A361565 -id:A361565 - cp's OEIS frontend

A361566 a(n) is the denominator of the median of divisors of n.

1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 2

Offset: 1

Stefano Spezia, Mar 15 2023

a(n) = 2 if n is in A139710, otherwise a(n) = 1. - Robert Israel, Mar 15 2023

			a(9) = 1 since the divisors of 9 are 1, 3, 9, and their median is 3.
a(12) = 2 since the divisors of 12 are 1, 2, 3, 4, 6, 12, and their median is 7/2.

Winston de Greef, Table of n, a(n) for n = 1..10000
Index entries for sequences related to divisors of numbers

Cf. A027750, A033676, A033677, A071090, A139710, A139711, A361565 (numerator).

a[n_]:=Denominator[Median[Divisors[n]]]; Array[a,88]

a(n) = my(d=divisors(n), m=#d+1); denominator((d[m\2] + d[m-m\2])/2); \\ Michel Marcus, Mar 16 2023

a(n) = denominator((A033676(n) + A033677(n))/2).

Except for p = 2 and k = 1, a(p^k) = 1, if p is a prime.

A361632 a(n) is the numerator of the median of the prime factors of n with repetition.

Original entry on oeis.org

2, 3, 2, 5, 5, 7, 2, 3, 7, 11, 2, 13, 9, 4, 2, 17, 3, 19, 2, 5, 13, 23, 2, 5, 15, 3, 2, 29, 3, 31, 2, 7, 19, 6, 5, 37, 21, 8, 2, 41, 3, 43, 2, 3, 25, 47, 2, 7, 5, 10, 2, 53, 3, 8, 2, 11, 31, 59, 5, 61, 33, 3, 2, 9, 3, 67, 2, 13, 5, 71, 2, 73, 39, 5, 2, 9, 3, 79

Offset: 2

Stefano Spezia, Mar 18 2023

			a(12) = 2 since 12 = 2*2*3, and the median of the factors is equal to 2.
a(36) = 5 since 30 = 2*2*3*3, and the median of the factors is equal to 5/2.

Cf. A001222, A027746, A079879, A323171, A361565, A361630 (without repetition), A361633 (denominator), A361725.

a[n_]:=Numerator[Median[Flatten[Table[#[[1]], {#[[2]]}] & /@ FactorInteger[n]]]]; Array[a,78,2]

For p a prime, a(p^k) = p.

a(n) = numerator((A079879(n) + A361725(n))/2).

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

Original entry on oeis.org

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

Charles Kusniec, Oct 30 2024

nonn

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)

			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)

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.

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

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

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)

New name from Rémi Guillaume, Feb 19 2025

A361630 a(n) is the numerator of the median of the distinct prime factors of n.

Original entry on oeis.org

2, 3, 2, 5, 5, 7, 2, 3, 7, 11, 5, 13, 9, 4, 2, 17, 5, 19, 7, 5, 13, 23, 5, 5, 15, 3, 9, 29, 3, 31, 2, 7, 19, 6, 5, 37, 21, 8, 7, 41, 3, 43, 13, 4, 25, 47, 5, 7, 7, 10, 15, 53, 5, 8, 9, 11, 31, 59, 3, 61, 33, 5, 2, 9, 3, 67, 19, 13, 5, 71, 5, 73, 39, 4, 21, 9, 3

Offset: 2

Stefano Spezia, Mar 18 2023

			a(12) = 5 since the distinct prime factors of 12 are 2 and 3, of median equal to 5/2.
a(30) = 3 since the distinct prime factors of 30 are 2, 3, and 5, of median equal to 3.

Winston de Greef, Table of n, a(n) for n = 2..10000

Cf. A001221, A027748, A323171, A361565, A361631 (denominator), A361632 (with multiplicity).

a[n_]:=Numerator[Median[FactorInteger[n][[All, 1]]]]; Array[a,77,2]

a(n)=my(f=factor(n)[,1]~, i=length(f)); numerator(if(i%2, f[i\2+1], (f[i/2]+f[i/2+1])/2)) \\ Winston de Greef, Mar 23 2023

For p a prime, a(p^k) = p.

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A361566 a(n) is the denominator of the median of divisors of n.

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A361632 a(n) is the numerator of the median of the prime factors of n with repetition.

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

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A361630 a(n) is the numerator of the median of the distinct prime factors of n.

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