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A074639
a(n)=Sum_h (hh'-1)/n with h and h' in [1,n], (h,n)=1, hh'=1 (mod n).
10
0, 0, 0, 1, 2, 5, 4, 11, 10, 15, 12, 31, 16, 39, 28, 36, 34, 75, 32, 91, 52, 64, 60, 145, 64, 115, 88, 141, 84, 225, 76, 241, 146, 160, 152, 250, 104, 319, 204, 272, 172, 419, 152, 447, 280, 286, 228, 599, 208, 501, 252, 440, 348, 727
OFFSET
0,5
COMMENTS
For a given n a(n) is the sum for h ranging over the set of least nonnegative residues coprimes with n of (hh'-1)/n, where h' is the (unique) number in the same set such that hh'=1 (mod n).
The summand is also the least nonnegative residue of (-1/n) mod h. - Robert Israel, May 18 2014
LINKS
M. Dondi, Plot of A074639(n)/phi(n) (Euler's totient function) against the line y=x/4 in the range [0,100].
M. Dondi, Plot of A074639(n)/phi(n) (Euler's totient function) against the line y=x/4 in the range [0,1000].
M. Dondi, Plot of A074639(n)/phi(n) (Euler's totient function) against the line y=x/4 in the range [0,10000].
M. Dondi, Plot of A074639(n)/phi(n) (Euler's totient function) against the line y=x/4 in the range [0,10000] showing only one point out of every 5.
EXAMPLE
(1,n)=1 for all n, 1*1=1, so 1 contributes 0 to the sum. (n-1,n)=1 for all n, (n-1)^2=1 (mod n), so n-1 contributes n-2. Thus a(6)=4, in fact only 1 and 5 are coprime with 6 in {1,...,6}; a(5)=2*1+(5-2), in fact 2*3=6=1 (mod 5) and 6=5+1.
MAPLE
seq(add((i*(i^(-1) mod m)-1)/m, i = select(t -> igcd(t, m)=1, [$1..m-1])), m=0..100); # Robert Israel, May 18 2014
CROSSREFS
KEYWORD
nonn
AUTHOR
Michele Dondi (bik.mido(AT)tiscalinet.it), Sep 12 2002
STATUS
approved