A396254 Irregular table read by rows: let p = prime(n) and chi be any Dirichlet character modulo p of order p-1. Then T(n,k), 0 <= k <= p-2, is the algebraic degree of tau(chi^k) over Q, where tau is the Gauss sum.

1, 1, 2, 1, 8, 2, 8, 1, 12, 6, 2, 6, 12, 1, 40, 20, 40, 20, 2, 20, 40, 20, 40, 1, 48, 12, 8, 6, 48, 2, 48, 6, 8, 12, 48, 1, 128, 32, 128, 8, 128, 32, 128, 2, 128, 32, 128, 8, 128, 32, 128, 1, 108, 54, 12, 54, 108, 6, 108, 54, 2, 54, 108, 6, 108, 54, 12, 54, 108, 1, 220, 110, 220, 110, 220, 110, 220, 110, 220, 110, 2, 110, 220, 110, 220, 110, 220, 110, 220, 110, 220

Offset

1,3

Comments

If chi has order d, then Q(tau(chi)) is the compositum of Q(zeta_d) and Q(zeta_p)^G, where G is the unique index-d subgroup of (Z/pZ)*. See the Math Stack Exchange link below.

Links

Jianing Song, Table of n, a(n) for n = 1..10123 (Rows 1..68)

Math Stack Exchange, Let chi be a nontrivial Dirichlet charcater of order d modulo p. Is it true that zeta_d is in Q(tau(chi))?.

Formula

T(n,k) = d*EulerPhi(d), where d = (prime(n)-1) / gcd(k, prime(n)-1).

Example

The first rows:
  1;
  1, 2;
  1, 8, 2, 8;
  1, 12, 6, 2, 6, 12;
  1, 40, 20, 40, 20, 2, 20, 40, 20, 40;
  1, 48, 12, 8, 6, 48, 2, 48, 6, 8, 12, 48;
  1, 128, 32, 128, 8, 128, 32, 128, 2, 128, 32, 128, 8, 128, 32, 128;
  1, 108, 54, 12, 54, 108, 6, 108, 54, 2, 54, 108, 6, 108, 54, 12, 54, 108;
  1, 220, 110, 220, 110, 220, 110, 220, 110, 220, 110, 2, 110, 220, 110, 220, 110, 220, 110, 220, 110, 220;
  1, 336, 84, 336, 42, 336, 84, 8, 42, 336, 84, 336, 42, 336, 2, 336, 42, 336, 84, 336, 42, 8, 84, 336, 42, 336, 84, 336.
Let n = 3, prime(3) = 5. Let chi = [1->1, 2->i, 3->-i, 4->-1] be a Dirichlet character modulo 5 of order 4.
- tau(chi^0) = 0, which has degree 1, so T(3,0) = 1;
- tau(chi^1) = 2*sin(Pi/5) + 2*i*sin(2*Pi/5), which has degree 8 (in fact Q(2*sin(Pi/5) + 2*i*sin(2*Pi/5)) = Q(zeta_20)), so T(3,1) = 8;
- tau(chi^2) = sqrt(5), which has degree 8, so T(3,2) = 2;
- tau(chi^3) = -2*sin(Pi/5) + 2*i*sin(2*Pi/5), which has degree 8 (in fact Q(-2*sin(Pi/5) + 2*i*sin(2*Pi/5)) = Q(zeta_20)), so T(3,3) = 8.
Let n = 4, prime(4) = 7. Let chi = [1->1, 2->w^2, 3->w, 4->w^4, 5->w^5, 6->-1] be a Dirichlet character modulo 7 of order 6, where w = zeta_6.
- tau(chi^0) = 0, which has degree 1, so T(4,0) = 1;
- tau(chi^1) is a root to x^12 - 497*x^6 + 117649 = 0, which has degree 12 (in fact Q(tau(chi^1)) = Q(zeta_42)), so T(4,1) = 12;
- tau(chi^2) is a root to x^6 - 7*x^3 + 343 = 0, which has degree 6 (in fact Q(tau(chi^2)) = Q(zeta_42+zeta_42^{-1})), so T(4,2) = 6;
- tau(chi^3) = sqrt(-7), which has degree 2, so T(4,3) = 2;
- tau(chi^4) and tau(chi^2) have the same minimal polynomial, so T(4,4) = 6;
- tau(chi^5) and tau(chi^1) have the same minimal polynomial, so T(4,5) = 12.

Prog

(PARI) row(n) = my(p=prime(n), d); vector(p-1, i, d=(p-1)/gcd(i-1, p-1); d*eulerphi(d))

Crossrefs

Cf. A006093 (row lengths), A002618 ({d*EulerPhi(d)}), A104039 (column k=1).

Cf. A396255 (another version).

Sequence in context: A152250 A154175 A385448 * A257777 A011208 A001281

Adjacent sequences: A396250 A396251 A396253 * A396255 A396256 A396257

Keyword

nonn,tabf,easy

Author

Jianing Song, May 20 2026

Status

approved