cp's OEIS Frontend

A337713 Irregular triangle T read by rows: row n gives the inverse elements of row n of A216319 Modd(n), for n >= 1.

%I A337713 #11 Sep 13 2023 13:49:40
%S A337713 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 A337713 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 A337713 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 A337713 Irregular triangle T read by rows: row n gives the inverse elements of row n of A216319 Modd(n), for n >= 1.
%C A337713 The length of row n is A055034(n), called here delta(n), for n >= 1.
%C A337713 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 A337713 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 A337713 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 A337713 T(n, k) = Inverse of A216319(n, k) (Modd n), for n >= 1. For Modd n see the comment above.
%e A337713 The irregular triangle T(n, k) begins:
%e A337713 n\k 1  2  3  4  5  6  7  8 9 ...
%e A337713 1:  1
%e A337713 2:  1
%e A337713 3:  1
%e A337713 4:  1  3
%e A337713 5:  1  3
%e A337713 6:  1  5
%e A337713 7:  1  5  3
%e A337713 8:  1  5  3  7
%e A337713 9:  1  7  5
%e A337713 10: 1  7  3  9
%e A337713 11: 1  7  9  3  5
%e A337713 12: 1  5  7 11
%e A337713 13: 1  9  5 11  3  7
%e A337713 14: 1  9 11  3  5 13
%e A337713 15: 1 13 11  7
%e A337713 16: 1 11 13  9  7  3  5 15
%e A337713 17: 1 11  7  5 15  3 13  9
%e A337713 18: 1  7  5 13 11 17
%e A337713 19: 1 13 15 11 17  7  3  5 9
%e A337713 20: 1 13 17  9 11  3  7 19
%e A337713 ...
%e A337713 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 A337713 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 A337713 (PARI) rowa(n) = select(x->(((x%2)==1) && (gcd(n, x)==1)), [1..n]); \\ A216319
%o A337713 Modd(x, n) = if ((x\n)%2, Mod(-x,n), Mod(x,n));
%o A337713 findinvm(k, n) = for (i=1, n, if (Modd(k*i, n) == 1, return(i)));
%o A337713 row(n) = my(ra=rowa(n)); vector(#ra, k, findinvm(ra[k], n)); \\ _Michel Marcus_, Sep 13 2023
%Y A337713 Cf. A055034, A203571, A187360, A216319.
%K A337713 nonn,tabf,easy
%O A337713 1,5
%A A337713 _Wolfdieter Lang_, Oct 20 2020