A366502 Let q = A246655(n) for n >= 2, then a(n) = (q - Kronecker(-4,q))/4 - 1.

0, 0, 0, 1, 1, 1, 2, 2, 3, 3, 4, 5, 5, 6, 6, 7, 7, 8, 9, 10, 11, 11, 12, 14, 14, 15, 16, 17, 17, 19, 19, 20, 21, 23, 24, 25, 26, 26, 27, 29, 30, 31, 31, 32, 33, 34, 36, 37, 38, 40, 41, 41, 42, 44, 44, 47, 47, 48, 49, 52

Offset: 2

Author

Jianing Song, Oct 12 2023

Keywords

• nonn
• easy

Comments

If q is not a power of 2, then a(n) is the number of pairs of consecutive nonzero squares in the finite field F_q. In other words, a(n) is the number of solutions to x^((q-1)/2) = (x+1)^((q-1)/2) = 1 in F_q. This can be proved by generalizing the argument of Jack D'Aurizio to the Math Stack Exchange question "Existence of Consecutive Quadratic residues" to the case of F_q.

Examples

			a(5) = 1 because there is one pair of consecutive nonzero squares in the finite field F_q with q = A246655(5) = 7, namely {1, 2}.
a(7) = 1 because there is one pair of consecutive nonzero squares in the finite field F_q with q = A246655(7) = 9, namely {1, 2} (note that 2 = -1 = i^2 in F_9 = F_3(i)).
		

Links

• Jianing Song, Table of n, a(n) for n = 2..10000
• Mathematics Stack Exchange, Existence of Consecutive Quadratic residues.

Crossrefs

Cf. A246655, A101455 ({kronecker(-4,n)}), A024698.

A015518(n)-1 and A003463(n)-1 are respectively the number of consecutive nonzero squares in F_{3^n} and in F_{5^n}.

Programs

PARI

lim_A366502(N) = for(n=3, N, if(isprimepower(n), print1((n - kronecker(-4,n))/4 - 1, ", ")))

Python

from sympy import primepi, integer_nthroot, kronecker_symbol
def A366502(n):
    def bisection(f,kmin=0,kmax=1):
        while f(kmax) > kmax: kmax <<= 1
        while kmax-kmin > 1:
            kmid = kmax+kmin>>1
            if f(kmid) <= kmid:
                kmax = kmid
            else:
                kmin = kmid
        return kmax
    def f(x): return int(n+x-sum(primepi(integer_nthroot(x,k)[0]) for k in range(1,x.bit_length())))
    return ((m:=bisection(f,n,n))-kronecker_symbol(-4,m)>>2)-1 # Chai Wah Wu, Jan 19 2025