

A152304


MarsagliaZaman type binet solution rationalized: f(n)=(11/20  Sqrt[512]/20)^n/3 + (2/3)*(11/20 + Sqrt[512]/20)^n; a(n)=Mod[Floor[f(n)],10].


0



1, 0, 1, 3, 5, 8, 5, 5, 2, 1, 0, 2, 0, 2, 2, 7, 9, 2, 8, 7, 5, 6, 7, 6, 3, 6, 8, 2, 2, 7, 9, 2, 4, 5, 5, 9, 2, 3, 7, 5, 4, 2, 5, 0, 2, 9, 5, 0, 0, 7, 6, 8, 7, 8, 2, 5, 8, 5, 0, 3, 5, 2, 7, 4, 3, 2, 6, 9, 7, 8, 3, 1, 9, 9, 3, 1
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

0,4


COMMENTS

The binet solution came from Mathematica:
f[n_Integer] = Module[{a}, a[n] /. RSolve[{a[n] == a[n  1] + a[n  2] + a[n  1]/10,a[0] == 1, a[1] == 1}, a[n], n][[1]] // FullSimplify].
I used coefficients{1/2,2/3} on the golden ration like roots to get my function.


REFERENCES

Ivars Peterson, The Jungles of Randomness, 1998, John Wiley and Sons, Inc., page 207


LINKS

Table of n, a(n) for n=0..75.


FORMULA

f(n)=(11/20  Sqrt[512]/20)^n/3 + (2/3)*(11/20 + Sqrt[512]/20)^n;
a(n)=Mod[Floor[f(n)],10].


MATHEMATICA

g[n_] := (11/20  Sqrt[512]/20)^n/3 + (2/3)*(11/20 + Sqrt[512]/20)^n;
Table[Mod[Floor[FullSimplify[Expand[g[n]]]], 10], {n, 0, 76}]


CROSSREFS

Sequence in context: A021283 A212224 A020864 * A021902 A136188 A073334
Adjacent sequences: A152301 A152302 A152303 * A152305 A152306 A152307


KEYWORD

nonn


AUTHOR

Roger L. Bagula, Dec 02 2008


STATUS

approved



