A091943 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).

2, 9, 2, 20, 64, 17, 37, 29, 293, 9, 59, 228, 559, 869, 49, 88, 17, 9, 317, 2039, 20, 121, 353, 755, 1375, 3002, 4072, 100, 159, 127, 1219, 143, 2597, 1315, 7183, 37, 200, 755, 146, 2443, 3922, 647, 10139, 12349, 164, 248, 54, 2159, 989, 188, 529, 10331, 3717

Offset

1,1

Comments

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.

Links

Table of n, a(n) for n=1..53.

Example

Rows begin:
{2},
{9,2},
{20,64,17},
{37,29,293,9},
{59,228,559,869,49},
{88,17,9,317,2039,20},
{121,353,755,1375,3002,4072,100},
{159,127,1219,143,2597,1315,7183,37},
{200,755,146,2443,3922,647,10139,12349,164},
{248,54,2159,989,188,529,10331,3717,19189,59},
{302,853,1909,4801,6214,8917,12755,15738,24803,29368,247},
{365,293,317,20,6239,37,15871,1363,2227,8759,41375,88},...

Prog

(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}

Crossrefs

Cf. A091941, A091942, A091944.

Sequence in context: A228375 A188966 A248433 * A318511 A345299 A276048

Adjacent sequences: A091940 A091941 A091942 * A091944 A091945 A091946

Keyword

nonn,tabl

Author

Paul D. Hanna, Feb 16 2004

Status

approved