OFFSET
1,3
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
LINKS
R. H. Hardin, Table of n, a(n) for n = 1..10000
EXAMPLE
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).
CROSSREFS
KEYWORD
nonn
AUTHOR
R. H. Hardin, formula from Max Alekseyev in the Sequence Fans Mailing List, Sep 20 2010
STATUS
approved