This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A318898 #15 Sep 06 2018 04:43:33 %S A318898 -1,-5,15,-565,-7085,25575,-1266205,-17602325,941362695,197665011735, %T A318898 2901803883615,-11147523830125,-165269711096165,637677823344495, %U A318898 2154364271382137415,-126774939137440139965,-1925041114036033717685,-447232673152232758272805,-6839447730858454557453725,410508614063545790640124095,-1608693655111966245554191885 %N A318898 a(n) = ((-4)^((p-1)/4) - 1)/p, where p is the n-th prime congruent to 1 mod 4. %C A318898 a(n) is always an integer. If p == 1 (mod 8), then (-4)^((p-1)/4) == 4^((p-1)/4) == 2^((p-1)/2) (mod p). 2 is a quadratic residue modulo p so 2^((p-1)/2) == 1 (mod p). If p == 5 (mod 8), then (-4)^((p-1)/4) == -4^((p-1)/4) == -2^((p-1)/2) (mod p). 2 is a quadratic nonresidue modulo p so 2^((p-1)/2) == -1 (mod p). Furthermore, for n > 1, a(n) is always an odd multiple of 5. %C A318898 (-4)^((p-1)/4) == 1 (mod p) implies -4 is always a quartic residue modulo p. Note that x^4 + 4 = (x^2 + 2)^2 - (2x)^2 = (x^2 + 2x + 2)(x^2 - 2x + 2) = ((x + 1)^2 + 1)*((x - 1)^2 + 1), so the solutions to x^4 == -4 (mod p) are x == ((p - 1)/2)! + 1, ((p - 1)/2)! - 1, -((p - 1)/2)! + 1 and -((p - 1)/2)! - 1 (mod p). %e A318898 The second prime congruent to 1 mod 4 is 13, so a(2) = ((-4)^3 - 1)/13 = (-65)/13 = -5. Also, the four solutions to x^4 == -4 (mod 13) are x == 4, 6, 7 and 9 (mod 13). %o A318898 (PARI) forstep(p=5, 100, 4, if(isprime(p), print1(((-4)^((p-1)/4)-1)/p, ", "))) %Y A318898 Cf. A002144 (primes of the form 4n + 1). %Y A318898 Cf. A270698 (composite k == 1 (mod 4) that divides (-4)^((k-1)/4) - 1). %K A318898 sign %O A318898 1,2 %A A318898 _Jianing Song_, Sep 05 2018