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%I #14 Oct 25 2021 11:09:10
%S 1,2,1,2,1,2,2,2,2,1,2,2,2,2,2,2,2,2,2,4,4,2,4,2,2,2,2,2,2,2,4,2,4,2,
%T 4,2,2,2,2,4,2,4,2,2,4,4,2,2,2,4,2,2,4,4,2,4,2,4,2,2,2,4,2,4,2,2,4,2,
%U 4,2,2,2,4,4,4,4,2,2
%N Number of solutions to the congruence (x+1)*x + 4 == 0 (mod A347831(n)).
%C A347832 gives the representatives of these residue classes.
%F a(n) equals the length of row n of A347832(n).
%o (PARI) isok(m) = {my(f=factor(m)); for (k=1, #f~, my(p=f[k,1]); if ((p==3) || (p==5), if (f[k,2] > 1, return (0)), if (kronecker(p, 15) != 1, return(0)));); return (1);} \\ A347831
%o f(n) = sum(x=0, n-1, Mod(x*(x+1), n) == -4);
%o lista(nn) = apply(f, select(isok, [1..nn])); \\ _Michel Marcus_, Oct 23 2021
%Y Cf. A347831, A347832.
%K nonn,easy
%O 1,2
%A _Wolfdieter Lang_, Sep 15 2021