A377499 For odd numbers 2n-1, half the sum of the least divisor not less than the square root and the largest divisor not exceeding the square root.

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 06 2024: (Start)

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

If 2n-1 = m^2, i.e., if n is a centered square (A001844), then the two "median divisors" coincide and their mean a(n) is m. (End)

Links

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

Formula

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

a(n) = (A033677(2n-1) + A033676(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), and in which case A219695(n) = n-1.

From Rémi Guillaume, Nov 21 2024: (Start)

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

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

Example

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

Mathematica

{1}~Join~Table[N = 2*n + 1; (# + N/#)/2 &@ #[[Floor[Length[#]/2] ]] &@ Divisors[N], {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. A063655, A033676, A033677, A219695, A377864.

Cf. A001844, A006254.

Sequence in context: A347123 A318510 A048276 * A127463 A076618 A116550

Adjacent sequences: A377496 A377497 A377498 * A377500 A377501 A377502

Keyword

nonn,changed

Author

Charles Kusniec, Oct 30 2024

Status

approved