OFFSET
1,12
COMMENTS
a(n) is also the number of linearly dependent diagonal/side length ratios R(n,k), in the regular n-gon. The following will explain this. In the regular n-gon inscribed in a circle the number of distinct diagonals including the side is floor(n/2). Not all of the corresponding length ratios R(n,k) = d(n,k)/d(n,1), k = 1..floor(n/2), with d(n,1) = s(n) (the length of the side), d(n,2) the length of the smallest diagonal, etc., are linearly independent because C(n,R(n,2)) = 0, where C is the minimal polynomial of R(n,2) = 2*cos(Pi/n) (see A187360) with degree delta(n) = A055034(n). Thus every ratio R(n,j), with j = delta(n)+1, ..., floor(n/2) can be expressed as a linear combination of the independent R(n,k), k=1, ..., delta(n). See the comment from Sep 21 2013 on A053121 for powers of R(n,2) (called there rho(N)). Therefore, a(n) = floor(n/2) - delta(n) is, for n>=2, the number of linearly dependent ratios R(n,k) in the regular n-gon. - Wolfdieter Lang, Sep 23 2013
From Wolfdieter Lang, Nov 23 2020: (Start)
This sequence gives the difference between the number of odd numbers in the smallest nonnegative residue system modulo n (called here RS(n)) and the smallest nonnegative restricted residue system (called here RRS(n), but RRS(1) = {1}, not {0}).
LINKS
Lei Zhou, Table of n, a(n) for n = 1..10000
FORMULA
a(n) = floor(n/2) - delta(n), with floor(n/2) = A004526 and delta(n) = A055034(n) = phi(2*n)/2, for n >= 2, with Euler's phi A000010. See the Aug 17 2011 comment on A055034. For n = 1 this would be -1, not 0, because delta(1) = 1. - Wolfdieter Lang, Sep 23 2013
Sum_{k=1..n} a(k) ~ c*n^2, where c = 1/4 - 2/Pi^2 = 0.04735763... (A190357). - Amiram Eldar, Feb 23 2025
EXAMPLE
n=1: there is no odd number greater than 2 but smaller than 1-1=0, so a(1)=0.
Same for n=2,3.
n=4: 3 is the only odd number in 2..(4-1), and GCD(3,4)=1, so a(4)=0.
For any prime numbers and numbers in the form of 2^n, no odd number in 2..(n-1) has common factor with n, so a(p)=0 and a(2^n)=0, n>0.
n=6: 3,5 are odd numbers in 2..(6-1), and GCD(3,6)=3>1 and GCD(5,6)=1, so a(6)=1.
n=15: candidates are 3,5,7,9,11,13. 3, 5, and 9 have greater than 1 common factors with 15, so a(15)=3
From Wolfdieter Lang, Sep 23 2013: (Start)
Example n = 15 for a(n) = floor(n/2) - delta(n): 1, 3, 5, 7, 9, 11, 13 take out 1, 7, 9, 11, leaving 3, 5, 13. Therefore, a(15) = 7 - 4 = 3. See the formula above for delta.
In the regular 15-gon the 3 (= a(15)) diagonal/side ratios R(15, 5), R(15, 6) and R(15,7) can be expressed as linear combinations of the R(15,j), j=1..4. See the n-gon comment above. (End)
From Wolfdieter Lang, Nov 23 2020: (Start)
n = 1: RS(1) = {0}, RRS(1) = {1}, hence a(1) = 0 - 1 = 0. Here RRS(1) is not {0}(standard) because delta(1) := 1 (the degree of minimal polynomial for 2*cos(Pi//1) = -2 which is x+2, see A187360).
MATHEMATICA
Table[Count[Range[3, n - 1, 2], _?(GCD[n, #] > 1 &)], {n, 100}] (* T. D. Noe, Nov 30 2012 *)
a[1] = 0; a[n_] := Floor[n/2] - EulerPhi[2*n]/2; Array[a, 80] (* Amiram Eldar, Nov 28 2020 *)
PROG
(PARI) a(n) = sum(i=2, n-1, (i%2) && (gcd(i, n)!=1)); \\ Michel Marcus, Aug 07 2018
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Lei Zhou, Nov 29 2012
STATUS
approved
