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

Author

T. D. Noe, Dec 07 2011

Keywords

• nonn
• tabf

Comments

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).

Examples

			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
		

Links

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

Crossrefs

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).

Programs

GAP

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

Mathematica

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]]}]]