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%I #41 Aug 27 2023 09:12:18
%S 521,190699
%N Primes p such that p^2 divides P(p), where P = Perrin sequence A001608.
%C It is not known if this sequence is infinite.
%C The squares are in A013998.
%C No other terms below 10^10. - _Max Alekseyev_, Aug 27 2023
%H G. L. Honaker, Jr. and Chris Caldwell, <a href="https://t5k.org/curios/cpage/16432.html">Prime Curios! 521</a>
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Perrin_pseudoprime">Perrin number</a>
%e 521 is in the sequence since its square 271441 is the factor of A001608(521).
%t lst = {}; a = 3; b = 0; c = 2; Do[P = b + a; If[PrimeQ[n] && Divisible[P, n^2], AppendTo[lst, n]]; a = b; b = c; c = P, {n, 3, 2*10^5}]; lst
%t lst = {}; PowerMod2[mat_, n_, m_] := Mod[Fold[Mod[If[#2 == 1, #1.#1.mat, #1.#1], m] &, mat, Rest@IntegerDigits[n, 2]], m]; LinearRecurrence2[coeffs_, init_, n_, m_] := Mod[First@PowerMod2[Append[RotateRight /@ Most@IdentityMatrix@Length[coeffs], coeffs], n, m].init, m] /; n >= Length[coeffs]; Do[n = Power[p, 2]; If[PrimeQ[p] && LinearRecurrence2[{1, 1, 0}, {3, 0, 2}, n, n] == 0, AppendTo[lst, p]], {p, 1, 2*10^5, 2}]; lst
%o (PARI)
%o /* Assuming b13998 containing second column of b013998.txt */
%o A013998 = readvec(b13998);
%o for (k=1,#A013998,if (issquare(A013998[k])==1,print(k," ",A013998[k])));
%o /* _Hugo Pfoertner_, Sep 01 2017 */
%Y Cf. A001608, A013998.
%K bref,hard,more,nonn
%O 1,1
%A _Arkadiusz Wesolowski_, Aug 15 2012
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