A203168 Positions of 1 in the continued fraction expansion of Pi.

4, 6, 7, 8, 10, 12, 15, 16, 21, 24, 25, 29, 35, 41, 42, 45, 47, 51, 53, 54, 56, 57, 58, 60, 61, 63, 64, 66, 68, 69, 74, 79, 82, 84, 87, 89, 92, 94, 96, 98, 99, 104, 108, 113, 115, 116, 121, 125, 126, 134, 136, 138, 141, 144, 148, 149, 150, 154, 157, 158, 160

Offset: 1

Author

Ben Branman, Dec 29 2011

Keywords

• nonn
• nice

Comments

In the Gauss-Kuzmin distribution, 1 appears with probability log_2(4/3) = 41.5037...%. Thus the n-th appearance of 1 in the continued fraction of a real number chosen uniformly from [0, 1) will be, with probability 1, n / (log_2(4/3)) + O(sqrt(n)). Does this sequence have the same asymptotic? - Charles R Greathouse IV, Dec 30 2011

Links

• T. D. Noe, Table of n, a(n) for n = 1..1000
• Eric Weisstein's World of Mathematics, Pi Continued Fraction
• Index entries for continued fractions for constants
• Index entries for sequences related to the number Pi

Crossrefs

Cf. A001203, A033089, A000796.

Programs

Mathematica

Flatten[Position[ContinuedFraction[Pi, 160], 1]]

PARI

v=contfrac(Pi);for(i=1,#v,if(v[i]==1,print1(i", "))) \\ Charles R Greathouse IV, Dec 30 2011

Formula

A001203(a(n)) = 1.