # Chai Wah Wu, Nov 07 2024

Crossrefs

Cf. A219695 (associated subtrahend square base forming 2n-1), A001844 (solutions of a(n)=sqrt(2n-1)), A006254 (indices of record highs).

See also A063655, A033676, A033677, A361565, A377864, A377865.

Sequence in context: A318510 A048276 A382723 * A127463 A076618 A116550

Adjacent sequences: A377496 A377497 A377498 * A377500 A377501 A377502

Keyword

nonn

Author

Charles Kusniec, Oct 30 2024

Extensions

New name from Rémi Guillaume, Feb 19 2025

Status

approved