OFFSET
0,1
COMMENTS
In other words, this constant satisfies x = Sum_{n>=0} ( floor(10*n*x) (mod 10) ) / 10^n.
FORMULA
a(n) = floor(10*n*x) (mod 10), where x = Sum_{k>0} a(k)/10^k.
a(n) = 9 - A071876(n).
EXAMPLE
x=.24792479247924792479247914691469146914691469146913...
a(8) = 9 since floor(10*8*x) (mod 10) = 9.
The multiples of this constant x begin:
1*x = 0.2479247924792479247924791469146914691469...
2*x = 0.4958495849584958495849582938293829382938...
3*x = 0.7437743774377437743774374407440744074407...
4*x = 0.9916991699169916991699165876587658765877...
5*x = 1.239623962396239623962395734573457345735...
6*x = 1.487548754875487548754874881488148814881...
7*x = 1.735473547354735473547354028402840284028...
8*x = 1.983398339833983398339833175317531753175...
9*x = 2.231323132313231323132312322232223222322...
10*x = 2.479247924792479247924791469146914691469...
11*x = 2.727172717271727172717270616061606160616...
12*x = 2.975097509750975097509749762976297629763...
wherein the tenths place of n*x yields the n-th digit of x.
MATHEMATICA
tenth[x_] := Floor[10*FractionalPart[x]]; xx[n_] := xx[n] = Catch[ For[x = xx[n-1], True, x += 10^(-n), If[tenth[n*x] == tenth[10^(n-1)*x], Throw[x]]]]; xx[1] = 2/10; Scan[xx, Range[100]]; RealDigits[xx[100]][[1]] (* Jean-François Alcover, Nov 30 2012 *)
Clear[a]; a[1] = 2; a[2] = 4; a[n0 = 3] = 7; a[_] = 0; digits = 10^(n0-1); Do[a[n] = Mod[Floor[10*n*Sum[a[k]/10^k, {k, 1, n}]], 10], {n, n0+1, digits}]; Table[a[n], {n, 1, digits}] (* Jean-François Alcover, May 11 2015 *)
CROSSREFS
KEYWORD
AUTHOR
Paul D. Hanna, Jun 06 2002
STATUS
approved