login
Numbers k such that -tau(k)^2 == tau(k) mod k where tau = A000005.
0

%I #17 Feb 26 2020 19:34:02

%S 1,2,3,4,10,24,156,600

%N Numbers k such that -tau(k)^2 == tau(k) mod k where tau = A000005.

%e 24 is in this sequence because tau(24) = 8 and -8^2 mod 24 = 8.

%p q:= n-> (t-> irem(t^2+t, n)=0)(numtheory[tau](n)):

%p select(q, [$1..1000])[]; # _Alois P. Heinz_, Feb 26 2020

%t Select[Range[1000], Divisible[(d = DivisorSigma[0, #]) + d^2, #] &] (* _Amiram Eldar_, Feb 26 2020 *)

%o (PARI) isok(m) = my(nd=numdiv(m)); Mod(-nd^2, m) == nd; \\ _Michel Marcus_, Feb 26 2020

%Y Cf. A000005.

%K nonn,fini,full

%O 1,2

%A _Juri-Stepan Gerasimov_, Feb 26 2020