A081846 Maximal element in the continued fraction for 1/n*sum(k>=0,1/2^(2^k)).

6, 12, 19, 25, 33, 39, 46, 52, 60, 66, 72, 79, 85, 93, 99, 106, 112, 120, 126, 132, 139, 145, 153, 159, 166, 172, 180, 186, 192, 199, 205, 206, 219, 226, 232, 240, 246, 252, 259, 265, 273, 279, 286, 292, 300, 306, 313, 319, 326, 333, 339, 346, 352, 360, 366, 373, 379, 386, 393, 399, 406, 412, 420, 413, 433, 439, 446, 453, 459, 466, 472, 480, 486, 493, 499, 506, 513, 519, 526, 532, 540, 546, 553, 559, 567, 573, 579, 586, 593, 600, 606, 613, 619, 627, 633, 619, 646, 653, 660, 666

Offset: 1

Author

Benoit Cloitre, Apr 10 2003

Keywords

• nonn

Comments

It seems that a(n)=20n/3 for infinitely many values of n.

Crossrefs

Cf. A007400, A007404, A078816 (erroneous version).

Cf. A384939.

Programs

Mathematica

s = N[Sum[1/2^(2^k), {k, 0, Infinity}], 1000000]; Table[Max[ContinuedFraction[s/n]], {n, 1, 100}] (* Vaclav Kotesovec, Jul 22 2025 *)

Extensions

Corrected and extended by Vaclav Kotesovec, Jul 22 2025