OFFSET
1,1
COMMENTS
For n >= 2, a(n) is divisible by 4 but not by 8. See my link "Terms of A396868" below for a proof.
For n >= 0, the least D such that the narrow class group of real quadratic field with discriminant D has 2-rank n (A317991) is a(n+1).
For n >= 0, the least D such that the class group of real quadratic field with discriminant D has 2-rank n (A391436) is 5 for n = 0, 40 for n = 1, 520 for n = 2, and a(n+2) for n >= 3. Proof: Note that this 2-rank is omega(D) - 1 if D has no prime factors congruent to 3 modulo 4, and omega(D) - 2 otherwise. Let D be the smallest fundamental discriminant such that Cl(D) = (C_2)^n, n >= 1, then D <= 8*p_2*...*p_{n+2}, where p_i = prime(i). If D has no prime factor congruent to 3 modulo 4, then D = 8*q_1*...*q_n, where q_i = A002144(i) is the i-th prime congruent to 1 modulo 4; but this is greater than 8*p_2*...*p_{n+2} for n >= 4. Since D must have a prime factor congruent to 3 modulo 4 for n >= 4, the minimal value of D is a(n+2).
For n >= 0, the least D such that the Pólya group of real quadratic field with discriminant D has 2-rank n (A396865) is also 5 for n = 0, 40 for n = 1, 520 for n = 2, and a(n+2) for n >= 3. Proof: This 2-rank is omega(D) - 1 if the real quadratic field with discriminant D has fundamental unit norm -1, omega(D) - 2 otherwise. It turns out the proof in the previous paragraph for the class groups also works here for the Pólya groups.
LINKS
Jianing Song, Table of n, a(n) for n = 1..350
Jianing Song, Table of n, a(n)/(4*p_2*...*p_n) for n = 2..10000.
Jianing Song, Terms of A396868.
FORMULA
Let p_i = prime(i), q_i = A002144(i) be the i-th prime congruent to 1 modulo 4, and r_i = A002145(i) be the i-th prime congruent to 3 modulo 4. Suppose that {p_2, ..., p_n} = {q_1, ..., q_{n-1-m}} U {r_1, ..., r_m}. Write N = 4*p_2*...*p_n.
- If m is odd, then a(n) = N.
- If m is even, then a(n) is either (q_{n-m}/r_m)*N or (r_{m+1}/q_{n-1-m})*N (i.e., obtained by either removing the largest prime <= prime(n) congruent to 1 modulo 4 and adding the least prime > prime(n) congruent to 3 modulo 4, or removing the largest prime <= prime(n) congruent to 3 modulo 4 and adding the least prime > prime(n) congruent to 1 modulo 4).
See my link "Terms of A396868" above.
EXAMPLE
For n = 4, we have {3, 5, 7} = {q_1=5} U {r_1=3, r_2=7}. Since the number of primes congruent to 3 modulo 4 is even, a(n) is the minimum between 4*5*13*3 and 4*3*7*11, which is 4*5*13*3 = 780.
For n = 8, we have {3, 5, 7, 11, 13, 17, 19} = {q_1=5, q_2=13, q_3=17} U {r_1=3, r_2=7, r_3=11, r_4=19}. Since the number of primes congruent to 3 modulo 4 is even, a(n) is the minimum between 4*5*13*17*29*3*7*11 and 4*5*13*3*7*11*19*23, which is 4*5*13*3*7*11*19*23 = 26246220.
PROG
(PARI) a(n) = {
if(n==1, return(5));
my(N=4, res=1, plim=prime(n), q, r);
forprime(p=3, plim, N=N*p; res=(res*p)%4; if(p%4==1, q=p, r=p)); \\ N = 4*p_2*...*p_n, q is the largest prime <= p_n congruent to 1 mod 4, r is the largest prime <= p_n congruent to 3 mod 4
if(res==3, return(N)); \\ if the number of primes <= p_n congruent to 3 mod 4 is odd, then a(n) = N
forprime(p=plim+1, oo, if(p%4==3, q=p/q; break())); \\ smallest prime > p_n congruent to 3 mod 4
forprime(p=plim+1, oo, if(p%4==1, r=p/r; break())); \\ smallest prime > p_n congruent to 1 mod 4
return(N * min(q, r));
}
CROSSREFS
KEYWORD
nonn
AUTHOR
Jianing Song, Jun 08 2026
STATUS
approved