cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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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

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Author

Charles Kusniec, Oct 30 2024

Keywords

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)

Examples

			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)
		

Crossrefs

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

Programs

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

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)

Extensions

New name from Rémi Guillaume, Feb 19 2025
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