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

2,7

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.

Links

Jianing Song, Table of n, a(n) for n = 2..10000

Math Stack Exchange, Existence of Consecutive Quadratic residues.

Example

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)).

Prog

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

Crossrefs

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

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

Sequence in context: A085268 A214656 A194979 * A098294 A195119 A077467

Adjacent sequences: A366499 A366500 A366501 * A366503 A366504 A366505

Keyword

nonn,easy

Author

Jianing Song, Oct 12 2023

Status

approved