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Least x > 0 such that (x^2-1)/(y^2-1) = n, where y=A130280(n), or 0 if no such x exists.
1

%I #8 May 23 2019 08:28:29

%S 2,7,5,0,4,7,13,5,0,9,23,17,14,41,11,7,16,55,39,31,8,23,91,19,0,25,

%T 109,15,59,49,61,79,10,169,29,17,36,191,131,11,83,71,85,1121,19,47,

%U 281,41,0,49,35,79,160,3079,21,13,37,175,3541,209,62,433,55,31,14,23,401,239

%N Least x > 0 such that (x^2-1)/(y^2-1) = n, where y=A130280(n), or 0 if no such x exists.

%F a(n) = sqrt(n*(A130280(n)^2-1) + 1).

%F a(A130283(n)) = 0.

%o (PARI) a(n, L=10^15) = if(issquare(n), L=2+n^2); for(k=2, L, if((k^2-1)%n==0,if(issquare(1+(k^2-1)/n), return(k))));

%Y Cf. A130280, A130283.

%K nonn

%O 1,1

%A _Jinyuan Wang_, Apr 13 2019