A173656 Primes p such that p^2 divides P(p), where P = Perrin sequence A001608.

521, 190699

Offset: 1

Author

Arkadiusz Wesolowski, Aug 15 2012

Keywords

• bref
• hard
• more
• nonn

Comments

It is not known if this sequence is infinite.

The squares are in A013998.

No other terms below 10^10. - Max Alekseyev, Aug 27 2023

Examples

			521 is in the sequence since its square 271441 is the factor of A001608(521).
		

Links

• G. L. Honaker, Jr. and Chris Caldwell, Prime Curios! 521
• Wikipedia, Perrin number

Crossrefs

Cf. A001608, A013998.

Programs

Mathematica

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
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

PARI

/* Assuming b13998 containing second column of b013998.txt */
A013998 = readvec(b13998);
for (k=1,#A013998,if (issquare(A013998[k])==1,print(k," ",A013998[k])));
/* Hugo Pfoertner, Sep 01 2017 */