A098290 Recurrence sequence based on positions of digits in decimal places of Zeta(3) (Apery's constant).

0, 2, 1, 10, 208, 380, 394, 159, 10, 208, 380, 394, 159, 10, 208, 380, 394, 159, 10, 208, 380, 394, 159, 10, 208, 380, 394, 159, 10, 208, 380, 394, 159, 10, 208, 380, 394, 159, 10, 208, 380, 394, 159, 10, 208, 380, 394, 159, 10, 208, 380, 394, 159, 10

Offset: 0

Author

Mark Hudson (mrmarkhudson(AT)hotmail.com), Sep 02 2004

Keywords

• nonn
• base

Comments

This recurrence sequence starts to repeat quite quickly because 1 appears at the 10th digit of Zeta(3), which is also where 159 starts.

Can the transcendental numbers such that recurrence relations of this kind eventually repeat be characterized? - Nathaniel Johnston, Apr 30 2011

Examples

			Zeta(3) = 1.2020569031595942853997...
a(0)=0, a(1)=2 because 2nd decimal = 0, a(2)=1 because first digit = 2, etc.
		

Crossrefs

Cf. A002117 for digits of Zeta(3). Other recurrence sequences: A097614 for Pi, A098266 for e, A098289 for log(2), A098290 for Zeta(3), A098319 for 1/Pi, A098320 for 1/e, A098321 for gamma, A098322 for G, A098323 for 1/G, A098324 for Golden Ratio (phi), A098325 for sqrt(Pi), A098326 for sqrt(2), A120482 for sqrt(3), A189893 for sqrt(5), A098327 for sqrt(e), A098328 for 2^(1/3).

Programs

Maple

with(StringTools): Digits:=400: G:=convert(evalf(Zeta(3)-1), string): a[0]:=0: for n from 1 to 50 do a[n]:=Search(convert(a[n-1], string), G)-1:printf("%d, ", a[n-1]):od: # Nathaniel Johnston, Apr 30 2011

Formula

a(0)=0, p(i)=position of first occurrence of a(i) in decimal places of Zeta(3), a(i+1)=p(i).