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%I #10 Mar 13 2020 12:56:00
%S 1,0,1,1,2,2,3,2,4,4,4,5,5,4,7,7,6,6,9,6,10,10,6,11,11,8,13,10,10,14,
%T 15,8,12,16,12,17,18,10,16,19,14,20,16,14,22,18,16,18,24,14,25,25,12,
%U 26,27,18,28,22,18,24,28,20,25,31,22,32,28,18,34,34,24
%N The number of even numbers <= n of the smallest nonnegative reduced residue system modulo 2*n + 1, for n >= 0.
%C For the smallest positive reduced residue system modulo N see the array A038566. Here the nonnegative residue system [0, 1, ..., N-1] is considered, differing only for N = 1 from A038566, with [0] (instead of [1]).
%C This sequence gives the complement of A332435 (with 0 for n = 0 included) relative to the number of positive numbers <= n of the smallest nonnegative reduced residue system modulo (2*n+1). Thus a(n) + A332435(n) = phi(n)/2, for n >= 1, with phi = A000010. For n = 0 one has 1 + 0 = 1.
%C a(n) gives also the number of even numbers appearing in the complete modified doubling sequence system (name it MDS(b)), for b = 2*n + 1, with n >= 1, proposed in a comment from _Gary W. Adamson_, Aug 24 2019, in the example section of A135303 for prime b.
%F a(n) = A000010(n)/2 - A332435(n), for n >= 1, and a(0) = 1.
%e n = 4, b = 9: the even numbers <= 4 in RRS(9) := [1, 2, 4, 5, 7, 8] are {2, 4}, hence a(4) = 2.
%e The complete MDS(9) system has one cycles of length 3: Cy*(9, 1) = (2, 4, 1), with the even numbers {2, 4}.
%e n = 8, b = 17: the even numbers <= 8 in RRS(17) := [1, 2, ..., 16] are {2, 4, 6 ,8}, hence a(8) = 4.
%e The complete MDS(17) system has two cycles of length 4: Cy*(17, 1) = (2, 4, 8, 1) and Cy*(17, 2) = (6, 5, 7, 3) and the even numbers are {2, 4, 6 ,8}.
%Y Cf. A000010, A038566, A135303, A332435.
%K nonn,easy
%O 0,5
%A _Wolfdieter Lang_, Feb 29 2020
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