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%I #6 Mar 30 2012 17:38:10
%S 7,70,38,41,117,378,500,682,776,3861,4052,515,5744,1710,6613,1744,
%T 11018,13241,3458,5099,1393,16610,26884,15006,2072,13637,31361,4443,
%U 26508,7850,37520,31152,39922,37107,6072,4005,32491,4030,43211,12238
%N Smallest number m such that m^2+1 is divisible by A002144(n)^2 (= squares of primes congruent to 1 mod 4).
%C a(2) = 70 since A002144(2)=13, 70^2+1 = 4091 = 13^2 * 29 and for no k<70 does 13^2 divide k^2+1. Related to period-1 continued fractions.
%H Klaus Brockhaus, <a href="/A059321/b059321.txt">Table of n, a(n) for n=1..2000</a>
%Y Cf. A002144, A059591, A059592.
%K easy,nonn
%O 1,1
%A _Marc LeBrun_, Jan 26 2001
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