A384939 Maximum element in the continued fraction for (1/n) * Sum_{k>=0} 1/3^(2^k).

5, 11, 17, 24, 29, 35, 42, 48, 53, 60, 66, 72, 77, 84, 90, 96, 102, 108, 114, 121, 126, 132, 139, 145, 151, 156, 163, 169, 175, 181, 187, 193, 200, 205, 211, 218, 224, 229, 235, 242, 248, 253, 260, 266, 272, 279, 284, 290, 297, 303, 308, 314, 321, 327, 332, 339, 345, 351, 357, 363, 369, 375, 381, 387, 393, 400, 406, 411, 418, 424, 430, 436, 442, 448, 454, 460, 466, 472, 479, 484, 326, 497, 503, 508, 515, 521, 527, 532, 539, 545, 551, 558, 563, 569, 576, 582, 587, 594, 600, 606

Offset: 1

Author

Vaclav Kotesovec, Jul 22 2025

Keywords

• nonn

Links

• Jeffrey O. Shallit, Simple continued fractions for some irrational numbers, J. Number Theory 11 (1979), no. 2, 209-217.

Crossrefs

Cf. A004200, A078885, A081846.

Programs

Mathematica

m = 10; terms = 100; t1 = ConstantArray[0, 2*terms]; t2 = ConstantArray[1, 2*terms]; Until[t1 == t2, m++; PrintTemporary["m=",m]; s = Sum[1/3^(2^k), {k, 0, m}]; t1 = t2; t2 = Table[Max[ContinuedFraction[s/n]], {n, 1, 2*terms}]]; Take[t2, terms]