%I #36 Aug 29 2016 14:28:14
%S 4,14791044,143014853,253149265,490434564,600606332,993861182,
%T 3279563483
%N Composite n which divide s(n)+1, where s is the linear recurrence sequence s(n) = -s(n-1) + s(n-2) - s(n-3) + s(n-5) with initial terms (5, -1, 3, -7, 11).
%C The pseudoprimes derived from the fifth-order linear recurrence A225984(n) are analogous to the Perrin pseudoprimes A013998, and the Lucas pseudoprimes A005845.
%C For prime p, A225984(p) == p - 1 (mod p). The pseudoprimes are composite numbers satisfying the same relation. 4 = 2^2; 14791044 = 2^2 * 3 * 19 * 29 * 2237; 143014853 = 907 * 157679.
%C Like the Perrin test, the modular sequence is periodic so simple pre-tests can be performed. Numbers divisible by 2, 3, 4, 5, 9, and 25 have periods 31, 11, 62, 24, 33, and 120 respectively. - _Dana Jacobsen_, Aug 29 2016
%C a(9) > 1.4*10^11. - _Dana Jacobsen_, Aug 29 2016
%H K. Brown, <a href="http://www.mathpages.com/home/kmath346/kmath346.htm">Proof of Generalized Little Theorem of Fermat</a>, proves that for prime p, a(p) == a(1) (mod p) for recurrences of the form of A225984.
%H R. Holmes, <a href="https://plus.google.com/u/0/100452847199780289157/posts/aR498doNE4k">comments to M. McIrvin's post on Google+</a> (found terms 4 through 7)
%e A225984(4) = 11, and 11 == 3 (mod 4). Since 4 is composite, it is a pseudoprime with respect to A225984.
%o (PARI)
%o N=10^10;
%o default(primelimit, N);
%o M = [0, 1, 0, 0, 0; 0, 0, 1, 0, 0; 0, 0, 0, 1, 0; 0, 0, 0, 0, 1; 1, 0, -1, 1, -1];
%o a(n)=lift( trace( Mod(M, n)^n ) );
%o ta(n)=lift( trace( Mod(M, n) ) );
%o { for (n=2, N,
%o if ( isprime(n), next() );
%o if ( a(n)==ta(n), print1(n, ", "); );
%o ); }
%o /* _Matt McIrvin_, after _Joerg Arndt_'s program for A013998, May 23 2013 */
%K nonn,hard,more
%O 1,1
%A _Matt McIrvin_, May 23 2013
%E Terms 4 through 7 found by Richard Holmes, added by _Matt McIrvin_, May 27 2013
%E a(8) from _Dana Jacobsen_, Aug 29 2016
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