A145571 Numerators of partial sums for Liouville's constant.

1, 11, 110001, 110001000000000000000001, 110001000000000000000001000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001

Offset: 1

Author

Wolfdieter Lang, Mar 06 2009

Keywords

• nonn
• frac
• easy

Comments

The denominators are 10^(n!).

In a(n) the 1's appear at positions j!, j=1..n. Therefore Liouville's constant c:=Sum_{k>=1} 1/10^(k!) is the number 0.a(n) with n -> infinity.

Liouville's constant c is transcendental. See, e.g., the proof in the Rosenberger-Fine reference.

The number of digits of a(n) is n! = A000142(n). The number of 0s is 0 for n = 1 and 2, and Sum_{k=3..n} (z(n) - 1), for n >= 3, where z(n) = n! - (n-1)! = A001563(n-1). This number is n! - n, for n >= 1. - Wolfdieter Lang, Apr 09 2024

Examples

			a(2)=11 because c(2)=1/10 + 1/100 = 11/100.
a(6) has 1's at positions 1,2,6,24,120,720 (A000142, factorials) and 0's in between.
		

References

• B. Fine and G. Rosenberger, Number theory: an introduction via the distribution of primes, Birkhäuser, Boston, Basel, Berlin, 2007. Th. 6.3.2.3., p. 286.

Crossrefs

Cf. A000142, A001563, A145572 (a(n) read as base 2 representation).

Programs

Mathematica

Numerator[Accumulate[1/10^Range[6]!]] (* Paolo Xausa, Jun 25 2024 *)
Block[{k = 0}, NestList[#*10^(++k*k!) + 1 &, 1, 5]] (* Paolo Xausa, Jun 26 2024 *)

Formula

a(n) = numerator(c(n)), with c(n):= Sum_{k=1..n} 1/10^(k!).

a(1) = 1, and a(n) = a(n-1)*10^(z(n)) + 1, for n >= 2, where z(n) = A001563(n-1) = n! - (n-1)! = (n-1)!*(n-1). - Wolfdieter Lang, Apr 09 2024 [Corrected by Paolo Xausa, Jun 26 2024 ]