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The number of solutions x from {0, 1, ..., A343238(n)-1} of the congruence x^2 + 5 == 0 (mod A343238(n)) is given by a(n).
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%I #12 Sep 17 2023 06:10:57

%S 1,1,2,1,2,2,2,1,2,2,2,4,2,2,2,2,2,2,4,2,2,2,2,2,2,2,2,4,2,4,2,2,2,2,

%T 2,4,2,2,2,2,2,2,4,2,2,2,2,4,4,2,4,2,2,4,4,2,4,2,4,2,2,2,2,4,2,2,4,4,

%U 4,2,4,2,2,4

%N The number of solutions x from {0, 1, ..., A343238(n)-1} of the congruence x^2 + 5 == 0 (mod A343238(n)) is given by a(n).

%C Row length of irregular triangle A343239.

%F a(n) = row length of A343239(n), for n >= 1.

%F a(1) = a(2) = a(4) = a(8) = 1, and otherwise a(n) = 2^{number of distinct primes from A139513}, that is, primes congruent to {1, 3, 7, 9} (mod 20), appearing in the prime factorization of A343238(n).

%e a(19) = 4 because A343238(19) = 42 = 2*3*7 has 2^(1+1) = 4 solutions from the primes 3 and 7.

%o (PARI) isok(k) = issquare(Mod(-5, k)); \\ A343238

%o lista(nn) = my(list = List()); for (n=1, nn, if (issquare(Mod(-5, n)), listput(list, sum(i=0, n-1, Mod(i,n)^2 + 5 == 0)););); Vec(list); \\ _Michel Marcus_, Sep 17 2023

%Y Cf. A343238, A343239.

%K nonn,easy

%O 1,3

%A _Wolfdieter Lang_, May 16 2021