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A021030
Decimal expansion of 1/26.
1
0, 3, 8, 4, 6, 1, 5, 3, 8, 4, 6, 1, 5, 3, 8, 4, 6, 1, 5, 3, 8, 4, 6, 1, 5, 3, 8, 4, 6, 1, 5, 3, 8, 4, 6, 1, 5, 3, 8, 4, 6, 1, 5, 3, 8, 4, 6, 1, 5, 3, 8, 4, 6, 1, 5, 3, 8, 4, 6, 1, 5, 3, 8, 4, 6, 1, 5, 3, 8, 4, 6, 1, 5, 3, 8, 4, 6, 1, 5, 3, 8, 4, 6, 1, 5, 3, 8, 4, 6, 1, 5, 3, 8, 4, 6, 1, 5, 3, 8
OFFSET
0,2
COMMENTS
A tool code breakers sometimes use is the index of coincidence, I_c. According to Swenson (2008), the theoretically perfect I_c is if all characters occur exactly the same number of times, so that none is more likely than any other to be repeated. For cypher text encrypted from English text (using an alphabet of 26 letters) of infinite length, this has the limit (n - 1)/(26n - 1), which by L'Hopital's rule is 1/26. - Alonso del Arte, Sep 13 2011
Also continued fraction expansion of (sqrt(5317635) - 2067)/746. - Bruno Berselli, Sep 13 2011
REFERENCES
Christopher Swenson, Modern Cryptanalysis: Techniques for Advanced Code Breaking. Indianopolis, Indiana: Wiley Publishing Inc. (2008): 12 - 15
FORMULA
Contribution by Bruno Berselli, Sep 13 2011: (Start)
G.f.: x*(3+5*x-4*x^2+5*x^3)/((1-x)*(1+x)*(1-x+x^2)).
a(n) = a(n-1) - a(n-3) + a(n-4) for n > 4.
a(n) = (1/30)*(-11*(n mod 6)+34*((n+1) mod 6) - ((n+2) mod 6) + 29*((n+3) mod 6) - 16*((n+4) mod 6) + 19*((n+5) mod 6)) for n > 0. (End)
EXAMPLE
0.03846153846153846153846153846...
MATHEMATICA
Join[{0}, RealDigits[1/26, 10, 120][[1]]] (* or *) PadRight[{0}, 120, {5, 3, 8, 4, 6, 1}] (* Harvey P. Dale, Dec 19 2012 *)
CROSSREFS
Sequence in context: A242030 A105722 A103559 * A276682 A303215 A240242
KEYWORD
nonn,cons,easy
AUTHOR
STATUS
approved