1,2

Factoring n^2-k = p*q for k > 1 gives p+q never divisible by k*a. The remainders p+q mod k*a have minimum > 0 for k > 1. This sequence shows the minima for k=1, where there are zero values, e.g. a(8)=0. The restriction 1 < p < n-1 is due to the fact that (n-1)(n+1) = n^2-1 and (n-1)+(n+1)=2n is zero mod n. Excluding this trivial case as well as 1*(n^2-1) with 1+(n^2-1)=n^2=0 (mod n) gives the more interesting elements.

IMO 1988 #6

If n^2-1 is prime then set a(n) = n. Otherwise check (p+q mod n) for all natural numbers p, q which satisfy: p*q = n^2-1 and 1 < p < n-1 and 1 < q < n-1.

a(7)=2 since 7^2-1 = 48 = 2*24 = 3*16 = 4*12 where the sums of the factors (mod 7) are: [2+24]=5, [3+16]=5, [4+12]=2.

Sequence in context: A124406 A065331 A066262 this_sequence A071439 A100941 A082119

Adjacent sequences: A069814 A069815 A069816 this_sequence A069818 A069819 A069820

easy,nice,nonn

Rainer Rosenthal (r.rosenthal(AT)web.de), Apr 29 2002