A180783 Number of distinct solutions of Sum_{i=1..1} (x(2i-1)*x(2i)) == 1 (mod n), with x() in {1,2,...,n-1}.

0, 1, 2, 2, 3, 2, 4, 4, 4, 3, 6, 4, 7, 4, 6, 6, 9, 4, 10, 6, 8, 6, 12, 8, 11, 7, 10, 8, 15, 6, 16, 10, 12, 9, 14, 8, 19, 10, 14, 12, 21, 8, 22, 12, 14, 12, 24, 12, 22, 11, 18, 14, 27, 10, 22, 16, 20, 15, 30, 12, 31, 16, 20, 18, 26, 12, 34, 18, 24, 14, 36, 16, 37, 19, 22, 20, 32, 14, 40, 20, 28

Offset: 1

Author

R. H. Hardin, formula from Max Alekseyev in the Sequence Fans Mailing List, Sep 20 2010

Keywords

• nonn

Comments

Except for the first term, this appears to be the number of pairs of integers i,j with 1 <= i <= n, 1 <= j <= i, such that i+j == i*j (mod n), for n=1,2,3,... - John W. Layman, Oct 19 2011

Layman's observation holds since i+j == i*j (mod n) is equivalent to (i-1)*(j-1) == 1 (mod n). - Max Alekseyev, Oct 22 2011

For i > 1, equal to the number of elements x relatively prime to n such that x mod n >= x^(-1) mod n. - Jeffrey Shallit, Jun 14 2018

Differs from A007897 for n = 1, 35, 45 etc. - Georg Fischer, Sep 20 2020

Examples

			Solutions for product of a single 1..10 pair = 1 (mod 11) are (1*1) (2*6) (3*4) (5*9) (7*8) (10*10).
		

Links

• R. H. Hardin, Table of n, a(n) for n = 1..10000

Crossrefs

Column 1 of A180793.

Cf. A000010, A007897, A060594.

Formula

a(n) = (A000010(n) + A060594(n)) / 2.