%I #11 Sep 13 2023 13:49:40

%S 1,1,1,1,3,1,3,1,5,1,5,3,1,5,3,7,1,7,5,1,7,3,9,1,7,9,3,5,1,5,7,11,1,9,

%T 5,11,3,7,1,9,11,3,5,13,1,13,11,7,1,11,13,9,7,3,5,15,1,11,7,5,15,3,13,

%U 9,1,7,5,13,11,17,1,13,15,11,17,7,3,5,9,1,13,17,9,11,3,7,19,1,17,19,13,5,11,1,15,9,19,5,17,3,13,7,21

%N Irregular triangle T read by rows: row n gives the inverse elements of row n of A216319 Modd(n), for n >= 1.

%C The length of row n is A055034(n), called here delta(n), for n >= 1.

%C For the modified modular equivalence relation Modd n see a comment in A203571, and the W. Lang link, Definition 4. p. 25. For Modd(a, n) one has to consider the parity of floor(a/n). If it is even then Modd(a, n) = mod(a, n), otherwise it is mod(-a, n).

%C The rows of A216319 are the smallest positive restricted residue system mod n with only odd members (RRSodd(n)). This is not a group mod n, but a group Modd n, called here G(rho(n)). This group is isomorphic to the Galois group Gal(Q(rho(n))/Q), where the algebraic number of degree delta(n) is rho(n) = 2*cos(Pi/n), for n >= 1. See A187360 for the minimal polynomials of rho(n), called C(n, x).

%H Wolfdieter Lang, <a href="http://arxiv.org/abs/1210.1018">The field Q(2cos(pi/n)), its Galois group and length ratios in the regular n-gon</a>, arXiv:1210.1018 [math.GR], 2012.

%F T(n, k) = Inverse of A216319(n, k) (Modd n), for n >= 1. For Modd n see the comment above.

%e The irregular triangle T(n, k) begins:

%e n\k 1 2 3 4 5 6 7 8 9 ...

%e 1: 1

%e 2: 1

%e 3: 1

%e 4: 1 3

%e 5: 1 3

%e 6: 1 5

%e 7: 1 5 3

%e 8: 1 5 3 7

%e 9: 1 7 5

%e 10: 1 7 3 9

%e 11: 1 7 9 3 5

%e 12: 1 5 7 11

%e 13: 1 9 5 11 3 7

%e 14: 1 9 11 3 5 13

%e 15: 1 13 11 7

%e 16: 1 11 13 9 7 3 5 15

%e 17: 1 11 7 5 15 3 13 9

%e 18: 1 7 5 13 11 17

%e 19: 1 13 15 11 17 7 3 5 9

%e 20: 1 13 17 9 11 3 7 19

%e ...

%e T(7, 2) = 5 because A216319(7, 2) = 3 and Modd(3*5, 7) = 1 since floor(15/7) = 2 is even, hence Modd(3*5, 7) = mod(15, 7) = 1. The residue classes Modd 7 for 1, 3, 5 are shown in the array given in A113807 (including the negative numbers) [3]*[5] = [1] (Modd 7).

%e T(9, 2) = 7 because A216319(9, 2) = 5 and Modd(7*5, 9) = 1, since floor(35/9) = 3 is odd, hence Moddn(35, 9) = mod(-35, 9) = 1.

%o (PARI) rowa(n) = select(x->(((x%2)==1) && (gcd(n, x)==1)), [1..n]); \\ A216319

%o Modd(x, n) = if ((x\n)%2, Mod(-x,n), Mod(x,n));

%o findinvm(k, n) = for (i=1, n, if (Modd(k*i, n) == 1, return(i)));

%o row(n) = my(ra=rowa(n)); vector(#ra, k, findinvm(ra[k], n)); \\ _Michel Marcus_, Sep 13 2023

%Y Cf. A055034, A203571, A187360, A216319.

%K nonn,tabf,easy

%O 1,5

%A _Wolfdieter Lang_, Oct 20 2020