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%I #34 Apr 20 2024 09:59:41
%S 0,1,2,3,0,5,6,1,8,9,2,11,0,3,14,15,4,1,18,5,20,21,2,23,0,7,26,3,8,29,
%T 30,1,4,33,10,35,36,5,2,39,0,41,6,13,44,3,14,7,48,1,50,51,4,53,54,17,
%U 56,9,2,5,0,19,10,63,20,65,6,3,68,69,22,1,12,7,74,75,4,13,78,25,8,81,2,83,0,5,86,9,28,89
%N For odd numbers 2n - 1, half the difference between the largest divisor not exceeding the square root, and the least divisor not less than the square root.
%C We consider 2n - 1 which has only odd divisors, so any difference among them is always even.
%H Seiichi Manyama, <a href="/A219695/b219695.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = A056737(2n - 1)/2 = (A033677(2n - 1) - A033676(2n - 1))/2.
%F a(n) = 0 if and only if 2n - 1 is a square.
%e For n = 2, consider divisors of 2n - 1 = 3 which are {1, 3}. The least one greater than or equal to sqrt(3) is 3, the largest one less than or equal to sqrt(3) is 1; whence a(2) = (3 - 1)/2 = 1.
%e For n = 14, consider divisors of 2n - 1 = 27 which are {1, 3, 9, 27}. The least one greater than or equal to sqrt(27) is 9, the largest one less than or equal to sqrt(27) is 3; whence a(14) = (9 - 3)/2 = 3.
%e For n = 1, 5, 13, 25,..., the number 2n - 1 equals the square 1, 9, 25, 49, ...; so the two beforementioned "median divisors" coincide with the square root, and a(n) = 0/2 = 0.
%t Table[(Divisors[n][[(Length[Divisors[n]] - Boole[IntegerQ[Sqrt[n]]])/2 + 1]] - Divisors[n][[(Length[Divisors[n]] + Boole[IntegerQ[Sqrt[n]]])/2]])/2, {n, 1, 199, 2}] (* Alonso del Arte, Nov 25 2012, corrected March 21, 2024 with help from _Giorgos Kalogeropoulos_ *)
%t A219695[n_] := (d = Divisors[2n - 1]; l = Floor[Length@d/2 + 1]; (d[[l]] - d[[-l]])/2); Array[A219695, 100] (* _Giorgos Kalogeropoulos_, Mar 15 2024 *)
%o (PARI)
%o A056737(n)={n=divisors(n); n[(2+#n)\2]-n[(1+#n)\2]}
%o A219695(n)=A056737(2*n-1)/2 \\ _M. F. Hasler_, Nov 25 2012
%Y Cf. A033676, A033677.
%K nonn,changed
%O 1,3
%A _M. F. Hasler_, Nov 25 2012
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