login
A091941
a(n) equals the least k that produces the maximum number of partial quotients in the simple continued fraction expansion of (1/n + 1/k).
3
2, 9, 20, 37, 59, 88, 121, 159, 200, 248, 302, 365, 428, 493, 574, 654, 738, 827, 898, 1029, 1133, 1205, 1342, 1459, 1592, 1740, 1831, 1991, 2168, 2339, 2485, 2757, 2734, 2991, 3072, 3307, 3546, 3745, 3943, 4037, 4261, 4576, 4727, 4889, 5182, 5491, 5733
OFFSET
1,1
COMMENTS
The maximum number of partial quotients in CF(1/n+1/k) equals A091942(n). Limit of a(n)/n^2 = (3+sqrt(5))/2 = 2.618...
EXAMPLE
a(100001)=26174739625; 26174739625/100001^2 = 2.61742...
a(1000001)=2617923148538; 2617923148538/1000001^2 = 2.61791...
PROG
(PARI) {a(n)=local(A); M=0; for(k=2*n^2-1, 3*n^2, L=length(contfrac(1/k+1/n)); if(L>M, M=L; A=k)); A}
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Feb 15 2004
STATUS
approved