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.
%I A377499 #112 Apr 27 2025 14:47:17
%S A377499 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,
%T A377499 30,31,8,9,34,13,36,37,10,9,40,9,42,11,16,45,10,17,12,49,10,51,52,11,
%U A377499 54,55,20,57,14,11,12,11,22,15,64,23,66,13,12,69,70,25,12,17,14,75,76
%N A377499 a(n) is the median of the divisors of 2n-1.
%C A377499 From _Rémi Guillaume_, Nov 26 2024 and Dec 05 2024: (Start)
%C A377499 2n-1 has only odd divisors; so the sum of any two of them is even.
%C A377499 a(n) and A219695(n) have opposite parity.
%C A377499 a(n) and n have the same parity.
%C A377499 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.
%C A377499 More generally, with s a nonnegative integer:
%C A377499 If j >= s and n is the centered square A001844(j), then a(n-2s^2) <= 2j+1 and A219695(n-2s^2) <= 2s.
%C A377499 If j > (s^2)/2 and n = A001844(j), then a(n-2s^2) = 2j+1 and A219695(n-2s^2) = 2s. (P)
%C A377499 Basis of the proofs: 2(n-2s^2)-1 = (2j+1)^2-(2s)^2.
%C A377499 If j = s and n = A001844(j), then n-2s^2 = 2j+1 and 2(n-2s^2)-1 = 4j+1.
%C A377499 (End)
%F A377499 a(n) = (A033677(2n-1) + A033676(2n-1))/2.
%F A377499 a(n) = A063655(2n-1)/2.
%F A377499 a(n) = sqrt((2n-1) + A219695(n)^2).
%F A377499 a(n) = n iff 2n-1 is 1 or prime (n is 1 or in A006254); in this case, A219695(n) = n-1.
%F A377499 From _Rémi Guillaume_, Nov 21 2024: (Start)
%F A377499 a(n) = A361565(2n-1).
%F A377499 sqrt(2n-1) <= a(n) <= n.
%F A377499 a(n) = (A377865(n) + A377864(n))/2.
%F A377499 a(n) = A377864(n) + A219695(n).
%F A377499 a(n) = A377865(n) - A219695(n). (End)
%e A377499 From _Michael De Vlieger_, Nov 01 2024: (Start)
%e A377499 Let u = 2*n-1, let factor d <= sqrt(u) be the largest such, and let D = u/d.
%e A377499 For n = 2, u = 2*2-1 = 3, d = 1, D = 3, so a(2) = (1+3)/2 = 2.
%e A377499 For n = 5, u = 2*5-1 = 9 is a perfect square and d = D = 3, so a(5) = (3+3)/2 = 3.
%e A377499 For n = 8, u = 2*8-1 = 15, d = 3, D = 5, so a(8) = (3+5)/2 = 4, etc. (End)
%t A377499 {1}~Join~Table[u = 2*n + 1; (# + u/#)/2 &@ #[[Floor[Length[#]/2] ]] &@ Divisors[u], {n, 2, 120}] (* _Michael De Vlieger_, Nov 01 2024 *)
%o A377499 (Python)
%o A377499 from sympy import divisors
%o A377499 def A377499(n): return (d:=(f:=divisors(m:=(n<<1)-1))[len(f)-1>>1])+m//d>>1 # _Chai Wah Wu_, Nov 07 2024
%Y A377499 Cf. A219695 (associated subtrahend square base forming 2n-1), A001844 (solutions of a(n)=sqrt(2n-1)), A006254 (indices of record highs).
%Y A377499 See also A063655, A033676, A033677, A361565, A377864, A377865.
%K A377499 nonn
%O A377499 1,2
%A A377499 _Charles Kusniec_, Oct 30 2024
%E A377499 New name from _Rémi Guillaume_, Feb 19 2025