A201911 Irregular triangle of 7^k mod prime(n).

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

Offset: 1

Author

T. D. Noe, Dec 07 2011

Keywords

• nonn
• tabf

Comments

Except for the fourth row, the first term of each row is 1. Many sequences are in this one: starting at A036132 (mod 71) and A070404 (mod 11).

Examples

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

Links

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

Crossrefs

Cf. A201908 (2^k), A201909 (3^k), A201910 (5^k).

Cf. A070404 (11), A070405 (13), A070407 (17), A070409 (23), A070413 (29), A070415 (31), A070420 (37), A070422 (39), A070424 (41), A070425 (43), A070429 (47), A036132 (71).

Programs

GAP

P:=Filtered([1..350],IsPrime);;
R:=List([1..Length(P)],n->OrderMod(7,P[n]));;
Flat(Concatenation([1,1,1,2,4,3,0],List([5..10],n->List([0..R[n]-1],k->PowerMod(7,k,P[n]))))); # Muniru A Asiru, Feb 01 2019

Mathematica

nn = 10; p = 7; 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]]}]]