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%I #33 Nov 29 2024 23:50:53
%S 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,
%T 19,19,20,21,23,24,25,26,26,27,29,30,31,31,32,33,34,36,37,38,40,41,41,
%U 42,44,44,47,47,48,49,52
%N Let q = A246655(n) for n >= 2, then a(n) = (q - Kronecker(-4,q))/4 - 1.
%C 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.
%H Jianing Song, <a href="/A366502/b366502.txt">Table of n, a(n) for n = 2..10000</a>
%H Mathematics Stack Exchange, <a href="https://math.stackexchange.com/q/164864">Existence of Consecutive Quadratic residues</a>.
%e 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}.
%e 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)).
%o (PARI) lim_A366502(N) = for(n=3, N, if(isprimepower(n), print1((n - kronecker(-4,n))/4 - 1, ", ")))
%Y Cf. A246655, A101455 ({kronecker(-4,n)}), A024698.
%Y A015518(n)-1 and A003463(n)-1 is respectively the number of consecutive nonzero squares in F_{3^n} and in F_{5^n}.
%K nonn,easy
%O 2,7
%A _Jianing Song_, Oct 12 2023