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A036117
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a(n) = 2^n mod 11.
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10
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1, 2, 4, 8, 5, 10, 9, 7, 3, 6, 1, 2, 4, 8, 5, 10, 9, 7, 3, 6, 1, 2, 4, 8, 5, 10, 9, 7, 3, 6, 1, 2, 4, 8, 5, 10, 9, 7, 3, 6, 1, 2, 4, 8, 5, 10, 9, 7, 3, 6, 1, 2, 4, 8, 5, 10, 9, 7, 3, 6, 1, 2, 4, 8, 5, 10, 9, 7, 3, 6, 1, 2, 4, 8, 5, 10, 9, 7
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OFFSET
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0,2
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COMMENTS
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a(k) = k has only one solution, namely k=7. - Jon Perry, Oct 30 2014
As 2 is a primitive root of 11, all integers 1 through 10 are present. - Jon Perry, Oct 30 2014
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REFERENCES
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H. Cohn, A Second Course in Number Theory, Wiley, NY, 1962, p. 256.
I. M. Vinogradov, Elements of Number Theory, pp. 220 ff.
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LINKS
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FORMULA
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a(n) = a(n-1) - a(n-5) + a(n-6). - R. J. Mathar, Apr 13 2010
G.f.: (1+x+2*x^2+4*x^3-3*x^4+6*x^5)/ ((1-x) * (1+x) * (x^4-x^3+x^2-x+1)). - R. J. Mathar, Apr 13 2010
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EXAMPLE
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2^6 = 64 = 66 - 2 == -2 mod 11 == 9 mod 11, so a(6) = 9.
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MAPLE
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i := 5: [ seq(numtheory[primroot](ithprime(i))^j mod ithprime(i), j=0..100) ];
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MATHEMATICA
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PROG
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(Sage) [power_mod(2, n, 11) for n in range(0, 78)] # Zerinvary Lajos, Nov 03 2009
(Magma) [Modexp(2, n, 11): n in [0..100]]; // G. C. Greubel, Oct 16 2018
(GAP) List([0..70], n->PowerMod(2, n, 11)); # Muniru A Asiru, Oct 18 2018
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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