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A343240
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).
6
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, 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, 4, 2, 4, 2, 2, 4
OFFSET
1,3
COMMENTS
Row length of irregular triangle A343239.
FORMULA
a(n) = row length of A343239(n), for n >= 1.
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).
EXAMPLE
a(19) = 4 because A343238(19) = 42 = 2*3*7 has 2^(1+1) = 4 solutions from the primes 3 and 7.
PROG
(PARI) isok(k) = issquare(Mod(-5, k)); \\ A343238
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
CROSSREFS
Sequence in context: A353929 A297770 A330617 * A145866 A103318 A197775
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, May 16 2021
STATUS
approved