A262161 a(n) is the largest term in the continued fraction for a(n-1) + n^2/a(n-1), where a(1)=1.

1, 5, 6, 8, 11, 14, 17, 20, 24, 28, 32, 36, 40, 44, 49, 54, 59, 64, 69, 74, 79, 85, 91, 97, 103, 109, 115, 121, 127, 134, 141, 148, 155, 162, 169, 176, 183, 190, 198, 206, 214, 222, 230, 238, 246, 254, 262, 270, 278, 286, 295, 304, 313, 322, 331, 340, 349, 358, 367, 376, 385, 394, 404, 414, 424, 434, 444, 454, 464, 474, 484, 494, 504, 514, 524, 535, 546, 557, 568, 579, 590, 601, 612, 623, 634, 645, 656, 667, 678, 689, 701, 713, 725, 737, 749, 761, 773, 785, 797, 809

Offset

1,2

Comments

There is an asymptotic formula for this sequence (see formula line).

Links

Table of n, a(n) for n=1..100.

Benoit Cloitre, Plot of (a(n)-n*(sqrt(2*n/3)-1/5))/sqrt(n) for 10000<n<20000

Formula

a(n) = n*(sqrt(2*n/3)-1/5) + O(sqrt(n)) and this is sharp (see link).

Example

We have a(7)=17 and the continued fraction for a(7) + 8^2/a(7) is [20, 1, 3, 4], hence a(8)=20.

Mathematica

a = {1}; Do[AppendTo[a, Max@ ContinuedFraction[a[[n - 1]] + (n^2)/a[[n - 1]]]], {n, 2, 100}]; a (* Michael De Vlieger, Sep 14 2015 *)

Prog

(PARI) a=1; for(n=2, 100, a=vecmax(contfrac(a+n^2/a)); print1(a, ", "))

Crossrefs

Sequence in context: A088766 A332549 A091091 * A290744 A120151 A022937

Adjacent sequences: A262158 A262159 A262160 * A262162 A262163 A262164

Keyword

nonn

Author

Benoit Cloitre, Sep 13 2015

Status

approved