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A088225 Solutions to x^n == 7 (mod 11).

%I A088225 #29 Feb 24 2024 01:07:11
%S A088225 2,6,7,8,13,17,18,19,24,28,29,30,35,39,40,41,46,50,51,52,57,61,62,63,
%T A088225 68,72,73,74,79,83,84,85,90,94,95,96,101,105,106,107,112,116,117,118,
%U A088225 123,127,128,129,134,138,139,140,145,149,150,151,156,160,161,162,167,171
%N A088225 Solutions to x^n == 7 (mod 11).
%C A088225 Also, numbers congruent to {2, 6, 7, 8} mod 11. - _Bruno Berselli_, Jan 20 2016
%D A088225 E. Grosswald, Topics From The Theory of Numbers, 1966, pp. 62-63.
%H A088225 Colin Barker, <a href="/A088225/b088225.txt">Table of n, a(n) for n = 1..1000</a>
%H A088225 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,1,-1).
%F A088225 a(n) = a(n-1) + a(n-4) - a(n-5) for n > 5. - _Harvey P. Dale_, Jan 30 2015
%F A088225 From _Colin Barker_, Oct 16 2015: (Start)
%F A088225 a(n) = (-9 - (-1)^n - (7-i)*(-i)^n - (7+i)*i^n + 22*n)/8, where i=sqrt(-1).
%F A088225 G.f.: x*(3*x^4+x^3+x^2+4*x+2) / ((x-1)^2*(x+1)*(x^2+1)). (End)
%e A088225 2^7 - 7 = 121 = 11*11. Thus 2 is in the sequence.
%e A088225 13^7 - 7 = 11*5704410. Thus 13 is in the sequence.
%t A088225 LinearRecurrence[{1,0,0,1,-1},{2,6,7,8,13},70] (* _Harvey P. Dale_, Jan 30 2015 *)
%o A088225 (PARI) conxkmap(a,p,n) = { for(x=1,n, for(j=1,n, y=x^j-a; if(y%p==0,print1(x","); break) ) ) }
%o A088225 (PARI) a(n) = (-9 - (-1)^n - (7-I)*(-I)^n - (7+I)*I^n + 22*n)/8 \\ _Colin Barker_, Oct 16 2015
%o A088225 (PARI) Vec(x*(3*x^4+x^3+x^2+4*x+2)/((x-1)^2*(x+1)*(x^2+1)) + O(x^100)) \\ _Colin Barker_, Oct 16 2015
%K A088225 nonn,easy
%O A088225 1,1
%A A088225 _Cino Hilliard_, Nov 03 2003