A070373 -id:A070373 - cp's OEIS frontend

A201910 Irregular triangle of 5^k mod prime(n).

1, 1, 2, 0, 1, 5, 4, 6, 2, 3, 1, 5, 3, 4, 9, 1, 5, 12, 8, 1, 5, 8, 6, 13, 14, 2, 10, 16, 12, 9, 11, 4, 3, 15, 7, 1, 5, 6, 11, 17, 9, 7, 16, 4, 1, 5, 2, 10, 4, 20, 8, 17, 16, 11, 9, 22, 18, 21, 13, 19, 3, 15, 6, 7, 12, 14, 1, 5, 25, 9, 16, 22, 23, 28, 24, 4

Offset: 1

T. D. Noe, Dec 07 2011

Except for the third row, the first term of each row is 1. Many sequences are in this one: starting at A036121 (mod 23) and A070365 (mod 7).

			The first 9 rows are:
1
1, 2
0
1, 5, 4, 6, 2, 3
1, 5, 3, 4, 9
1, 5, 12, 8
1, 5, 8, 6, 13, 14, 2, 10, 16, 12, 9, 11, 4, 3, 15, 7
1, 5, 6, 11, 17, 9, 7, 16, 4
1, 5, 2, 10, 4, 20, 8, 17, 16, 11, 9, 22, 18, 21, 13, 19, 3, 15, 6, 7, 12, 14

T. D. Noe, Rows n = 1..60, flattened

Cf. A201908 (2^k), A201909 (3^k), A201911 (7^k).

Cf. A070365 (7), A070367 (11), A070368 (13), A070371 (17), A070373 (19), A036121 (23), A070379 (29), A070384 (37), A070387 (41), A070389 (43), A036127 (47), A036133 (73), A036137 (97), A036139 (103), A036149 (157), A036151 (167), A036156 (193).

P:=Filtered([1..350],IsPrime);;
R:=List([1..Length(P)],n->OrderMod(5,P[n]));;
Flat(Concatenation([1,1,2,0],List([3..10],n->List([0..R[n]-1],k->PowerMod(5,k,P[n]))))); # Muniru A Asiru, Feb 02 2019

nn = 10; p = 5; t = p^Range[0,Prime[nn]]; Flatten[Table[If[Mod[n, p] == 0, {0}, tm = Mod[t, n]; len = Position[tm, 1, 1, 2][[-1,1]]; Take[tm, len-1]], {n, Prime[Range[nn]]}]]

A271378 a(n) = 5^n mod 31.

Original entry on oeis.org

1, 5, 25, 1, 5, 25, 1, 5, 25, 1, 5, 25, 1, 5, 25, 1, 5, 25, 1, 5, 25, 1, 5, 25, 1, 5, 25, 1, 5, 25, 1, 5, 25, 1, 5, 25, 1, 5, 25, 1, 5, 25, 1, 5, 25, 1, 5, 25, 1, 5, 25, 1, 5, 25, 1, 5, 25, 1, 5, 25, 1, 5, 25, 1, 5, 25, 1, 5, 25, 1, 5, 25, 1, 5, 25, 1, 5, 25

Offset: 0

Vincenzo Librandi, Apr 06 2016

Period 3: repeat [1, 5, 25].

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

Cf. similar sequences of the type 5^n mod p, where p is a prime: A070365 (p=7), A070367 (p=11), A070368 (p=13), A070371 (p=17), A070373 (p=19), A036121 (p=23), A070379 (p=29), this sequence (p=31), A070384 (p=37), A070387 (p=41), A070389 (p=43), A036127 (p=47), A036133 (p=73), A036137 (p=97), A271379 (p=101), A036139 (p=103), A036149 (p=157), A271380 (p=163) A036151 (p=167), A036156 (p=193).

Magma
```
[Modexp(5, n, 31): n in [0..100]];
```

&cat [[1,5,25]^^30]; // Bruno Berselli, Apr 07 2016

seq(op([1, 5, 25]), n=0..50); # Wesley Ivan Hurt, Jun 30 2016

Mathematica
```
PowerMod[5, Range[0, 100], 31]
```

x='x+O('x^99); Vec((1+5*x+25*x^2)/(1-x^3)) \\ Altug Alkan, Apr 06 2016

G.f.: (1+5*x+25*x^2)/(1-x^3).
a(n) = a(n-3) for n>2.
a(n) = 5^(n mod 3).
a(n) = (31 - 28*cos(2*n*Pi/3) - 20*sqrt(3)*sin(2*n*Pi/3))/3. - Wesley Ivan Hurt, Jun 30 2016

Edited by Bruno Berselli, Apr 07 2016

A187532 a(n) = 4^n mod 19.

Original entry on oeis.org

1, 4, 16, 7, 9, 17, 11, 6, 5, 1, 4, 16, 7, 9, 17, 11, 6, 5, 1, 4, 16, 7, 9, 17, 11, 6, 5, 1, 4, 16, 7, 9, 17, 11, 6, 5, 1, 4, 16, 7, 9, 17, 11, 6, 5, 1, 4, 16, 7, 9, 17, 11, 6, 5, 1, 4, 16, 7, 9, 17, 11, 6, 5, 1, 4, 16, 7, 9, 17, 11, 6, 5, 1, 4, 16, 7, 9, 17, 11, 6, 5

Offset: 0

M. F. Hasler, Mar 10 2011

Period 9: repeat (1,4,16,7,9,17,11,6,5).

Also continued fraction expansion of (13140908+sqrt(323139488118562))/24969762. - Bruno Berselli, Sep 09 2011

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

Cf. A036120, A070342, A070373, A070395, A070421.

[4^n mod 19 : n in [0..80]]; // Vincenzo Librandi, Sep 09 2011

PowerMod[4, Range[0, 100], 19]  (* or *)
PadRight[{}, 100, {1, 4, 16, 7, 9, 17, 11, 6, 5}] (* Paolo Xausa, Mar 17 2024 *)

a(n)=lift(Mod(4,19)^n) \\ Charles R Greathouse IV, Mar 22 2016

a(n+9) = a(n).

G.f.: (1 + 4*x + 16*x^2 + 7*x^3 + 9*x^4 + 17*x^5 + 11*x^6 + 6*x^7 + 5*x^8)/((1-x)*(1+x+x^2)*(1+x^3+x^6)). - Bruno Berselli, Sep 09 2011

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);););}

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A201910 Irregular triangle of 5^k mod prime(n).

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A271378 a(n) = 5^n mod 31.

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A187532 a(n) = 4^n mod 19.

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Magma

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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