login
A396486
Irregular triangle read by rows: T(1,1) = T(2,1) = 1; for n >= 3, the n-th row lists the number of Dirichet characters chi modulo n such that Q(chi) = Q(zeta_(2d)) for each divisor d of psi(n)/2, where Q(chi) is the field generated by the values of chi, and psi = A002322 is the reduced totient function.
3
1, 1, 2, 2, 2, 2, 2, 2, 4, 4, 2, 4, 2, 2, 2, 8, 4, 2, 2, 4, 4, 2, 4, 4, 4, 4, 4, 2, 2, 4, 8, 2, 4, 2, 4, 12, 4, 4, 4, 8, 2, 8, 2, 20, 8, 2, 2, 8, 8, 2, 2, 4, 4, 2, 4, 12, 4, 8, 2, 2, 12, 12, 4, 4, 2, 4, 8, 16, 4, 4, 8, 4, 16, 2, 2, 4, 8, 4, 4, 8, 8, 4, 8, 2, 2, 4, 4, 12, 12
OFFSET
1,3
COMMENTS
For n >= 3, let d is the k-th divisor of psi(n)/2. If d is odd, then T(n,k) is the number of elements in (Z/nZ)* having order d or 2*d; otherwise T(n,k) is the number of elements in (Z/nZ)* having order 2*d.
EXAMPLE
Table begins
1,
1,
2,
2,
2, 2,
2,
2, 4,
4,
2, 4,
2, 2,
2, 8,
4,
2, 2, 4, 4,
2, 4,
4, 4,
4, 4,
2, 2, 4, 8,
2, 4,
2, 4, 12,
4, 4.
PROG
(PARI) power_solution(invariant_factors, k) = vecprod(apply(x->gcd(x, k), invariant_factors)); \\ number of solutions to x^k = 1 in C_{k_1} X ... C_{k_r}
count_order(invariant_factors, k) = sumdiv(k, d, moebius(k/d)*power_solution(invariant_factors, d)); \\ number of elements of order k
row(n) = if(n<=2, [1], my(invariant_factors=znstar(n)[2], k=lcm(invariant_factors)/2); apply(x->count_order(invariant_factors, 2*x)+if(x%2==1, count_order(invariant_factors, x)), divisors(k)));
CROSSREFS
Cf. A060594 (first column), A395473 (row lengths), A395683 (rightmost elements of rows).
Sequence in context: A036263 A307378 A339049 * A279401 A168514 A326353
KEYWORD
nonn,tabf
AUTHOR
Jianing Song, May 28 2026
STATUS
approved