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A323874
Irregular triangle of 13^k mod prime(n).
1
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
1,4
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.
EXAMPLE
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
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 *)
PROG
(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);
CROSSREFS
Cf. A000040.
Cf. A201908 (2^k), A201909 (3^k), A201910 (5^k), A201911 (7^k), A323873 (11^k), this sequence (13^k).
Sequence in context: A303869 A081246 A264928 * A326765 A096411 A228340
KEYWORD
nonn,tabf
AUTHOR
Muniru A Asiru, Feb 04 2019
STATUS
approved