A168429 -id:A168429 - cp's OEIS frontend

A036117 a(n) = 2^n mod 11.

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

Offset: 0

N. J. A. Sloane

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

			2^6 = 64 = 66 - 2 == -2 mod 11 == 9 mod 11, so a(6) = 9.

H. Cohn, A Second Course in Number Theory, Wiley, NY, 1962, p. 256.
I. M. Vinogradov, Elements of Number Theory, pp. 220 ff.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Wikipedia, Primitive roots
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,-1,1).

Cf. A000079 (2^n), A008830, A168429.

List([0..70],n->PowerMod(2,n,11)); # Muniru A Asiru, Oct 18 2018

[Modexp(2, n, 11): n in [0..100]]; // G. C. Greubel, Oct 16 2018

i := 5: [ seq(numtheory[primroot](ithprime(i))^j mod ithprime(i),j=0..100) ];

Table[Mod[2^n,11],{n,0,6!}] (* Vladimir Joseph Stephan Orlovsky, Apr 29 2010 *)

a(n)=lift(Mod(2,11)^n) \\ Charles R Greathouse IV, Jul 02 2013

[power_mod(2,n,11) for n in range(0, 78)] # Zerinvary Lajos, Nov 03 2009

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
a(n+10) = a(n). - Jon Perry, Oct 30 2014
a(n+5) = 11 - a(n) for all n in Z. - Michael Somos, Oct 17 2018

A010691 Period 2: repeat (1,10).

Original entry on oeis.org

1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10

Offset: 0

N. J. A. Sloane

Regular continued fraction of (5+sqrt 35)/10. - R. J. Mathar, Nov 21 2011

Sequence is an infinite palindrome in two ways (numbers and English names): ONE, TEN, ONE, TEN, ONE, TEN, ONE, ... . - Eric Angelini, Sep 16 2023

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,1).

Cf. A036117, A070341, A168429, A070367, A070392, A070404, A048271, A187466.

[10^n mod 11: n in [0..80]]; // Vincenzo Librandi, Aug 24 2011

g:=(1+10*z)/((1-z^2)): gser:=series(g, z=0, 66): seq((coeff(gser, z, n)), n=0..65); # Zerinvary Lajos, Feb 25 2009

PadRight[{},100,{1,10}] (* Harvey P. Dale, Aug 27 2013 *)

a(n) = -9/2*(-1)^n + 11/2.
G.f.: (1+10*z)/(1-z^2). - Zerinvary Lajos, Feb 25 2009
a(n) = 10^n mod 11. - M. F. Hasler, Mar 10 2011
From Nicolas Bělohoubek, Nov 11 2021: (Start)
a(n) = 10/a(n-1). See also A010695.
a(n) = 11 - a(n-1). See also A010712. (End)

A187466 a(n) = 9^n mod 11.

Original entry on oeis.org

1, 9, 4, 3, 5, 1, 9, 4, 3, 5, 1, 9, 4, 3, 5, 1, 9, 4, 3, 5, 1, 9, 4, 3, 5, 1, 9, 4, 3, 5, 1, 9, 4, 3, 5, 1, 9, 4, 3, 5, 1, 9, 4, 3, 5, 1, 9, 4, 3, 5, 1, 9, 4, 3, 5, 1, 9, 4, 3, 5, 1, 9, 4, 3, 5, 1, 9, 4, 3, 5, 1, 9, 4, 3, 5, 1, 9, 4, 3, 5, 1, 9, 4, 3, 5, 1

Offset: 0

M. F. Hasler, Mar 10 2011

Period 5: repeat [1, 9, 4, 3, 5].

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,1).

Cf. A010691, A036117, A048271, A070341, A070367, A070392, A070404, A168429.

A187466:=n->9^n mod 11: seq(A187466(n), n=0..100); # Wesley Ivan Hurt, Jun 11 2016

PowerMod[9, Range[0, 100], 11] (* Wesley Ivan Hurt, Jun 11 2016 *)

a(n)=lift(Mod(9,11)^n) \\ Charles R Greathouse IV, Mar 22 2016

G.f.: (5*x^4 + 3*x^3 + 4*x^2 + 9*x + 1)/(1 - x^5). - Chai Wah Wu, Jun 04 2016

a(n) = a(n-5) for n>4. - Wesley Ivan Hurt, Jun 11 2016

A266973 a(n) = 4^n mod 17.

Original entry on oeis.org

1, 4, 16, 13, 1, 4, 16, 13, 1, 4, 16, 13, 1, 4, 16, 13, 1, 4, 16, 13, 1, 4, 16, 13, 1, 4, 16, 13, 1, 4, 16, 13, 1, 4, 16, 13, 1, 4, 16, 13, 1, 4, 16, 13, 1, 4, 16, 13, 1, 4, 16, 13, 1, 4, 16, 13, 1, 4, 16, 13, 1, 4, 16, 13, 1, 4, 16, 13, 1, 4, 16, 13, 1, 4, 16

Offset: 0

Vincenzo Librandi, Apr 06 2016

Period 4: repeat [1, 4, 16, 13].

Index entries for linear recurrences with constant coefficients, signature (0,0,0,1).

Cf. similar sequences of the type 4^n mod p, where p is a prime: A010685 (5), A153727 (7), A168429 (11), A168430 (13), this sequence (17), A187532 (19).

Magma
```
[Modexp(4, n, 17): n in [0..100]];
```

A266973:=n->power(4,n) mod 17: seq(A266973(n), n=0..100); # Wesley Ivan Hurt, Jun 29 2016

Mathematica
```
PowerMod[4, Range[0, 100], 17]
```

G.f.: (1+4*x+16*x^2+13*x^3)/(1-x^4).
a(n) = a(n-4) for n>3.
From Wesley Ivan Hurt, Jun 29 2016: (Start)
a(n) = (34 - 3*(5 + 3*I)*I^(-n) - 3*(5 - 3*I)*I^n)/4 where I=sqrt(-1).
a(n) = (17 - 15*cos(n*Pi/2) - 9*sin(n*Pi/2))/2. (End)

A268271 Primes p such that there is a Fibonacci-type sequence (mod p) that begins with (1,b) and encounters all quadratic residues of p in the first (p-1)/2 iterations (for some b).

Original entry on oeis.org

11, 19, 29, 31, 59, 71, 79, 89, 101, 131, 179, 181, 191, 229, 239, 251, 271, 311, 349, 359, 379, 401, 419, 431, 439, 479, 491, 499, 509, 571, 599, 631, 659, 719, 739, 751, 761, 839, 941, 971, 1019, 1021, 1039, 1051, 1061, 1091, 1109, 1171, 1229, 1249, 1259, 1319, 1361, 1399

Offset: 1

Michel Marcus, Mar 02 2016

nonn

			p=11 is a term since, modulo 11, the sequence 1, 4, 5, 9, 3 satisfies 5=4+1, 9=5+4, 3=9+5, 1=9+3, ..., with a period of (11-1)/2 = 5.

Alexandru Gica, Quadratic Residues in Fibonacci Sequences, Fibonacci Quart. 46/47 (2008/2009), no. 1, 68-72.

Subsequence of A045468.
Cf. A003147 (similar sequence for a different period).
Cf. A168429, A070373 (examples of such Fibonacci-type sequences).

findr(p) = {for (k=1, (p-1)/2, if ((k^2 % p) == 5, return(k)););}
isok(p) = {if ((p % 2) && isprime(p), pm = p % 5; if ((pm == 1) || (pm == 4), rf = findr(p);(znorder(Mod((1+rf)/2, p)) == (p-1)/2) || (znorder(Mod((1-rf)/2, p)) == (p-1)/2);););}

cp's OEIS Frontend

A036117 a(n) = 2^n mod 11.

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A010691 Period 2: repeat (1,10).

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A187466 a(n) = 9^n mod 11.

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A266973 a(n) = 4^n mod 17.

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A268271 Primes p such that there is a Fibonacci-type sequence (mod p) that begins with (1,b) and encounters all quadratic residues of p in the first (p-1)/2 iterations (for some b).

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PARI