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

2, 6, 7, 8, 13, 17, 18, 19, 24, 28, 29, 30, 35, 39, 40, 41, 46, 50, 51, 52, 57, 61, 62, 63, 68, 72, 73, 74, 79, 83, 84, 85, 90, 94, 95, 96, 101, 105, 106, 107, 112, 116, 117, 118, 123, 127, 128, 129, 134, 138, 139, 140, 145, 149, 150, 151, 156, 160, 161, 162, 167, 171

Offset: 1

Author

Cino Hilliard, Nov 03 2003

Keywords

• nonn
• easy

Comments

Also, numbers congruent to {2, 6, 7, 8} mod 11. - Bruno Berselli, Jan 20 2016

Examples

			2^7 - 7 = 121 = 11*11. Thus 2 is in the sequence.
13^7 - 7 = 11*5704410. Thus 13 is in the sequence.
		

References

• E. Grosswald, Topics From The Theory of Numbers, 1966, pp. 62-63.

Links

• Colin Barker, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Programs

Mathematica

LinearRecurrence[{1,0,0,1,-1},{2,6,7,8,13},70] (* Harvey P. Dale, Jan 30 2015 *)

PARI

conxkmap(a,p,n) = { for(x=1,n, for(j=1,n, y=x^j-a; if(y%p==0,print1(x","); break) ) ) }

PARI

a(n) = (-9 - (-1)^n - (7-I)*(-I)^n - (7+I)*I^n + 22*n)/8 \\ Colin Barker, Oct 16 2015

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

Formula

a(n) = a(n-1) + a(n-4) - a(n-5) for n > 5. - Harvey P. Dale, Jan 30 2015

From Colin Barker, Oct 16 2015: (Start)

a(n) = (-9 - (-1)^n - (7-i)*(-i)^n - (7+i)*i^n + 22*n)/8, where i=sqrt(-1).

G.f.: x*(3*x^4+x^3+x^2+4*x+2) / ((x-1)^2*(x+1)*(x^2+1)). (End)