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%I #3 Mar 30 2012 18:36:40
%S 2,9,2,20,64,17,37,29,293,9,59,228,559,869,49,88,17,9,317,2039,20,121,
%T 353,755,1375,3002,4072,100,159,127,1219,143,2597,1315,7183,37,200,
%U 755,146,2443,3922,647,10139,12349,164,248,54,2159,989,188,529,10331,3717
%N Triangle, read by rows, where T(n,k) equals the least m>0 that produces the maximum number of partial quotients in the simple continued fraction expansion of (1/n + 1/k + 1/m).
%C First column is A091941. Number of partial quotients forms triangle A091944. Maximum term in n-th row is T(n,n-1); what does limit of T(n,n-1)/n^4 equal? Limit T(n,n-1)/n^4 appears to exist, with a value less than 2.5 for n<=100.
%e Rows begin:
%e {2},
%e {9,2},
%e {20,64,17},
%e {37,29,293,9},
%e {59,228,559,869,49},
%e {88,17,9,317,2039,20},
%e {121,353,755,1375,3002,4072,100},
%e {159,127,1219,143,2597,1315,7183,37},
%e {200,755,146,2443,3922,647,10139,12349,164},
%e {248,54,2159,989,188,529,10331,3717,19189,59},
%e {302,853,1909,4801,6214,8917,12755,15738,24803,29368,247},
%e {365,293,317,20,6239,37,15871,1363,2227,8759,41375,88},...
%o (PARI) {T(n,k)=local(t);M=0;for(m=1,3*n^4, L=length(contfrac(1/n+1/k+1/m));if(L>M,M=L;t=m));t}
%Y Cf. A091941, A091942, A091944.
%K nonn,tabl
%O 1,1
%A _Paul D. Hanna_, Feb 16 2004
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