A091944 Triangle, read by rows, where T(n,k) equals the maximum number of partial quotients in the simple continued fraction expansion of (1/n + 1/k + 1/m) for m>=1.

2, 6, 2, 7, 9, 7, 8, 8, 16, 6, 9, 13, 15, 16, 9, 9, 7, 6, 16, 19, 7, 11, 15, 15, 17, 17, 19, 11, 11, 13, 16, 13, 15, 16, 21, 8, 12, 17, 12, 18, 18, 15, 20, 20, 12, 11, 11, 17, 16, 11, 15, 21, 17, 24, 9, 12, 16, 18, 18, 21, 20, 22, 22, 21, 23, 12, 14, 16, 16, 7, 19, 8, 23, 16, 16, 19

Offset

1,1

Comments

First column is A091942. Least m that produce maximum number of partial quotients of CF(1/n + 1/k + 1/m) forms triangle A091943.

Links

Table of n, a(n) for n=1..76.

Formula

T(n, k) = length(contfrac(1/n + 1/k + 1/A091943(n, k))).

Example

Rows begin:
{2},
{6,2},
{7,9,7},
{8,8,16,6},
{9,13,15,16,9},
{9,7,6,16,19,7},
{11,15,15,17,17,19,11},
{11,13,16,13,15,16,21,8},
{12,17,12,18,18,15,20,20,12},
{11,11,17,16,11,15,21,17,24,9},
{12,16,18,18,21,20,22,22,21,23,12},
{14,16,16,7,19,8,23,16,16,19,24,9},...

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, A091943.

Sequence in context: A086356 A242435 A086359 * A069936 A043294 A375805

Adjacent sequences: A091941 A091942 A091943 * A091945 A091946 A091947

Keyword

nonn,tabl

Author

Paul D. Hanna, Feb 16 2004

Status

approved