A323874 Irregular triangle of 13^k mod prime(n).

1, 1, 1, 3, 4, 2, 1, 6, 1, 2, 4, 8, 5, 10, 9, 7, 3, 6, 0, 1, 13, 16, 4, 1, 13, 17, 12, 4, 14, 11, 10, 16, 18, 6, 2, 7, 15, 5, 8, 9, 3, 1, 13, 8, 12, 18, 4, 6, 9, 2, 3, 16, 1, 13, 24, 22, 25, 6, 20, 28, 16, 5, 7, 4, 23, 9, 1, 13, 14, 27, 10, 6, 16, 22, 7, 29, 5

Offset: 1

Author

Muniru A Asiru, Feb 04 2019

Keywords

• nonn
• tabf

Comments

Length of the n-th row (n != 6) is the order of 13 modulo the n-th prime.

Except for the sixth row, the first term of each row is 1.

Examples

			The first 10 rows are:
  1
  1
  1, 3, 4, 2
  1, 6
  1, 2, 4, 8, 5, 10, 9, 7, 3, 6
  0
  1, 13, 16, 4
  1, 13, 17, 12, 4, 14, 11, 10, 16, 18, 6, 2, 7, 15, 5, 8, 9, 3
  1, 13, 8, 12, 18, 4, 6, 9, 2, 3, 16
  1, 13, 24, 22, 25, 6, 20, 28, 16, 5, 7, 4, 23, 9
		

Crossrefs

Cf. A000040.

Cf. A201908 (2^k), A201909 (3^k), A201910 (5^k), A201911 (7^k), A323873 (11^k), this sequence (13^k).

Programs

GAP

A000040:=Filtered([1..350],IsPrime);; p:=6;;
R:=List([1..Length(A000040)],n->OrderMod(A000040[p],A000040[n]));;
a1:=List([1..p-1],n->List([0..R[n]-1],k->PowerMod(A000040[p],k,A000040[n])));;
a:=Flat(Concatenation(a1,[0],List([p+1..2*p],n->List([0..R[n]-1],k->PowerMod(A000040[p],k,A000040[n])))));; Print(a);

Maple

T:= n-> (p-> `if`(p=13, 0, seq(13&^k mod p,
         k=0..numtheory[order](13, p)-1)))(ithprime(n)):
seq(T(n), n=1..15);  # Alois P. Heinz, Feb 06 2019

Mathematica

With[{q = 13}, Table[If[p == q, {0}, Array[PowerMod[q, #, p] &, MultiplicativeOrder[q, p], 0]], {p, Prime@ Range@ 11}]] // Flatten (* Michael De Vlieger, Feb 25 2019 *)