A309534 - OEIS

%I #15 Aug 07 2019 04:29:23

%S 1,5,34,610,698870988,1134903170

%N Numbers k such that A001414(k^2+1) is divisible by k.

%C If A001519(i-1) and A001519(i+1) are prime (or 1), then since A001519(i)^2+1=A001519(i-1)*A001519(i+1) and A001519(i-1)+A001519(i+1)=3*A001519(i), A001519(i) is in the sequence. 1, 5, 34, 610 and 1134903170 arise in this way.

%C a(7) > 10^10. - _Giovanni Resta_, Aug 07 2019

%e 34 is a member because 34^2+1=13*89 and 13+89=3*34.

%p filter:= proc(n) local F,t,y;

%p   F:= ifactors(n^2+1)[2];

%p   y:= add(t[1]*t[2],t=F);

%p   y mod n = 0

%p end proc:

%p select(filter, [$1..10^6]);

%o (PARI) f(n) = (n=factor(n))[, 1]~*n[, 2]; \\ A001414

%o isok(k) = !(f(k^2+1) % k); \\ _Michel Marcus_, Aug 07 2019

%Y Cf. A001414, A001519.

%K nonn,more

%O 1,2

%A _J. M. Bergot_ and _Robert Israel_, Aug 06 2019

%E a(5)-a(6) from _Giovanni Resta_, Aug 07 2019