A159243 Number of elements in the continued fraction for Sum_{k=0..n} 1/(1+2^(2^k)).

2, 4, 8, 15, 24, 41, 85, 159, 314, 651, 1267, 2496, 4977, 9889, 19731, 38945, 77356, 154693, 308051, 615768, 1229080, 2456328, 4908126, 9815038, 19620985, 39237465, 78466413, 156910438, 313788371, 627528817

Offset: 0

Author

Enrique Pérez Herrero, Apr 06 2009

Keywords

• nonn
• more

Comments

Number of terms in the n-th partial sum of the Fermat number reciprocals.

Examples

			The partial sum for n = 3 (four terms) is: 1/3 + 1/5 + 1/17 + 1/257 = 39062/65535 expressed in continued fraction gives: {0,1,1,2,9,1,2,1,1,2,2,1,2,1,5} that has 15 elements so: a(3) = 15.
		

Links

• Daniel Duverney, Irrationality of Fast Converging Series of Rational Numbers, J. Math. Sci. Univ. Tokyo, 8 (2001), 275-316.
• N. J. A. Sloane and James A. Sellers, On non-squashing partitions, Discrete Mathematics, Vol. 294, No. 3 (2005), pp. 259-274; arXiv preprint, arXiv:math/0312418 [math.CO], 2003.

Crossrefs

Cf. A056469, A051158.

Programs

Mathematica

Table[Length[ContinuedFraction[Sum[1/(1 + 2^2^k), {k, 0, v}]]], {v, 0, 20}]

Extensions

Offset corrected and a(21)-a(29) added by Amiram Eldar, May 05 2024